NCERT solutions for Class 12 Maths Chapter 4 Determinants

NCERT solutions for Class 12 Maths Chapter 4 - Determinants

Updated on Mar 24, 2025 06:12 PM IST | #CBSE Class 12th

Determinants is a significant Class 12 NCERT Maths topic that helps us in solving the system of linear equations, calculating the area of a triangle, and for clear knowledge of the theory of matrices. In this chapter, we handle the concepts of determinants, i.e., properties, how to calculate it, and application. This article helps in making the topic easy for Class 12 students by giving clear and detailed solutions, examples, and NCERT solutions. We also handle the questions such as 'What is a determinant?', 'How to calculate the determinant of a matrix?', and 'What are the various properties of determinants?'.

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NCERT Class 12 Maths Chapter 4 Question Answer - Important Formulae
NCERT Solutions for Class 12 Maths: Chapter Wise
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NCERT solutions for Class 12 Maths Chapter 4 - Determinants

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The NCERT Solutions for Class 12 Maths Chapter 4 Determinants are prepared by Careers360 experts and offer concise and straightforward solutions for students sitting for their CBSE Class 12 board exams. The NCERT solutions for class 12 cover all topics of the NCERT textbook, providing necessary steps for explanations of a wide range of problems. These solutions include all NCERT textbook topics and elaborate on every step so that the fundamental concepts of determinants are understood by the students. The NCERT solutions for Class 12 Maths Chapter 4 are a helpful tool for studying determinants. Students are encouraged to thoroughly study the Class 12 Maths Chapter 4 Determinants Notes to cover all the concepts present in this chapter. Also, go through chapter-wise NCERT solutions for math, and for more practice, you can use NCERT Exemplar Solutions For Class 12 Maths Chapter 4 Determinants.

NCERT Determinants Class 12 Questions And Answers PDF Free Download

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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 $a_{i j}$ of a determinant is a determinant obtained by deleting the ith row and jth column in which element $a_{i j}$ lies.

The cofactor of an element $a_{i j}$ of a determinant, denoted by $A_{i j}$ or $C_{i j}$ , is defined as $A_{i j} = (- 1)^{i + j} \cdot M_{i j}$ , where $M_{i j}$ is the minor of element $a_{i j}$ .

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

For a 2x2 matrix A: $| A | = a_{11} \cdot a_{22} - a_{12} \cdot a_{21}$

For a 3x3 matrix A: $| A | = a_{11} \cdot | A_{11} | - a_{12} \cdot | A_{12} | + a_{13} \cdot | A_{13} |$

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., $| A B | = | A | \cdot | B |$ .

Adjoint of a Matrix:

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:

$A \cdot adj (A) = adj (A) \cdot A = | A | I_{n}$ (Identity Matrix)

$| adj (A) | = | A |^{n - 1}$

$adj (A^{T}) = (adj (A))^{T}$ (Transpose of the adjoint)

Finding the Area of a Triangle Using Determinants:

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

1694619308727
The inverse of a Square Matrix:

For a non-singular matrix A ( $| A | \neq 0$ ), the inverse $A^{- 1}$ is defined as $A^{- 1} = \frac{1}{| A |} \cdot adj (A)$ .

Properties of an Inverse Matrix:

${(A^{- 1})}^{- 1} = A$

${(A^{T})}^{- 1} = {(A^{- 1})}^{T}$

$(A B)^{- 1} = B^{- 1} A^{- 1}$

$(A B C)^{- 1} = C^{- 1} B^{- 1} A^{- 1}$

$adj (A^{- 1}) = (adj (A))^{- 1}$

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

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

Case I: If $| A | \neq 0$ , the system is consistent, and $X = A^{- 1} B$ has a unique solution.

Case II: If $| A | = 0$ and $(adj A) B \neq 0$ , 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 (Exercise)

Class 12 Maths chapter 4 solutions - Exercise: 4.1
Page number: 81-82
Total questions: 8

Question:1 Evaluate the following determinant- $| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} |$

Answer:

The determinant is evