NCERT solutions for Class 12 Maths Chapter 4 Determinants

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# NCERT solutions for Class 12 Maths Chapter 4 Determinants

Edited By Ramraj Saini | Updated on Sep 13, 2023 09:18 PM IST | #CBSE Class 12th

## NCERT Determinants Class 12 Questions And Answers

NCERT solutions for Class 12 Maths Chapter 4 Determinants are proved here. These NCERT solutions are created by expert team at Careers360 keeping align the latest syllabus of CBSE 2023-24. In this chapter, students will be able to understand the Class 12 Maths Chapter 4 NCERT solutions. If you multiply a matrix with the coordinates of a point, it will give a new point in the space which is explained in NCERT class 12 chapter 4 maths Determinants solutions. In this sense, the matrix is a linear transformation. The determinant of the matrix is the factor by which its volume blows up. You will be familiar with these points after going through ch 4 maths class 12. Interested students can visit chapter wise NCERT solution for math.

The important topics of class 12 maths ch 4 are determinants and their properties, finding the area of the triangle, minor and cofactors, adjoint and the inverse of the matrix, and applications of determinants like solving the system of linear equations, etc are covered in NCERT solutions for Class 12 Maths Chapter 4 Determinants. If you are looking for determinants class 12 solutions then check all NCERT solutions at a single place which will help the students to learn CBSE maths. Here you will get NCERT solutions for class 12 also. Read further to know more about NCERT solutions for Class 12 Maths Chapter 4 PDF download.

## NCERT Class 12 Maths Chapter 4 Question Answer - Important Formulae

>> Determinant of a Matrix: The determinant is the numerical value of a square matrix.

For a square matrix A of order n, the determinant is denoted by det A or |A|.

Minor and Cofactor of a Matrix:

Minor of an element aij of a determinant is a determinant obtained by deleting the ith row and jth column in which element aij lies.

The cofactor of an element aij of a determinant, denoted by Aij or Cij, is defined as Aij = (-1)(i+j) * Mij, where Mij is the minor of element aij.

Value of a Determinant (2x2 and 3x3 matrices):

For a 2x2 matrix A: |A| = a11 * a22 - a21 * a12

For a 3x3 matrix A: |A| = a11 * |A11| - a12 * |A12| + a13 * |A13|

Singular and Non-Singular Matrix:

If the determinant of a square matrix is zero, the matrix is said to be singular; otherwise, it is non-singular.

Determinant Theorems:

If A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.

The determinant of the product of matrices is equal to the product of their respective determinants, i.e., |AB| = |A| * |B|.

The adjoint of a square matrix A is the transpose of the matrix obtained by cofactors of each element of the determinant corresponding to A. It is denoted by adj(A).

In general, the adjoint of a matrix A = [aij]n×n is a matrix [Aji]n×n, where Aji is a cofactor of element aji.

Properties of Adjoint of a Matrix:

Finding Area of a Triangle Using Determinants:

The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by

Inverse of a Square Matrix:

For a non-singular matrix A (|A| ≠ 0), the inverse A-1 is defined as A-1 = (1/|A|) * adj(A).

Properties of an Inverse Matrix:

(A-1)-1 = A

(AT)-1 = (A-1)T

(AB)-1 = B-1A-1

(ABC)-1 = C-1B-1A-1

Solving a System of Linear Equations using Inverse of a Matrix:

Given a system of equations AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

Case I: If |A| ≠ 0, the system is consistent, and X = A-1B has a unique solution.

Case II: If |A| = 0 and (adj A)B ≠ 0, the system is inconsistent and has no solution.

Case III: If |A| = 0 and (adj A)B = 0, the system may be either consistent or inconsistent, depending on whether it has infinitely many solutions or no solutions.

## NCERT Class 12 Maths Chapter 4 Question Answer (Intext Questions and Exercise)

NCERT determinants class 12 questions and answers: Excercise- 4.1

The determinant is evaluated as follows

$\begin{vmatrix} 2 & 4\\ -5 & -1\end{vmatrix} = 2(-1) - 4(-5) = -2 + 20 = 18$

The given two by two determinant is calculated as follows

$\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix} = cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2\theta + \sin ^2 \theta = 1$

We have determinant $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

So, $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix} = (x^2-x+1)(x+1) - (x-1)(x+1)$

$= (x+1)(x^2-x+1-x+1) = (x+1)(x^2-2x+2)$

$=x^3-2x^2+2x +x^2-2x+2$

$= x^3-x^2+2$

Given determinant $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then we have to show that $\dpi{100} | 2 A |=4|A|$ ,

So, $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then, $\dpi{100} 2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}$

Hence we have $\dpi{100} \left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24$

So, L.H.S. = |2A| = -24

then calculating R.H.S. $\dpi{100} 4\left | A \right |$

We have,

$\dpi{100} \left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6$

hence R.H.S becomes $\dpi{100} 4\left | A \right | = 4\times(-6) = -24$

Therefore L.H.S. =R.H.S.

Hence proved.

Given Matrix $A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$

Calculating $3A =3\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} = \begin{bmatrix} 3 &0 &3 \\ 0& 3& 6\\ 0& 0 &12 \end{bmatrix}$

So, $\left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108$

calculating $\dpi{100} 27|A|$ ,

$\dpi{100} |A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4$

So, $\dpi{100} 27|A| = 27(4) = 108$

Therefore $\dpi{100} |3A|=27|A|$ .

Hence proved.

Question:5(i) Evaluate the determinants.

Given the determinant $\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$ ;

now, calculating its determinant value,

$\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix} = 3\begin{vmatrix} 0 &-1 \\ -5& 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 &0 \\ 3& -5 \end{vmatrix}$

$\dpi{100} = 3(0-5)+1(0+3) -2(0-0) = -15+3-0 = -12$ .

Question:5(ii) Evaluate the determinants.

Given determinant $\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$ ;

Now calculating the determinant value;

$\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix} = 3\begin{vmatrix} 1 &-2 \\ 3&1 \end{vmatrix} -(-4)\begin{vmatrix} 1 &-2 \\ 2& 1 \end{vmatrix}+5\begin{vmatrix} 1 & 1\\ 2& 3 \end{vmatrix}$

$= 3(1+6) +4(1+4) +5(3-2) = 21+20+5 = 46$ .

Question:5(iii) Evaluate the determinants.

Given determinant $\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$ ;

Now calculating the determinant value;

$\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix} = 0\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} -1\begin{vmatrix} -1 &-3 \\ -2& 0 \end{vmatrix}+2\begin{vmatrix} -1 &0 \\ -2& 3 \end{vmatrix}$

$= 0 - 1(0-6)+2(-3-0) = 6 -6 =0$

Question:5(iv) Evaluate the determinants.

Given determinant: $\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}$ ,

We now calculate determinant value:

$\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix} =2\begin{vmatrix} 2 &-1 \\ -5 & 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3 & 0 \end{vmatrix}+(-2)\begin{vmatrix} 0 &2 \\ 3&-5 \end{vmatrix}$

$=2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5$

Given the matrix $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ then,

Finding the determinant value of A;

$|A| = 1\begin{vmatrix} 1 &-3 \\ 4& -9 \end{vmatrix} -1\begin{vmatrix} 2 &-3 \\ 5& -9 \end{vmatrix}-2\begin{vmatrix} 2 &1 \\ 5& 4 \end{vmatrix}$

$= 1(-9+12)-1(-18+15)-2(8-5) =3+3-6 =0$

Question:7(i) Find values of x, if

Given that $\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

$\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =2-20 = -18$ and $\dpi{100} \begin{vmatrix}2x &4 \\6 &x \end{vmatrix} = 2x(x)-24 = 2x^2-24$

So, we have then,

$\dpi{100} -18= 2x^2-24$ or $\dpi{100} 3= x^2$ or $\dpi{100} x= \pm \sqrt{3}$

Question:7(ii) Find values of x, if

Given $\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$ ;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

$\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}= 10 - 12 = -2$

Similarly R.H.S. determinant value;

$\dpi{100} \begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x$

So, we have then;

$\dpi{100} -2 = -x$ or $\dpi{100} x =2$ .

Solving the L.H.S. determinant ;

$\dpi{100} \begin{vmatrix}x &2 \\18 &x \end{vmatrix}= x^2 - 36$

and solving R.H.S determinant;

$\dpi{100} \begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} = 36-36 = 0$

So equating both sides;

$\dpi{100} x^2 - 36 =0$ or $\dpi{100} x^2 = 36$ or $\dpi{100} x = \pm 6$

NCERT determinants class 12 questions and answers: Excercise - 4.2

We can split it in manner like;

$\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}= \begin{vmatrix} x &a &x \\ y & b &y \\ z &c &z \end{vmatrix} + \begin{vmatrix} x &a & a\\ y &b &b \\ z&c & c \end{vmatrix}$

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

$\therefore 0 + 0 = 0$

Hence the sum is zero.

Given determinant $\dpi{100} \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

Applying the rows addition $R_{1} \rightarrow R_{1}+R_{2}$ then we have;

$\dpi{100} \triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0$

$\dpi{100} =-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0$

So, we have two rows $\dpi{100} R_{1}$ and $\dpi{100} R_{2}$ identical hence we can say that the value of determinant = 0

Therefore $\dpi{100} \triangle = 0$ .

Given determinant $\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}$

So, we can split it in two addition determinants:

$\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix} = \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix} = \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} + \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} = 0$ [ $\dpi{100} \because$ Here two columns are identical ]

and $\dpi{100} \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix} = \begin{vmatrix} 2 & 7 &9(7) \\ 3& 8 &9(8) \\ 5 &9 & 9(9) \end{vmatrix} = 9 \begin{vmatrix} 2 & 7 &7 \\ 3& 8& 8\\ 5& 9&9 \end{vmatrix}$ [ $\dpi{100} \because$ Here two columns are identical ]

$\dpi{100} = 0$

Therefore we have the value of determinant = 0.

We have determinant:

$\triangle = \begin{vmatrix} 1 &bc &a(b+c) \\ 1& ca &b(c+a) \\ 1& ab &c(a+b) \end{vmatrix}$

Applying $C_{3} \rightarrow C_{2} + C_{3}$ we have then;

$\triangle = \begin{vmatrix} 1 &bc & ab+bc+ca \\ 1& ca &ab+bc+ca \\ 1& ab &ab+bc+ca \end{vmatrix}$

So, here column 3 and column 1 are proportional.

Therefore, $\triangle = 0$ .

Given determinant :

$\dpi{100} \triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}$

Splitting the third row; we get,

$\dpi{100} = \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that)$ .

Then we have,

$\dpi{100} \triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}$

On Applying row transformation $\dpi{100} R_{2} \rightarrow R_{2} - R_{3}$ and then $\dpi{100} R_{1} \rightarrow R_{1} - R_{2}$ ;

we get, $\dpi{100} \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}$

Applying Rows exchange transformation $\dpi{100} R_{1} \leftrightarrow R_{2}$ and $\dpi{100} R_{2} \leftrightarrow R_{3}$ , we have:

also $\dpi{100} \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$

On applying rows transformation, $\dpi{100} R_{1} \rightarrow R_{1} - R_{3}$ and then $\dpi{100} R_{2} \rightarrow R_{2} - R_{1}$

$\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$ and then $\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}$

Then applying rows exchange transformation;

$\dpi{100} R_{1} \leftrightarrow R_{2}$ and then $\dpi{100} R_{2} \leftrightarrow R_{3}$ . we have then;

$\dpi{100} \triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}$

So, we now calculate the sum = $\dpi{100} \triangle_{1} + \triangle _{2}$

$\dpi{100} \triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}$

Hence proved.

We have given determinant

$\dpi{100} \triangle = \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}$

Applying transformation, $\dpi{100} R_{1} \rightarrow cR_{1}$ we have then,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} 0 &ac &-bc \\-a &0 & -c\\b &c &0 \end{vmatrix}$

We can make the first row identical to the third row so,

Taking another row transformation: $\dpi{100} R_{1} \rightarrow R_{1}-bR_{2}$ we have,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} ab &ac &0 \\-a &0 & -c\\b &c &0 \end{vmatrix} = \frac{a}{c} \begin{vmatrix} b &c &0 \\-a &0 & -c\\b &c &0 \end{vmatrix}$

So, determinant has two rows $\dpi{100} R_{1}\ and\ R_{3}$ identical.

Hence $\dpi{100} \triangle = 0$ .

Given determinant : $\dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}$

$\triangle = \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}$

As we can easily take out the common factors a,b,c from rows $R_{1},R_{2},R_{3}$ respectively.

So, get then:

$=abc \begin{vmatrix} -a &b &c \\ a &-b &c \\ a & b & -c \end{vmatrix}$

Now, taking common factors a,b,c from the columns $C_{1},C_{2},C_{3}$ respectively.

$=a^2b^2c^2 \begin{vmatrix} -1 &1 &1 \\ 1 &-1 &1 \\ 1 & 1 & -1 \end{vmatrix}$

Now, applying rows transformations $R_{1} \rightarrow R_{1} + R_{2}$ and then $R_{3} \rightarrow R_{2} + R_{3}$ we have;

$\triangle = a^2b^2c^2\begin{vmatrix} 0 &0 &2 \\ 1&-1 &1 \\ 2& 0 &0 \end{vmatrix}$

Expanding to get R.H.S.

$\triangle = a^2b^2c^2 \left ( 2\begin{vmatrix} 1 &-1 \\ 2& 0 \end{vmatrix} \right ) = 2a^2b^2c^2(0+2) =4a^2b^2c^2$

Question:8(i) By using properties of determinants, show that:

We have the determinant $\dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}$

Applying the row transformations $R_{1} \rightarrow R_{1} -R_{2}$ and then $R_{2} \rightarrow R_{2} -R_{3}$ we have:

$\dpi{100} \triangle = \begin{vmatrix} 0 &a-b &a^2-b^2 \\ 0 &b-c &b^2-c^2 \\ 1 &c &c^2 \end{vmatrix}$

$\dpi{100} = \begin{vmatrix} 0 &a-b &(a-b)(a+b) \\ 0 &b-c &(b-c)(b+c) \\ 1 &c &c^2 \end{vmatrix} = (a-b)(b-c)\begin{vmatrix} 0 &1 &(a+b) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$

Now, applying $R_{1} \rightarrow R_{1} -R_{2}$ we have:

$\dpi{100} = (a-b)(b-c)\begin{vmatrix} 0 &0 &(a-c) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$ or $\dpi{100} = (a-b)(b-c)(a-c)\begin{vmatrix} 0 &0 &1 \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} =(a-b)(b-c)(a-c)\begin{vmatrix} 0 &1 \\ 1 & c \end{vmatrix}$

$\dpi{100} = (a-b)(b-c)(c-a)$

Hence proved.

Question:8(ii) By using properties of determinants, show that:

Given determinant :

$\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}$ ,

Applying column transformation $\dpi{100} C_{1} \rightarrow C_{1}-C_{3}$ and then $\dpi{100} C_{2} \rightarrow C_{2}-C_{3}$

We get,

$\dpi{100} \triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}$

$\dpi{100} =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

Now, applying column transformation $\dpi{100} C_{1} \rightarrow C_{1} - C_{2}$ , we have:

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)(a-b)(a+b+c)\begin{vmatrix} 0&1 \\ 1& c \end{vmatrix}$

$\dpi{100} =-(a-c)(b-c)(a-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)$

Hence proved.

We have the determinant:

$\triangle = \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}$

Applying the row transformations $R_{1} \rightarrow R_{1}- R_{3}$ and then $R_{2} \rightarrow R_{2}- R_{3}$ , we have;

$\triangle = \begin{vmatrix} x-z & x^2-z^2 & yz-xy\\ y-z & y^2-z^2 &zx-xy \\ z & z^2 & xy \end{vmatrix}$

$= \begin{vmatrix} x-z & (x-z)(x+z) & y(z-x)\\ y-z & (y-z)(y+z) &x(z-y) \\ z & z^2 & xy \end{vmatrix}$

$= (x-z)(y-z)\begin{vmatrix} 1 & (x+z) & -y\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}$

Now, applying $R_{1} \rightarrow R_{1} - R_{2}$ ; we have

$= (x-z)(y-z)\begin{vmatrix} 0 & (x-y) & (x-y)\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}$

$= (x-z)(y-z)(x-y)\begin{vmatrix} 0 & 1 & 1\\ 1 & (y+z) &-x \\ z & z^2 & xy \end{vmatrix}$

Now, expanding the remaining determinant;

$= (x-z)(y-z)(x-y) \left [ (xy+zx) + (z^2 - zy-z^2) \right]$

$= -(x-z)(y-z)(x-y) \left [ xy+zx + zy \right]$

$= (x-y)(y-z)(z-x) \left [ xy+zx + zy \right]$

Hence proved.

Question:10(i) By using properties of determinants, show that:

Given determinant:

$\begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}$

Applying row transformation: $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$ then we have;

$\triangle = \begin{vmatrix} 5x+4 &5x+4 &5x+4 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}$

Taking a common factor: 5x+4

$= (5x+4)\begin{vmatrix} 1 &1 &1 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}$

Now, applying column transformations $C_{1} \rightarrow C_{1}- C_{2}$ and $C_{2} \rightarrow C_{2}- C_{3}$

$= (5x+4)\begin{vmatrix} 0 &0 &1 \\ x-4 & 4-x & 2x\\ 0 & x-4 & x+4 \end{vmatrix}$

$= (5x+4)(4-x)(4-x)\begin{vmatrix} 0 &0 &1 \\ 1 & 1 & 2x\\ 0 & 1 & x+4 \end{vmatrix}$

$= (5x+4)(4-x)^2$

Question:10(ii) By using properties of determinants, show that:

Given determinant:

$\triangle = \begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$

Applying row transformation $R_{1} \rightarrow R_{1} +R_{2}+R_{3}$ we get;

$= \begin{vmatrix} 3y+k & 3y+k & 3y+k\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$

$=(3y+k) \begin{vmatrix}1 & 1 & 1\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$ [taking common (3y + k) factor]

Now, applying column transformation $C_{1} \rightarrow C_{1} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{3}$

$=(3y+k) \begin{vmatrix}0 & 0 & 1\\ -k & k &y \\ 0 & -k & y+k \end{vmatrix}$

$=(3y+k)(k^2) \begin{vmatrix}0 & 0 & 1\\ -1 & 1 &y \\ 0 & -1 & y+k \end{vmatrix}$

$=k^2 (3y+k)$

Hence proved.

Question:11(i) By using properties of determinants, show that:

Given determinant:

$\triangle = \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}$

We apply row transformation: $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$= \begin{vmatrix} a+b+c &a+b+c &a+b+c \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}$

Taking common factor (a+b+c) out.

$=(a+b+c) \begin{vmatrix} 1 &1 &1 \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}$

Now, applying column tranformation $C_{1} \rightarrow C_{1}- C_{2}$ and then $C_{2} \rightarrow C_{2}- C_{3}$

We have;

$=(a+b+c) \begin{vmatrix} 0 &0 &1 \\ b+c+a &-b-c-a &2b \\ 0 &c+a+b &c-a-b \end{vmatrix}$

$=(a+b+c)(a+b+c)(a+b+c) \begin{vmatrix} 0 &0 &1 \\ 1 &-1 &2b \\ 0 &1 &c-a-b \end{vmatrix}$

$=(a+b+c)(a+b+c)(a+b+c) = (a+b+c)^3$

Hence Proved.

Question:11(ii) By using properties of determinants, show that:

Given determinant

$\triangle =\begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}$

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ we get;

$=\begin{vmatrix} 2(x+y+z) &x &y \\ 2(z+y+x) & y+z+2x & y\\ 2(z+y+x) & x &z+x+2y \end{vmatrix}$

Taking 2(x+y+z) factor out, we get;

$=2(x+y+z)\begin{vmatrix} 1 &x &y \\ 1 & y+z+2x & y\\ 1 & x &z+x+2y \end{vmatrix}$

Now, applying row transformations, $R_{1} \rightarrow R_{1} -R_{2}$ and then $R_{2} \rightarrow R_{2} -R_{3}$ .

we get;

$=2(x+y+z)\begin{vmatrix} 0 &-x-y-z &0 \\ 0 & y+z+x & -y-z-x\\ 1 & x &z+x+2y \end{vmatrix}$

$=2(x+y+z)^3\begin{vmatrix} 0 &-1 &0 \\ 0 & 1 & -1\\ 1 & x &z+x+2y \end{vmatrix}$

$=2(x+y+z)^3\begin{vmatrix} -1 &0 \\ 1& -1 \end{vmatrix} = 2(x+y+z)^3$

Hence proved.

Give determinant $\begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}$

Applying column transformation $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ we get;

$\triangle = \begin{vmatrix} 1+x+x^2 &x &x^2 \\ x^2+1+x &1 &x \\ x+x^2+1 &x^2 &1 \end{vmatrix}$

$= (1+x+x^2)\begin{vmatrix} 1 &x &x^2 \\ 1 &1 &x \\ 1 &x^2 &1 \end{vmatrix}$ [ after taking the (1+x+x 2 ) factor common out.]

Now, applying row transformations, $R_{1} \rightarrow R_{1}-R_{2}$ and then $R_{2} \rightarrow R_{2}-R_{3}$ .

we have now,

$= (1+x+x^2)\begin{vmatrix} 0 &x-1 &x^2-x \\ 0 &1-x^2 &x-1 \\ 1 &x^2 &1 \end{vmatrix}$

$= (1+x+x^2)\begin{vmatrix} x-1 &x^2-x \\ 1-x^2 &x-1 \end{vmatrix}$

$= (1+x+x^2)((x-1)^2-x(x-1)(1-x^2))$

$= (1+x+x^2)(x-1)(x^3-1) = (x^3-1)^2$

As we know $\left [\because (1+x+x^2)(x-1) = (x^3-1) \right ]$

Hence proved.

We have determinant:

$\dpi{100} \triangle = \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Applying row transformations, $\dpi{100} R_{1} \rightarrow R_{1} +bR_{3}$ and $\dpi{100} R_{2} \rightarrow R_{2} -aR_{3}$ then we have;

$\dpi{100} = \begin{vmatrix} 1+a^2+b^2 &0 &-b(1+a^2+b^2) \\ 0 &1+a^2+b^2 &a(1+a^2+b^2) \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

taking common factor out of the determinant;

$\dpi{100} = (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Now expanding the remaining determinant we get;

$\dpi{100} = (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1-a^2-b^2+2a^2-b(-2b)\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3$

Hence proved.

Given determinant:

$\dpi{100} \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}$

Let $\triangle = \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}$

Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C 1, C 2, and C 3.

We then get;

$=abc \begin{vmatrix} \left ( a+\frac{1}{a} \right ) &a &a \\ b &(b+\frac{1}{b}) &b \\ c & c &(c+\frac{1}{c}) \end{vmatrix}$

Now, applying column transformations: $C_{1} \rightarrow C_{1} -C_{2}$ and $C_{2} \rightarrow C_{2} -C_{3}$

then we have;

$=abc \begin{vmatrix} \left ( \frac{1}{a} \right ) &0 &a \\ -\frac{1}{b} &(\frac{1}{b}) &b \\ 0 & -\frac{1}{c} &(c+\frac{1}{c}) \end{vmatrix}$

$=abc\times \frac{1}{abc} \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}$

$= \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}$

Now, expanding the remaining determinant:

$\triangle = 1\begin{vmatrix} 1&b^2 \\ -1&(c^2+1) \end{vmatrix} + a^2\begin{vmatrix} -1&1 \\ 0& -1 \end{vmatrix}$

$= 1[(c^2+1)+b^2] + a^2(1)=a^2+b^2+c^2+1$ .

Hence proved.

Assume a square matrix A of order of $3\times3$ .

$A = \begin{bmatrix} a_1 & b_1&c_1 \\ a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{bmatrix}$

Then we have;

$kA = \begin{bmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{bmatrix}$

( Taking the common factors k from each row. )

$|kA| = \begin{vmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{vmatrix} = k^3 \begin{vmatrix} a_1 & b_1&c_1 \\a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{vmatrix}$

$= k^3 |A|$

Therefore correct option is (C).

The answer is (C) Determinant is a number associated to a square matrix.

As we know that To every square matrix $A = [a_{ij}]$ of order n, we can associate a number (real or complex) called determinant of the square matrix A, where $a_{ij} = (i, j)^{th}$ element of A.

NCERT class 12 maths chapter 4 question answer: Excercise-4.3

We can find the area of the triangle with vertices $(1,0), (6,0), (4,3)$ by the following determinant relation:

$\triangle =\frac{1}{2} \begin{vmatrix} 1& 0 &1 \\ 6 & 0 &1 \\ 4& 3& 1 \end{vmatrix}$

Expanding using second column

$=\frac{1}{2} (-3) \begin{vmatrix} 1 &1 & \\ 6& 1 & \end{vmatrix}$

$= \frac{15}{2}\ square\ units.$

We can find the area of the triangle with given coordinates by the following method:

$\triangle = \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix}$

$=\frac{1}{2} \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix} = \frac{1}{2}\left [ 2(1-8)-7(1-10)+1(8-10) \right ]$

$= \frac{1}{2}\left [ 2(-7)-7(-9)+1(-2) \right ] = \frac{1}{2}\left [ -14+63-2 \right ] = \frac{47}{2}\ square\ units.$

Area of the triangle by the determinant method:

$Area\ \triangle = \frac{1}{2} \begin{vmatrix} -2 &-3 &1 \\ 3& 2 & 1\\ -1& -8 & 1 \end{vmatrix}$

$=\frac{1}{2}\left [ -2(2+8)+3(3+1)+1(-24+2) \right ]$

$=\frac{1}{2}\left [ -20+12-22 \right ] = \frac{1}{2}[-30]= -15$

Hence the area is equal to $|-15| = 15\ square\ units.$

If the area formed by the points is equal to zero then we can say that the points are collinear.

So, we have an area of a triangle given by,

$\triangle = \frac{1}{2} \begin{vmatrix} a &b+c &1 \\ b& c+a &1 \\ c& a+b & 1 \end{vmatrix}$

calculating the area:

$= \frac{1}{2}\left [ a\begin{vmatrix} c+a &1 \\ a+b& 1 \end{vmatrix} - (b+c)\begin{vmatrix} b & 1\\ c&1 \end{vmatrix}+1\begin{vmatrix} b &c+a \\ c&a+b \end{vmatrix} \right ]$

$= \frac{1}{2}\left [ a(c+a-a-b) - (b+c)(b-c)+1(b(a+b)-c(c+a)) \right ]$

$= \frac{1}{2}\left [ ac-ab - b^2+c^2+ab+b^2-c^2-ac \right ] = \frac{1}{2} \left [ 0 \right] = 0$

Hence the area of the triangle formed by the points is equal to zero.

Therefore given points $A (a, b+c), B (b,c+a), C (c,a+b)$ are collinear.

We can easily calculate the area by the formula :

$\triangle = \frac{1}{2} \begin{vmatrix} k &0 &1 \\ 4& 0& 1\\ 0 &2 & 1 \end{vmatrix} = 4\ sq.\ units$

$= \frac{1}{2}\left [ k\begin{vmatrix} 0 &1 \\ 2& 1 \end{vmatrix} -0\begin{vmatrix} 4 &1 \\ 0 & 1 \end{vmatrix}+1\begin{vmatrix} 4 &0 \\ 0& 2 \end{vmatrix} \right ]= 4\ sq.\ units$

$=\frac{1}{2}\left [ k(0-2)-0+1(8-0) \right ] = \frac{1}{2}\left [ -2k+8 \right ] = 4\ sq.\ units$

$\left [ -2k+8 \right ] = 8\ sq.\ units$ or $-2k +8 = \pm 8\ sq.\ units$

or $k = 0$ or $k = 8$

Hence two values are possible for k.

The area of the triangle is given by the formula:

$\triangle = \frac{1}{2} \begin{vmatrix} -2 &0 &1 \\ 0 & 4 & 1\\ 0& k & 1 \end{vmatrix} = 4\ sq.\ units.$

Now, calculating the area:

$= \frac{1}{2} \left | -2(4-k)-0(0-0)+1(0-0) \right | = \frac{1}{2} \left | -8+2k \right | = 4$

or $-8+2k =\pm 8$

Therefore we have two possible values of 'k' i.e., $k = 8$ or $k = 0$ .

As we know the line joining $\small (1,2)$ , $\small (3,6)$ and let say a point on line $A\left ( x,y \right )$ will be collinear.

Therefore area formed by them will be equal to zero.

$\triangle = \frac{1}{2}\begin{vmatrix} 1 &2 &1 \\ 3& 6 &1 \\ x & y &1 \end{vmatrix} = 0$

So, we have:

$=1(6-y)-2(3-x)+1(3y-6x) = 0$

or $6-y-6+2x+3y-6x = 0 \Rightarrow 2y-4x=0$

Hence, we have the equation of line $\Rightarrow y=2x$ .

We can find the equation of the line by considering any arbitrary point $A(x,y)$ on line.

So, we have three points which are collinear and therefore area surrounded by them will be equal to zero .

$\triangle = \frac{1}{2}\begin{vmatrix} 3 &1 &1 \\ 9& 3 & 1\\ x& y &1 \end{vmatrix} = 0$

Calculating the determinant:

$=\frac{1}{2}\left [ 3\begin{vmatrix} 3 &1 \\ y& 1 \end{vmatrix}-1\begin{vmatrix} 9 &1 \\ x& 1 \end{vmatrix}+1\begin{vmatrix} 9 &3 \\ x &y \end{vmatrix} \right ]$

$=\frac{1}{2}\left [ 3(3-y)-1(9-x)+1(9y-3x) \right ] = 0$

$\frac{1}{2}\left [ 9-3y-9+x+9y-3x \right ] = \frac{1}{2}[6y-2x] = 0$

Hence we have the line equation:

$3y= x$ or $x-3y = 0$ .

Area of triangle is given by:

$\triangle = \frac{1}{2} \begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 35\ sq.\ units.$

or $\begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 70\ sq.\ units.$

$2\begin{vmatrix} 4 &1 \\ 4& 1 \end{vmatrix}-(-6)\begin{vmatrix} 5 &1 \\ k &1 \end{vmatrix}+1\begin{vmatrix} 5 &4 \\ k&4 \end{vmatrix} = 70$

$2(4-4) +6(5-k)+(20-4k) = \pm70$

$50-10k = \pm70$

$k = 12$ or $k = -2$

Hence the possible values of k are 12 and -2.

Therefore option (D) is correct.

NCERT class 12 maths chapter 4 question answer: Excercise: 4.4

GIven determinant: $\begin{vmatrix}2 &-4 \\0 &3 \end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$ .

Therefore we have

$M_{11}$ = minor of element $a_{11}$ = 3

$M_{12}$ = minor of element $a_{12}$ = 0

$M_{21}$ = minor of element $a_{21}$ = -4

$M_{22}$ = minor of element $a_{22}$ = 2

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1)^{i+j}M_{ij}$ .

Therefore we have:

$A_{11} = (-1)^{1+1}M_{11} = (-1)^2(3) = 3$

$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(0) = 0$

$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(-4) = 4$

$A_{22} = (-1)^{2+2}M_{22} = (-1)^4(2) = 2$

GIven determinant: $\begin{vmatrix} a &c \\ b &d \end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$ .

Therefore we have

$M_{11}$ = minor of element $a_{11}$ = d

$M_{12}$ = minor of element $a_{12}$ = b

$M_{21}$ = minor of element $a_{21}$ = c

$M_{22}$ = minor of element $a_{22}$ = a

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1)^{i+j}M_{ij}$ .

Therefore we have:

$A_{11} = (-1)^{1+1}M_{11} = (-1)^2(d) = d$

$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(b) = -b$

$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(c) = -c$

$A_{22} = (-1)^{2+2}M_{22} = (-1)^4(a) = a$

Given determinant : $\begin{vmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 0 &1 \\ 0 &0 \end{vmatrix} = 0$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &0 \\ 1 &0 \end{vmatrix} = 0$ $M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 1$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = 0$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 0$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 0$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 1$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = 0$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = 0$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 0$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 1$ .

Given determinant : $\begin{vmatrix} 1 &0 &4 \\ 3 & 5 &-1 \\ 0 &1 &2 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 5 &-1 \\ 1 &2 \end{vmatrix} = 11$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 3 &-1 \\ 0 &2 \end{vmatrix} = 6$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 3 &5 \\ 0 &1 \end{vmatrix} = 3$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &4 \\ 1 &2 \end{vmatrix} = -4$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &4 \\ 0 &2 \end{vmatrix} = 2$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &4 \\ 5 &-1 \end{vmatrix} = -20$

$M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &4 \\ 3 &-1 \end{vmatrix} = -1-12=-13$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 3 &5 \end{vmatrix} = 5$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 11$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = -6$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 3$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 4$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 2$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = -1$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = -20$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 13$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 5$ .

Given determinant : $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

First finding Minors of the second rows by the definition,

$M_{21} =$ minor of $a_{21} = \begin{vmatrix} 3 &8 \\ 2 &3 \end{vmatrix} =9-16 = -7$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 5 &8 \\ 1 &3 \end{vmatrix} = 15-8=7$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 5 &3 \\ 1 &2 \end{vmatrix} = 10-3 =7$

Finding the Cofactors of the second row:

$A_{21}=$ Cofactor of $a_{21} = (-1)^{2+1}M_{21} = 7$

$A_{22}=$ Cofactor of $a_{22} = (-1)^{2+2}M_{22} = 7$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = -7$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

$\triangle = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) +0(7) +1(-7) =14-7=7$

Given determinant : $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

First finding Minors of the third column by the definition,

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 1 &y \\ 1 &z \end{vmatrix} =z-y$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &x \\ 1 &z \end{vmatrix} = z-x$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &x \\ 1 &y \end{vmatrix} =y-x$

Finding the Cofactors of the second row:

$A_{13}=$ Cofactor of $a_{13} = (-1)^{1+3}M_{13} = z-y$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = x-z$

$A_{33}=$ Cofactor of $a_{33} = (-1)^{3+3}M_{33} = y-x$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$

$= (z-y)yz + (x-z)zx +(y-x)xy$

$=yz^2-y^2z + zx^2-xz^2 + xy^2-x^2y$

$=z(x^2-y^2) + z^2(y-x) +xy(y-x)$

$= (x-y) \left [ zx+zy-z^2-xy \right ]$

$=(x-y)\left [ z(x-z) +y(z-x) \right ]$

$= (x-y)(z-x)[-z+y]$

$= (x-y)(y-z)(z-x)$

Thus, we have value of $\triangle = (x-y)(y-z)(z-x)$ .

Answer is (D) $\small a_1_1A_1_1+a_2_1A_2_1+a_3_1A_3_1$ by the definition itself, $\small \Delta$ is equal to the product of the elements of the row/column with their corresponding cofactors.

## NCERT determinants class 12 solutions: Excercise: 4.5

Question:1 Find adjoint of each of the matrices.

Given matrix: $\small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}= A$

Then we have,

$A_{11} = 4, A_{12}=-(1)3, A_{21} = -(1)2,\ and\ A_{22}= 1$

Hence we get:

$adjA = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} &A_{22} \end{bmatrix}^T = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} &A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 &1 \end{bmatrix}$

Question:2 Find adjoint of each of the matrices

Given the matrix: $\small A = \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}$

Then we have,

$A_{11} = (-1)^{1+1}\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix} =(3-0)= 3$

$A_{12} = (-1)^{1+2}\begin{vmatrix} 2 &5 \\ -2& 1 \end{vmatrix} =-(2+10)= -12$

$A_{13} = (-1)^{1+3}\begin{vmatrix} 2 &3 \\ -2& 0 \end{vmatrix} =0+6= 6$

$A_{21} = (-1)^{2+1}\begin{vmatrix} -1 &2 \\ 0& 1 \end{vmatrix} =-(-1-0)= 1$

$A_{22} = (-1)^{2+2}\begin{vmatrix} 1 &2 \\ -2& 1 \end{vmatrix} =(1+4)= 5$

$A_{23} = (-1)^{2+3}\begin{vmatrix} 1 &-1 \\-2& 0 \end{vmatrix} =-(0-2)= 2$

$A_{31} = (-1)^{3+1}\begin{vmatrix} -1 &2 \\ 3& 5 \end{vmatrix} =(-5-6)= -11$

$A_{32} = (-1)^{3+2}\begin{vmatrix} 1 &2 \\2& 5\end{vmatrix} =-(5-4)= -1$

$A_{33} = (-1)^{3+3}\begin{vmatrix} 1 &-1 \\ 2& 3 \end{vmatrix} =(3+2)= 5$

Hence we get:

$adjA = \begin{bmatrix} A_{11} &A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}&A_{23} &A_{33} \end{bmatrix} = \begin{bmatrix} 3 &1 &-11 \\ -12&5 &-1 \\ 6&2 &5 \end{bmatrix}$

Given the matrix: $\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Let $\small A = \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1}(-6) = -6$

$\small A_{12} = (-1)^{1+2}(-4) = 4$

$\small A_{21} = (-1)^{2+1}(3) = -3$

$\small A_{22} = (-1)^{2+2}(2) = 2$

Hence, $\small adjA = \begin{bmatrix} -6 &-3 \\ 4& 2 \end{bmatrix}$

Now,

$\small A (adj A) = \begin{bmatrix} 2 &3 \\ -4&-6 \end{bmatrix}\left ( \begin{bmatrix} -6 &-3 \\ 4 &2 \end{bmatrix} \right )$

$\small \begin{bmatrix} -12+12 &-6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

aslo,

$\small (adjA)A = \begin{bmatrix} -6 &-3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ -4& -6 \end{bmatrix}$

$\small = \begin{bmatrix} -12+12 &-18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = -12-(-12) = -12+12 = 0$

So, $\small |A|I = 0\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Hence we get

$\small A (adj A)=(adj A)A=|A|I$

Given matrix: $\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Let $\small A= \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1} \begin{vmatrix} 0 &-2 \\ 0& 3 \end{vmatrix} = 0$

$\small A_{12} = (-1)^{1+2} \begin{vmatrix} 3 &-2 \\1& 3 \end{vmatrix} = -(9+2) =-11$

$\small A_{13} = (-1)^{1+3} \begin{vmatrix} 3 &0 \\ 1& 0 \end{vmatrix} = 0$

$\small A_{21} = (-1)^{2+1} \begin{vmatrix} -1 &2 \\ 0& 3 \end{vmatrix} = -(-3-0)= 3$

$\small A_{22} = (-1)^{2+2} \begin{vmatrix} 1 &2 \\ 1& 3 \end{vmatrix} = 3-2=1$

$\small A_{23} = (-1)^{2+3} \begin{vmatrix} 1 &-1 \\ 1& 0 \end{vmatrix} = -(0+1) = -1$

$\small A_{31} = (-1)^{3+1} \begin{vmatrix} -1 &2 \\ 0& -2 \end{vmatrix} = 2$

$\small A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &2 \\ 3& -2 \end{vmatrix} = -(-2-6) = 8$

$\small A_{33} = (-1)^{3+3} \begin{vmatrix} 1 &-1 \\ 3& 0 \end{vmatrix} = 0+3 =3$

Hence, $\small adjA = \begin{bmatrix} 0 &3 &2 \\ -11 & 1& 8\\ 0 &-1 & 3 \end{bmatrix}$

Now,

$\small A (adj A) =\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+11+0 &3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 &3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

also,

$\small A (adj A) =\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+9+2 &0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 &0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = 1(0-0) +1(9+2) +2(0-0) = 11$

So, $\small |A|I = 11\begin{bmatrix} 1 &0&0 \\ 0& 1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 &0&0 \\ 0& 11&0\\ 0&0&11 \end{bmatrix}$

Hence we get,

$\small A (adj A)=(adj A)A=|A|I$ .

Given matrix : $\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (3) = 3$

$A_{12} = (-1)^{1+2} (4) = -4$

$A_{21} = (-1)^{2+1} (-2) = 2$

$A_{22} = (-1)^{2+2} (2) = 2$

So, we have $adjA = \begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{14}\begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{14} &\frac{1}{7} \\ \\ \frac{-2}{7} & \frac{1}{7} \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (2) = 2$

$A_{12} = (-1)^{1+2} (-3) = 3$

$A_{21} = (-1)^{2+1} (5) =-5$

$A_{22} = (-1)^{2+2} (-1) = -1$

So, we have $adjA = \begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix} = \begin{bmatrix} \frac{2}{13} &\frac{-5}{13} \\ \\ \frac{3}{13} & \frac{-1}{13} \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}= A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(10-0)-2(0-0)+3(0-0) = 10$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (10) = 10$ $A_{12} = (-1)^{1+2} (0) = 0$

$A_{13} = (-1)^{1+3} (0) =0$ $A_{21} = (-1)^{2+1} (10) = -10$

$A_{22} = (-1)^{2+2} (5-0) = 5$ $A_{23} = (-1)^{2+1} (0-0) = 0$

$A_{31} = (-1)^{3+1} (8-6) = 2$ $A_{32} = (-1)^{3+2} (4-0) =-4$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 10 &-10 &2 \\ 0& 5 &-4 \\ 0& 0 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{10}\begin{bmatrix} 10 &-10 &2 \\ 0 & 5& -4\\ 0 &0 &2 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-3-0)-0(-3-0)+0(6-15) = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-3-0) = -3$ $A_{12} = (-1)^{1+2} (-3-0) = 3$

$A_{13} = (-1)^{1+3} (6-15) =-9$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-1-0) = -1$ $A_{23} = (-1)^{2+1} (2-0) = -2$

$A_{31} = (-1)^{3+1} (0-0) = 0$ $A_{32} = (-1)^{3+2} (0-0) =0$

$A_{33} = (-1)^{3+3} (3-0) = 3$

So, we have $adjA = \begin{bmatrix} -3 &0 &0 \\ 3& -1 &0 \\ -9& -2 &3 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{-1}{3}\begin{bmatrix} -3 &0 &0 \\ 3 & -1& 0\\ -9 &-2 &3 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 2(-1-0)-1(4-0)+3(8-7) =-2-4+3 = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-1-0) = -1$ $A_{12} = (-1)^{1+2} (4-0) = -4$

$A_{13} = (-1)^{1+3} (8-7) =1$ $A_{21} = (-1)^{2+1} (1-6) = 5$

$A_{22} = (-1)^{2+2} (2+21) = 23$ $A_{23} = (-1)^{2+1} (4+7) = -11$

$A_{31} = (-1)^{3+1} (0+3) = 3$ $A_{32} = (-1)^{3+2} (0-12) =12$

$A_{33} = (-1)^{3+3} (-2-4) = -6$

So, we have $adjA = \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(8-6)+1(0+9)+2(0-6) =2+9-12 = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (8-6) = 2$ $A_{12} = (-1)^{1+2} (0+9) = -9$

$A_{13} = (-1)^{1+3} (0-6) =-6$ $A_{21} = (-1)^{2+1} (-4+4) = 0$

$A_{22} = (-1)^{2+2} (4-6) = -2$ $A_{23} = (-1)^{2+1} (-2+3) = -1$

$A_{31} = (-1)^{3+1} (3-4) = -1$ $A_{32} = (-1)^{3+2} (-3-0) =3$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -2 &0 &1 \\ 9& 2 &-3 \\ 6& 1 &-2 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-\cos^2 \alpha-\sin^2 \alpha)+0(0-0)+0(0-0)$

$=-(\cos^2 \alpha + \sin^2 \alpha) = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-\cos^2 \alpha - \sin^2 \alpha) = -1$ $A_{12} = (-1)^{1+2} (0-0) = 0$

$A_{13} = (-1)^{1+3} (0-0) =0$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-\cos \alpha-0) = -\cos \alpha$ $A_{23} = (-1)^{2+1} (\sin \alpha-0) = -\sin \alpha$

$A_{31} = (-1)^{3+1} (0-0) = 0$ $A_{32} = (-1)^{3+2} (\sin \alpha-0) =-\sin \alpha$

$A_{33} = (-1)^{3+3} (\cos \alpha - 0) = \cos \alpha$

So, we have $adjA = \begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1}\begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix} = \begin{bmatrix}1 &0 &0 \\ 0&\cos \alpha &\sin \alpha \\ 0& \sin \alpha &-\cos \alpha \end{bmatrix}$

We have $\small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}$ and $\small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}$ .

then calculating;

$AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$

$=\begin{bmatrix} 18+49 &24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}$

Finding the inverse of AB.

Calculating the cofactors fo AB:

$AB_{11}=(-1)^{1+1}(61) = 61$ $AB_{12}=(-1)^{1+2}(47) = -47$

$AB_{21}=(-1)^{2+1}(87) = -87$ $AB_{22}=(-1)^{2+2}(67) = 67$

$adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}$

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse:

$(AB)^{-1}=\frac{1}{|AB|}adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 &-87 \\ -47 & 67 \end{bmatrix}$

$= \begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$ .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1 and |B| = 54- 56 = -2

$adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix}$ and $adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}$

therefore we have

$A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$ and $B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}$

Now calculating $B^{-1}A^{-1}$ .

$B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$

$=\begin{bmatrix} \frac{-45}{2}-8 && \frac{63}{2}+12 \\ \\ \frac{35}{2}+6 && \frac{-49}{2}-9 \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} && \frac{87}{2} \\ \\ \frac{47}{2} && \frac{-67}{2} \end{bmatrix}$ ........................(2)

From (1) and (2) we get

$\small (AB)^-^1=B^{-1}A^{-1}$

Hence proved.

Given $\small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}$ then we have to show the relation $A^2-5A+7I=0$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}$

therefore $A^2-5A+7I$ ;

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - 5\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} + \begin{bmatrix} 7 &0 \\ 0 & 7 \end{bmatrix}$

$\begin{bmatrix} 8-15+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}$

Hence $A^2-5A+7I = 0$ .

$\therefore A.A -5A = -7I$

$\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1}$

[ Post multiplying by $A^{-1}$ , also $|A| \neq 0$ ]

$\Rightarrow A(AA^{-1}) - 5I = -7A^{-1}$

$\Rightarrow AI - 5I = -7A^{-1}$

$\Rightarrow -\frac{1}{7}(AI - 5I)= \frac{1}{7}(5I-A)$

$\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}$

Given $\small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix}$ then we have the relation $A^2+aA+bI=O$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}$

therefore $A^2+aA+bI=O$ ;

$=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

So, we have equations;

$11+3a+b = 0,\ 8+2a = 0$ and $4+a = 0,and\ \ 3+a+b = 0$

We get $a = -4\ and\ b= 1$ .

Given matrix: $\small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$ ;

To show: $\small A^3-6A^2+5A+11I=O$

Finding each term:

$A^{2} = \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 1+1+2 &&1+2-1 &&1-3+3 \\ 1+2-6 &&1+4+3 &&1-6-9 \\ 2-1+6 &&2-2-3 && 2+3+9 \end{bmatrix}$

$= \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}$

$A^{3} = \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$

$= \begin{bmatrix} 4+2+2 &4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 & -3-24-42 \\ 7-3+28&7-6-14 &7+9+42 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}$

So now we have, $\small A^3-6A^2+5A+11I$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-6\begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}+5\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+11\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-\begin{bmatrix} 24 &&12 &&6 \\ -18 &&48 &&-84 \\ 42 &&-18 && 84 \end{bmatrix}+\begin{bmatrix} 5 &5 &5 \\ 5 &10 &-15 \\ 10 &-5 &15 \end{bmatrix}+\begin{bmatrix} 11 &0 &0 \\ 0 &11 & 0\\ 0& 0& 11 \end{bmatrix}$

$= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}$

$\Rightarrow A^{2}-6A +5I=- 11A^{-1}$

$A^{-1} = \frac{-1}{11}(A^{2}-6A+5I)$ ...................(1)

Now,

From equation (1) we get;

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}-6\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+5\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}$

Given matrix: $\small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$ ;

To show: $\small A^3-6A^2+9A-4I$

Finding each term:

$A^{2} = \begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 4+1+1 &&-2-2-1 &&2+1+2 \\ -2-2-1 &&1+4+1 &&-1-2-2 \\ 2+1+2 &&-1-2-2 && 1+1+4 \end{bmatrix}$

$= \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}$

$A^{3} =\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 12+5+5 &-6-10-5 &6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}$

$= \begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}$

So now we have, $\small A^3-6A^2+9A-4I$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}-6 \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}+9\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}-4\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}- \begin{bmatrix} 36 &&-30 &&30 \\ -30 &&36 &&-30 \\30 &&-30 && 36 \end{bmatrix}+\begin{bmatrix} 18 &-9 &9 \\ -9 &18 &-9 \\ 9 &-9 &18 \end{bmatrix}-\begin{bmatrix} 4 &0 &0 \\ 0 &4 & 0\\ 0& 0& 4 \end{bmatrix}$

$= \begin{bmatrix} 22-36+18-4 &-21+30-9 &21-30+9 \\ -21+30-9&22-36+18-4 &-21+30-9 \\ 21-30+9&-21+30-9 & 22-36+18-4 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = O$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +9AA^{-1}-4IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +9(AA^{-1})=4IA^{-1}$

$\Rightarrow A^{2}-6A +9I=4A^{-1}$

$A^{-1} = \frac{1}{4}(A^{2}-6A+9I)$ ...................(1)

Now,

From equation (1) we get;

$A^{-1} = \frac{1}{4}(\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}-6\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}+9\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{1}{4} \begin{bmatrix} 6-12+9 &&-5+6 &&5-6 \\ -5+6 &&6-12+9 &&-5+6 \\ 5-6 &&-5+6 && 6-12+9 \end{bmatrix}$

Hence inverse of A is :

$A^{-1} = \frac{1}{4} \begin{bmatrix} 3 &&1 &&-1 \\ 1 &&3 &&1 \\ -1 &&1 && 3 \end{bmatrix}$

We know the identity $(adjA)A = |A| I$

Hence we can determine the value of $|(adjA)|$ .

Taking both sides determinant value we get,

$|(adjA)A| = ||A| I|$ or $|(adjA)||A| = ||A||| I|$

or taking R.H.S.,

$||A||| I| = \begin{vmatrix} |A| & 0&0 \\ 0&|A| &0 \\ 0&0 &|A| \end{vmatrix}$

$= |A| (|A|^2) = |A|^3$

or, we have then $|(adjA)||A| = |A|^3$

Therefore $|(adjA)| = |A|^2$

Hence the correct answer is B.

Given that the matrix is invertible hence $A^{-1}$ exists and $A^{-1} = \frac{1}{|A|}adjA$

Let us assume a matrix of the order of 2;

$A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}$ .

Then $|A| = ad-bc$ .

$adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix}$ and $|adjA| = ad-bc$

Now,

$A^{-1} = \frac{1}{|A|}adjA$

Taking determinant both sides;

$|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}$

$\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}$

Therefore we get;

$|A^{-1}| = \frac{1}{|A|}$

Hence the correct answer is B.

NCERT determinants class 12 ncert solutions: Excercise- 4.6

We have given the system of equations:18967

$\small x+2y=2$

$\small 2x+3y=3$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &2 \\ 2&3 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 2\\3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(3) -2(2) = -1 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

We have given the system of equations:

$\small 2x-y=5$

$\small x+y=4$

The given system of equations can be written in the form of matrix; $AX =B$

where $A= \begin{bmatrix} 2 &-1 \\ 1&1 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 2(1) -1(-1) = 3 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

We have given the system of equations:

$\small x+3y=5$

$\small 2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &3 \\ 2&6 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\8 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6) -2(3) = 0$

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}$

So, $(adjA)B= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 5\\8 \end{bmatrix} = \begin{bmatrix} 30-24\\-10+8 \end{bmatrix}=\begin{bmatrix} 6\\-2 \end{bmatrix} \neq 0$

As, $(adjA)B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

We have given the system of equations:

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 1& 1&1 \\ 2& 3& 2\\ a& a &2a \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 1\\2 \\ 4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6a-2a) -1(4a-2a)+1(2a-3a)$

$= 4a -2a-a = 4a -3a =a \neq 0$

[ If zero then it won't satisfy the third equation ]

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

We have given the system of equations:

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

The given system of equations can be written in the form of matrix; $AX =B$

where $A = \begin{bmatrix} 3& -1&-2 \\ 0& 2& -1\\ 3& -5 &0 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 3(0-5) -(-1)(0+3)-2(0-6)$

$= -15 +3+12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

$(adjA) = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}$

$\therefore (adjA)B = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}\begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15\\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5\\-3 \\ -6 \end{bmatrix} \neq 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

We have given the system of equations:

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 5& -1&4 \\ 2& 3& 5\\ 5& -2 &6 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 5\\2 \\ -1 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 5(18+10) +1(12-25)+4(-4-15)$

$= 140-13-76 = 51 \neq 0$

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

The given system of equations

$\small 5x+2y=4$

$\small 7x+3y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &4 \\ 7& 3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 4\\5 \end{bmatrix}$

we have,

$|A| = 15-14=1 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = (adjA) = \begin{bmatrix} 3 &-2 \\ -7& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 3 &-2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 12-10\\ -28+25 \end{bmatrix} = \begin{bmatrix} 2\\-3 \end{bmatrix}$

Hence the solutions of the given system of equations;

x = 2 and y =-3 .

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &-1 \\ 3& 4 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} -2\\3 \end{bmatrix}$

we have,

$|A| = 8+3=11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-5}{11} \ and\ y =\frac{12}{11}.$

The given system of equations

$\small 4x-3y=3$

$\small 3x-5y=7$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 4 &-3 \\ 3& -5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\7 \end{bmatrix}$

we have,

$|A| = -20+9=-11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{-1}{11}\begin{bmatrix} -5 &3 \\ -3& 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 &-3 \\ 3& -4 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 5 &-3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3\\7 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 15-21\\ 9-28 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -6\\-19 \end{bmatrix}= \begin{bmatrix} -\frac{6}{11}\\ \\-\frac{19}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-6}{11} \ and\ y =\frac{-19}{11}.$

The given system of equations

$\small 5x+2y=3$

$\small 3x+2y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\5 \end{bmatrix}$

we have,

$|A| = 10-6=4 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =-1 \ and\ y =4.$

The given system of equations

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ and $B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}$

we have,

$|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(10+3) = 13$ $A_{12} =(-1)^{1+2}(-5-0) = 5$

$A_{13} =(-1)^{1+3}(3-0) = 3$ $A_{21} =(-1)^{2+1}(-5-3) = 8$

$A_{22} =(-1)^{2+2}(-10-0) = -10$ $A_{23} =(-1)^{2+3}(6-0) = -6$

$A_{31} =(-1)^{3+1}(-1+2) = 1$ $A_{32} =(-1)^{3+2}(-2-1) = 3$

$A_{33} =(-1)^{3+3}(-4-1) = -5$

$(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.$

The given system of equations

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}.$

we have,

$|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(1+3) = 4$ $A_{12} =(-1)^{1+2}(2+3) = -5$

$A_{13} =(-1)^{1+3}(2-1) = 1$ $A_{21} =(-1)^{2+1}(-1-1) = 2$

$A_{22} =(-1)^{2+2}(1-1) = 0$ $A_{23} =(-1)^{2+3}(1+1) = -2$

$A_{31} =(-1)^{3+1}(3-1) = 2$ $A_{32} =(-1)^{3+2}(-3-2) = 5$

$A_{33} =(-1)^{3+3}(1+2) = 3$

$(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =-1,\ and\ \ z=1.$

The given system of equations

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}.$

we have,

$|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(4+1) = 5$ $A_{12} =(-1)^{1+2}(-2-3) = 5$

$A_{13} =(-1)^{1+3}(-1+6) = 5$ $A_{21} =(-1)^{2+1}(-6+3) = 3$

$A_{22} =(-1)^{2+2}(-4-9) = -13$ $A_{23} =(-1)^{2+3}(-2-9) = 11$

$A_{31} =(-1)^{3+1}(3+6) = 9$ $A_{32} =(-1)^{3+2}(2-3) = 1$

$A_{33} =(-1)^{3+3}(-4-3) = -7$

$(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=-1.$

The given system of equations

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}.$

we have,

$|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{12-5} = 7$ $A_{12} =(-1)^{1+2}(9+10) = -19$

$A_{13} =(-1)^{1+3}(-3-8) = -11$ $A_{21} =(-1)^{2+1}(-3+2) = 1$

$A_{22} =(-1)^{2+2}(3-4) = -1$ $A_{23} =(-1)^{2+3}(-1+2) = -1$

$A_{31} =(-1)^{3+1}(5-8) = -3$ $A_{32} =(-1)^{3+2}(-5-6) = 11$

$A_{33} =(-1)^{3+3}(4+3) = 7$

$(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =1,\ and\ \ z=3.$

The given system of equations

$\dpi{100} 2x-3y+5z=11$

$\dpi{100} 3x+2y-4z=-5$

$\dpi{100} x+y-2z=-3$

can be written in the matrix form of AX =B, where

$\dpi{100} A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}.$

we have,

$|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{-4+4} = 0$ $A_{12} =(-1)^{1+2}(-6+4) = 2$

$A_{13} =(-1)^{1+3}(3-2) = 1$ $A_{21} =(-1)^{2+1}(6-5) = -1$

$A_{22} =(-1)^{2+2}(-4-5) = -9$ $A_{23} =(-1)^{2+3}(2+3) = -5$

$A_{31} =(-1)^{3+1}(12-10) = 2$ $A_{32} =(-1)^{3+2}(-8-15) = 23$

$A_{33} =(-1)^{3+3}(4+9) = 13$

$(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = -1\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix} = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=3.$

So, let us assume the cost of onion, wheat, and rice be x , y and z respectively.

Then we have the equations for the given situation :

$4x+3y+2z = 60$

$2x+4y+6z = 90$

$6x+2y+3y = 70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$ .

We have;

$A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix},$ $X= \begin{bmatrix} x\\y \\ z \end{bmatrix}$ $and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.$

$|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0$

Now, we will find the cofactors of A;

$A_{11} = (-1)^{1+1}(12-12) = 0$ $A_{12} = (-1)^{1+2}(6-36) = 30$

$A_{13} = (-1)^{1+3}(4-24) = -20$ $A_{21} = (-1)^{2+1}(9-4) = -5$

$A_{22} = (-1)^{2+2}(12-12) = 0$ $A_{23} = (-1)^{2+3}(8-18) = 10$

$A_{31} = (-1)^{3+1}(18-8) = 10$ $A_{32} = (-1)^{3+2}(24-4) = -20$

$A_{33} = (-1)^{3+3}(16-6) = 10$

$adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$ s

So, the solutions can be found by $X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =5,\ y =8,\ and\ \ z=8.$

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

NCERT solutions for class 12 maths chapter 4 Determinants: Miscellaneous exercise

Calculating the determinant value of $\dpi{100} \begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ ;

$= x\begin{bmatrix} -x &1 \\ 1& x \end{bmatrix}-\sin \Theta\begin{bmatrix} -\sin \Theta &1 \\ \cos \Theta& x \end{bmatrix} + \cos \Theta \begin{bmatrix} -\sin \Theta &-x \\ \cos \Theta& 1 \end{bmatrix}$

$= x(-x^2-1)-\sin \Theta (-x\sin \Theta-\cos \Theta)+\cos\Theta(-\sin \Theta+x\cos\Theta)$

$= -x^3-x+x\sin^2 \Theta+ \sin \Theta\cos \Theta-\cos\Theta\sin \Theta+x\cos^2\Theta$

$= -x^3-x+x(\sin^2 \Theta+\cos^2\Theta)$

$= -x^3-x+x = -x^3$

Clearly, the determinant is independent of $\Theta$ .

We have the

$\dpi{100} L.H.S. = \begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}$

Multiplying rows with a, b, and c respectively.

$\dpi{100} R_{1} \rightarrow aR_{1}, R_{2} \rightarrow bR_{2},\ and\ R_{3} \rightarrow cR_{3}$

we get;

$\dpi{100} = \frac{1}{abc} \begin{vmatrix} a^2 &a^3 &abc \\ b^2& b^3 &abc \\ c^2& c^3 & abc \end{vmatrix}$

$\dpi{100} = \frac{1}{abc}.abc \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix}$ $\dpi{100} [after\ taking\ out\ abc\ from\ column\ 3].$

$\dpi{100} = \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix} = \begin{vmatrix} 1&a^2 & a^3\\ 1& b^2 &b^3 \\ 1& c^2 & c^3 \end{vmatrix}$ $\dpi{100} [Applying\ C_{1}\leftrightarrow C_{3}\ and\ C_{2} \leftrightarrow C_{3}]$

= R.H.S.

Hence proved. L.H.S. =R.H.S.

Given determinant $\dpi{100} \begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$ ;

$= \cos \alpha \cos \beta \begin{vmatrix} \cos \beta &0 \\ \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} - \cos \alpha \sin \beta \begin{vmatrix} -\sin \beta & 0 \\ \sin \alpha \cos \beta & \cos \alpha \end{vmatrix} -\sin \alpha \begin{vmatrix} -\sin \beta &\cos \beta \\ \sin \alpha \cos \beta& \sin \alpha \sin \beta \end{vmatrix}$ $= \cos \alpha \cos \beta (\cos \beta \cos \alpha -0 )- \cos \alpha \sin \beta (-\cos \alpha\sin \beta- 0) -\sin \alpha (-\sin \alpha\sin^2\beta - \sin \alpha \cos^2 \beta)$

$= \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta +\sin^2 \alpha \sin^2\beta + \sin^2 \alpha \cos^2 \beta$

$= \cos^2 \alpha(\cos^2 \beta+\sin^2 \beta) +\sin^2 \alpha(\sin^2\beta+\cos^2 \beta)$

$= \cos^2 \alpha(1) +\sin^2 \alpha(1) = 1$ .

We have given $\dpi{100} \Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Applying the row transformations; $R_{1} \rightarrow R_{1} +R_{2} +R_{3}$ we have;

$\dpi{100} \Delta =\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Taking out common factor 2(a+b+c) from the first row;

$\dpi{100} \Delta =2(a+b+c)\begin{vmatrix} 1 & 1 &1 \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Now, applying the column transformations; $\dpi{100} C_{1}\rightarrow C_{1} - C_{2}\ and\ C_{2} \rightarrow C_{2}- C_{3}$

we have;

$\dpi{100} =2(a+b+c)\begin{vmatrix} 0 & 0 &1 \\ c-b &a-c &b+c \\ a-c & b-a & c+a \end{vmatrix}$

$\dpi{100} =2(a+b+c)[(c-b)(b-a)-(a-c)^2]$

$\dpi{100} =2(a+b+c)[ab+bc+ca-a^2-b^2-c^2]$

and given that the determinant is equal to zero. i.e., $\dpi{100} \triangle = 0$ ;

$\dpi{100} (a+b+c)[ab+bc+ca-a^2-b^2-c^2] = 0$

So, either $\dpi{100} (a+b+c) = 0$ or $\dpi{100} [ab+bc+ca-a^2-b^2-c^2] = 0$ .

we can write $\dpi{100} [ab+bc+ca-a^2-b^2-c^2] = 0$ as;

$\dpi{100} \Rightarrow -2ab-2bc-2ca+2a^2+2b^2+2c^2 =0$

$\dpi{100} \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 =0$

$\dpi{100} \because (a-b)^2,(b-c)^2,\ and\ (c-a)^2$ are non-negative.

Hence $\dpi{100} (a-b)^2= (b-c)^2=(c-a)^2 = 0$ .

we get then $\dpi{100} a=b=c$

Therefore, if given $\dpi{100} \triangle$ = 0 then either $\dpi{100} (a+b+c) = 0$ or $\dpi{100} a=b=c$ .

Question:5 Solve the equation

Given determinant $\dpi{100} \begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Applying the row transformation; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$\dpi{100} \begin{vmatrix} 3x+a & 3x+a &3x+a \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Taking common factor (3x+a) out from first row.

$\dpi{100} (3x+a)\begin{vmatrix} 1 & 1 &1 \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Now applying the column transformations; $\dpi{100} C_{1} \rightarrow C_{1}-C_{2}$ and $\dpi{100} C_{2} \rightarrow C_{2}-C_{3}$ .

we get;

$\dpi{100} (3x+a)\begin{vmatrix} 0 & 0 &1 \\ -a &a &x\\ 0 & -a & x+a \end{vmatrix} =0$

$\dpi{100} =(3x+a)(a^2)=0$ as $\dpi{100} a^2 \neq 0$ ,

or $\dpi{100} 3x+a=0$ or $\dpi{100} x= -\frac{a}{3}$

Given matrix $\dpi{100} \begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}$

Taking common factors a,b and c from the column $\dpi{100} C_{1}, C_{2}, and\ C_{3}$ respectively.

we have;

$\dpi{100} \triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b & a\\ b &b+c &c \end{vmatrix}$

Applying $\dpi{100} R_{2} \rightarrow R_{2}-R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$ , we have;

$\dpi{100} \triangle = abc\begin{vmatrix} a &c &a+c \\ b & b-c &-c\\ b-a &b &-a \end{vmatrix}$

Then applying $\dpi{100} R_{2} \rightarrow R_{2}+R_{1}$ , we get;

$\dpi{100} \triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ b-a &b &-a \end{vmatrix}$

Applying $\dpi{100} R_{3} \rightarrow R_{3}+R_{2}$ , we have;

$\dpi{100} \triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 2b &2b &0 \end{vmatrix} = 2ab^2c\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 1 &1 &0 \end{vmatrix}$

Now, applying column transformation; $\dpi{100} C_{2} \rightarrow C_{2 }-C_{1}$ , we have

$\dpi{100} \triangle = 2ab^2c\begin{vmatrix} a &c-a &a+c \\ a+b & -a &a\\ 1 &0 &0 \end{vmatrix}$

So we can now expand the remaining determinant along $\dpi{100} R_{3}$ we have;

$\dpi{100} \triangle = 2ab^2c\left [ a(c-a)+a(a+c) \right ]$

$\dpi{100} = 2ab^2c\left [ ac-a^2+a^2+ac) \right ] = 2ab^2c\left [ 2ac \right ]$

$\dpi{100} = 4a^2b^2c^2$

Hence proved.

We know from the identity that;

$(AB)^{-1} = B^{-1}A^{-1}$ .

Then we can find easily,

Given $\dpi{100} A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

Then we have to basically find the $B^{-1}$ matrix.

So, Given matrix $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

$|B| = 1(3-0) -2(-1-0)-2(2-0) = 3+2-4 = 1 \neq 0$

Hence its inverse $B^{-1}$ exists;

Now, as we know that

$B^{-1} = \frac{1}{|B|} adjB$

So, calculating cofactors of B,

$B_{11} = (-1)^{1+1}(3-0) = 3$ $B_{12} = (-1)^{1+2}(-1-0) = 1$

$B_{13} = (-1)^{1+3}(2-0) = 2$ $B_{21} = (-1)^{2+1}(2-4) = 2$

$B_{22} = (-1)^{2+2}(1-0) = 1$ $B_{23} = (-1)^{2+3}(-2-0) = 2$

$B_{31} = (-1)^{3+1}(0+6) = 6$ $B_{32} = (-1)^{3+2}(0-2) = 2$

$B_{33} = (-1)^{3+3}(3+2) = 5$

$adjB = \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|} adjB = \frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

Now, We have both $A^{-1}$ as well as $B^{-1}$ ;

Putting in the relation we know; $(AB)^{-1} = B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$

$= \begin{bmatrix} 9-30+30 &-3+12-12 &3-10+12 \\ 3-15+10&-1+6-4 &1-5+4 \\ 6-30+25 &-2+12-10 &2-10+10 \end{bmatrix}$

$= \begin{bmatrix} 9 &-3 &5 \\ -2&1 &0 \\ 1 &0 &2\end{bmatrix}$

Given that $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$ matrix and $adjA = C$ then;

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

Finding the inverse of C;

$|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0$

Hence its inverse exists;

$C^{-1} = \frac{1}{|C|}adj C$

Now, finding the $adjC$ ;

$C_{11}= (-1)^{1+1}(-4-1) = -5$ $C_{12}= (-1)^{1+2}(9+1) = -10$

$C_{13}= (-1)^{1+3}(-9+4) = -5$ $C_{21}= (-1)^{2+1}(9+1) = -10$

$C_{22}= (-1)^{2+2}(-14-1) = -15$ $C_{23}= (-1)^{2+3}(14-9) = -5$

$C_{31}= (-1)^{3+1}(-9+4) = -5$ $C_{32}= (-1)^{3+2}(14-9) = -5$

$C_{33}= (-1)^{3+3}(56-81) = -25$

$adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}$

$C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

or $L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Now, finding the R.H.S.

$adj (A^{-1}) = adj B$

$A^{-1} =B= \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5}&& \frac{-4}{5}&& \frac{-1}{5}\\ \\ \frac{1}{5}&& \frac{-1}{5} &&\frac{1}{5}\end{bmatrix}$

Cofactors of B;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$

$B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$

$B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1$

$R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Hence L.H.S. = R.H.S. proved.

Given that $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

Finding the inverse of B ;

$|B| = \frac{-14}{5}(\frac{-4}{25}-\frac{1}{25})-\frac{9}{5}(\frac{9}{25}+\frac{1}{25})+\frac{1}{5}(\frac{-9}{25}+\frac{4}{25})$

$= \frac{70}{125}-\frac{90}{125}-\frac{5}{125} = \frac{-25}{125} = \frac{-1}{5} \neq 0$

Hence its inverse exists;

$B^{-1} = \frac{1}{|B|}adj B$

Now, finding the $adjB$ ;

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$ $B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) = \frac{-2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$ $B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$ $B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = \frac{-25}{25} =-1$

$adjB = \begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|}adjB = \frac{-5}{1}\begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}= \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$L.H.S. = B^{-1} = (A^{-1})^{-1} = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$R.H.S. = A = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1& 1 &5 \end{bmatrix}$

Hence proved L.H.S. =R.H.S. .

We have determinant $\triangle = \begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ , we have then;

$\triangle = \begin{vmatrix} 2(x+y) & 2(x+y) &2(x+y) \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Taking out the common factor 2(x+y) from the row first.

$= 2(x+y)\begin{vmatrix} 1 & 1 &1 \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Now, applying the column transformation; $C_{1} \rightarrow C_{1} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{1}$ we have ;

$= 2(x+y)\begin{vmatrix} 0 & 0 &1 \\ -x & y &x \\ y & x-y & y \end{vmatrix}$

Expanding the remaining determinant;

$= 2(x+y)(-x(x-y)-y^2) = 2(x+y)[-x^2+xy-y^2]$

$= -2(x+y)[x^2-xy+y^2] = -2(x^3+y^3)$ .

We have determinant $\triangle = \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ then we have then;

$\triangle = \begin{vmatrix} 0 & -y &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Taking out the common factor -y from the row first.

$\triangle = -y\begin{vmatrix} 0 & 1 &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Expanding the remaining determinant;

$-y[1(-x-o)] = xy$

Question:11 Using properties of determinants, prove that

Given determinant $\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}$

Applying Row transformations; and $R_{3} \rightarrow R_{3}-R_{1}$ , then we have;

$\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta -\alpha & \beta ^2 - \alpha^2 &\alpha - \beta \\ \gamma-\alpha &\gamma ^2-\alpha^2 &\alpha -\gamma \end{vmatrix}$

$= (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ 1 & \beta + \alpha &-1 \\ 0 &\gamma-\beta &0\end{vmatrix}$

Expanding the remaining determinant;

$= (\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$

$= (\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$

$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

hence the given result is proved.

Question:12 Using properties of determinants, prove that

Given the determinant $\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}$

Applying the row transformations; $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$ then we have;

$\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y-x& y^2-x^2& p(y^3-x^3)\\ z-x&z^2-x^2 & p(z^3-x^3) \end{vmatrix}$

Applying row transformation $R_{3} \rightarrow R_{3} - R_{2}$ we have then;

$\triangle =(y-x )(z-x)(z-y)\begin{vmatrix} x & x^2&1+px^3 \\ 1& y+x& p(y^2+x^2+xy)\\ 0&1 & p(x+y+z) \end{vmatrix}$

Now we can expand the remaining determinant to get the result;

$\triangle =(y-x )(z-x)(z-y)[(-1)(p)(xy^2+x^3+x^2y)+1+px^3+p(x+y+z)(xy)]$

$=(x-y)(y-z)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]$

$=(x-y)(y-z)(z-x)(1+pxyz)$

hence the given result is proved.

Question:13 Using properties of determinants, prove that

Given determinant $\dpi{100} \triangle = \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}$

Applying the column transformation, $\dpi{100} C_{1} \rightarrow C_{1} +C_{2}+C_{3}$ we have then;

$\dpi{100} \triangle = \begin{vmatrix} a+b+c &-a+b &-a+c \\ a+b+c &3b &-b+c \\ a+b+c &-c+b &3c \end{vmatrix}$

Taking common factor (a+b+c) out from the column first;

$\dpi{100} =(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 1 &3b &-b+c \\1 &-c+b &3c \end{vmatrix}$

Applying $\dpi{100} R_{2} \rightarrow R_{2}-R_{1}$ and $\dpi{100} R_{3} \rightarrow R_{3}-R_{1}$ , we have then;

$\dpi{100} \triangle=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 0 &2b+a &a-b \\0 &a-c &2c+a \end{vmatrix}$

Now we can expand the remaining determinant along $\dpi{100} C_{1}$ we have;

$\dpi{100} \triangle=(a+b+c) [(2b+a)(2c+a)-(a-b)(a-c)]$

$\dpi{100} =(a+b+c) [4bc+2ab+2ac+a^2-a^2+ac+ba-bc]$

$\dpi{100} =(a+b+c)(3ab+3bc+3ac)$

$\dpi{100} =3(a+b+c)(ab+bc+ac)$

Hence proved.

Question:14 Using properties of determinants, prove that

Given determinant $\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}$

Applying the row transformation; $R_{2} \rightarrow R_{2}-2R_{1}$ and $R_{3} \rightarrow R_{3}-3R_{1}$ we have then;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&3 & 7+3p \end{vmatrix}$

Now, applying another row transformation $R_{3}\rightarrow R_{3}-3R_{2}$ we have;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&0 & 1 \end{vmatrix}$

We can expand the remaining determinant along $C_{1}$ , we have;

$\triangle = 1\begin{vmatrix} 1 & 2+p\\ 0 &1 \end{vmatrix} = 1(1-0) =1$

Hence the result is proved.

Question:15 Using properties of determinants, prove that

Given determinant $\triangle = \begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}$

Multiplying the first column by $\sin \delta$ and the second column by $\cos \delta$ , and expanding the third column, we get

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \sin \alpha \sin \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \sin \beta\sin \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \sin \gamma\sin \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Applying column transformation, $C_{1} \rightarrow C_{1}+C_{3}$ we have then;

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \cos \alpha \cos \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \cos \beta\cos \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \cos \gamma\cos \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Here we can see that two columns $C_{1}\ and\ C_{2}$ are identical.

The determinant value is equal to zero. $\therefore \triangle = 0$

Hence proved.

Question:16 Solve the system of equations

We have a system of equations;

$\dpi{100} \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\dpi{100} \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\dpi{100} \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, $\frac{1}{x} = a$ , $\frac{1}{y} = b\ and\ \frac{1}{z} = c$

Then we have the equations;

$\dpi{100} 2a +3b+10c = 4$

$\dpi{100} 4a-6b+5c =1$

$\dpi{100} 6a+9b-20c = 2$

We can write it in the matrix form as $\dpi{100} AX =B$ , where

$\dpi{100} A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.$

Now, Finding the determinant value of A;

$\dpi{100} |A| = 2(120-45)-3(-80-30)+10(36+36)$

$\dpi{100} =150+330+720$

$\dpi{100} =1200 \neq 0$

Hence we can say that A is non-singular $\dpi{100} \therefore$ its invers exists;

Finding cofactors of A;

$\dpi{100} A_{11} = 75$ , $\dpi{100} A_{12} = 110$ , $\dpi{100} A_{13} = 72$

$\dpi{100} A_{21} = 150$ , $\dpi{100} A_{22} = -100$ , $\dpi{100} A_{23} = 0$

$\dpi{100} A_{31} =75$ , $\dpi{100} A_{31} =30$ , $\dpi{100} A_{33} =-24$

$\dpi{100} \therefore$ as we know $\dpi{100} A^{-1} = \frac{1}{|A|}adjA$

$\dpi{100} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}$

Now we will find the solutions by relation $\dpi{100} X = A^{-1}B$ .

$\dpi{100} \Rightarrow \begin{bmatrix} a\\b \\ c \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}\begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}$

$\dpi{100} = \frac{1}{1200}\begin{bmatrix} 300+150+150\\440-100+60 \\ 288+0-48 \end{bmatrix}$

$\dpi{100} = \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}$

Therefore we have the solutions $\dpi{100} a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.$

Or in terms of x, y, and z;

$\dpi{100} x =2,\ y =3,\ and\ z = 5$

Given determinant $\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$ and given that a, b, c are in A.P.

That means , 2b =a+c

$\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+(a+c)\\ x+4 & x+5 &x+2c \end{vmatrix}$

Applying the row transformations, $R_{1} \rightarrow R_{1} -R_{2}$ and then $R_{3} \rightarrow R_{3} -R_{2}$ we have;

$\triangle = \begin{vmatrix} -1 &-1 &a-c \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Now, applying another row transformation, $R_{1} \rightarrow R_{1} + R_{3}$ , we have

$\triangle = \begin{vmatrix} 0 &0 &0 \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Clearly we have the determinant value equal to zero;

Hence the option (A) is correct.

Given Matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$ ,

$|A| = x(yz-0) =xyz$

As we know,

$A^{-1} = \frac{1}{|A|}adjA$

So, we will find the $adjA$ ,

Determining its cofactor first,

$A_{11} = yz$ $A_{12} = 0$ $A_{13} = 0$

$A_{21} = 0$ $A_{22} = xz$ $A_{23} = 0$

$A_{31} = 0$ $A_{32} = 0$ $A_{33} = xy$

Hence $A^{-1} = \frac{1}{|A|}adjA = \frac{1}{xyz}\begin{bmatrix} yz &0 &0 \\ 0& xz & 0\\ 0& 0& xy \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{1}{x} &&0 &&0 \\ 0&& \frac{1}{y} && 0\\ 0&& 0&& \frac{1}{z} \end{bmatrix}$

Therefore the correct answer is (A)

Given determinant $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix}$

$|A| = 1(1+\sin^2 \Theta) -\sin \Theta(-\sin \Theta+\sin \Theta)+1(\sin^2 \Theta +1)$

$= 1+ \sin ^2 \Theta + \sin ^2 \Theta +1$

$= 2+2\sin ^2 \Theta = 2(1+\sin^2 \Theta)$

Now, given the range of $\Theta$ from $0\leq \Theta \leq 2\pi$

$\Rightarrow 0 \leq \sin \Theta \leq 1$

$\Rightarrow 0 \leq \sin^2 \Theta \leq 1$

$\Rightarrow 1 \leq 1+\sin^2 \Theta \leq 2$

$\Rightarrow 2 \leq 2(1+\sin^2 \Theta) \leq 4$

Therefore the $|A|\ \epsilon\ [2,4]$ .

Hence the correct answer is D.

If you are interested in Determinants Class 12 NCERT Solutions exercises then these are listed below.

### An insight to the NCERT solutions for Class 12 Maths Chapter 4 Determinants:

The six exercises of NCERT Class 12 Maths solutions chapter 4 Determinants covers the properties of determinants, co-factors and applications like finding the area of triangle, solutions of linear equations in two or three variables, minors, consistency and inconsistency of system of linear equations, adjoint and inverse of a square matrix, and solution of linear equations in two or three variables using inverse of a matrix. You can also check Determinants NCERT solutions if you are facing any problems during practice.

What are the Determinants?

To every square matrix of order n, we can associate a number (real or complex) called determinant of the square matrix A. Let's take a determinant (A) of order two-

If A is a then the determinant of A is written as |A|

matrix ,

The six exercises of this chapter determinants covers the properties of determinants, co-factors and applications like finding the area of triangle, solutions of linear equations in two or three variables, minors, consistency and inconsistency of system of linear equations, adjoint and inverse of a square matrix, and solution of linear equations in two or three variables using inverse of a matrix.

Topics and sub-topics of NCERT class 12 maths chapter 4 Determinants

4.1 Introduction

4.2 Determinant

4.2.1 Determinant of a matrix of order one

4.2.2 Determinant of a matrix of order two

4.2.3 Determinant of a matrix of order 3 × 3

4.3 Properties of Determinants

4.4 Area of a Triangle

4.5 Minors and Cofactors

4.6 Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

4.7.1 Solution of a system of linear equations using the inverse of a matrix

NCERT exemplar solutions class 12 maths chapter 4

Topics of NCERT Class 12 Maths Chapter Determinants

The main topics covered in chapter 4 maths class 12 are:

• Determinants

Ch 4 maths class 12 includes concepts of calculation of determinants with respect to their order one, two, three. Also class 12 NCERT topics discuss concepts related to the expansion of the matrix to calculate the determinant. there are good quality questions in Determinants class 12 solutions.

• Properties of determinants

This ch 4 maths class 12 comprehensively and elaborately discussed the properties of determinants, which are vastly used. To get a good hold of these concepts you can refer to NCERT solutions for class 12 maths chapter 4.

• Area of triangle

This ch 4 maths class 12 also includes concepts of the area of a triangle in which vertices are given. You can refer to class 12 NCERT solutions for questions about these concepts.

• Minors and Cofactors

Maths class 12 chapter 4 discussed the minors and cofactors. To get command of these concepts you can go through the NCERT solution for class 12 maths chapter 4.

• Adjoint and Inverse of a matrix

concepts related to adjoints and inverse of the matrix are detailed in maths class 12 chapter 4. And it also concerns conditions for the existence of the inverse of a matrix. Determinants class 12 solutions include quality questions to understand the concepts.

• Applications of determinants and matrix

ch 4 maths class 12 deliberately discussed the applications of determinants and matrices. it also includes the terms consistent system inconsistent system. concepts related to the solution of a system of linear equations using the inverse of a matrix. For questions on these concepts, you can browse NCERT solutions for class 12 chapter 4.

Topics mentioned in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 4

## NCERT solutions for class 12 Maths - Chapter wise

 Chapter 1 NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Chapter 2 NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions Chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices Chapter 4 NCERT solutions for class 12 maths chapter 4 Determinants Chapter 5 NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives Chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals Chapter 8 NCERT solutions for class 12 maths chapter 8 Application of Integrals Chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations Chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra Chapter 11 NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry Chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming Chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

## NCERT Solutions - Class Wise

### Benefits of NCERT solutions for Class 12 Maths Chapter 4 Determinants:

• NCERT Class 12 Maths solutions chapter 4 will assist the students in the exam preparation in a strategic way.

• Class 12 Maths Chapter 4 NCERT solutions are prepared by the experts, therefore, students can rely upon the same without any second thought .

• NCERT solutions for Class 12 Maths Chapter 4 provides the detailed solution for all the questions. This will help the students in analysing and understanding the questions in a better way.

## NCERT Books and NCERT Syllabus

1. What are the key themes discussed in chapter 4 maths class 12 ncert solutions?

NCERT Solutions for Class 12 Maths Chapter 4 primarily focuses on the topic of determinants. This chapter covers the following key themes:

1. Definition of determinants

2. Properties of determinants

3. Area of a parallelogram and a triangle

4. The inverse of a matrix

5. Adjoint and inverse of a matrix

6. Solutions of linear equations using matrices

7. Determinant as scaling factor

2. What is the weightage of the chapter determinants for CBSE board exam?

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination. students can prioritise their subjects according to respective weightage and study accordingly.

3. How are the NCERT solutions helpful in the board exam?

Only knowing the answer does not guarantee to score good marks in the exam. One should know how to answer in order to get good marks. NCERT solutions are provided by the experts who know how best to write answers in the board exam in order to get good marks in the board exam.

4. Which is the best book for CBSE class 12 Maths ?

NCERT textbook is the best book for CBSE class 12 maths. Most of the questions in CBSE class 12 board exam are directly asked from NCERT textbook. So you don't need to buy any supplementary books for CBSE class 12 maths.

5. What applications of determinants are discussed in ncert solutions class 12 maths chapter 4?

According to NCERT Solutions for Class 12 Maths Chapter 4, determinants play a crucial role in algebra and have multiple practical applications. The concept of determinants is valuable in solving systems of linear equations. With determinants, students can explore concepts such as changes in area, volume, and variables through integrals. Additionally, determinants can be used to determine the values of square matrices. Interested students can study determinants class 12 ncert pdf both online and offline.

6. Where can I find the complete solutions of NCERT class 12 Maths ?

Here you will get the detailed NCERT solutions for class 12 maths  by clicking on the link. also you can find these in official web page of careers360.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

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3 Jobs Available
##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available
##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available
##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Chemical Pathologist

Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid.

2 Jobs Available
##### Biochemical Engineer

A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Engineer

A career as Production Engineer is crucial in the manufacturing industry. He or she ensures the functionality of production equipment and machinery to improve productivity and minimize production costs in order to drive revenues and increase profitability.

2 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available
##### Product Designer

Individuals who opt for a career as product designers are responsible for designing the components and overall product concerning its shape, size, and material used in manufacturing. They are responsible for the aesthetic appearance of the product. A product designer uses his or her creative skills to give a product its final outlook and ensures the functionality of the design.

Students can opt for various product design degrees such as B.Des and M.Des to become product designers. Industrial product designer prepares 3D models of designs for approval and discusses them with clients and other colleagues. Individuals who opt for a career as a product designer estimate the total cost involved in designing.

2 Jobs Available
##### R&D Personnel

A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

2 Jobs Available
##### Commercial Manager

A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available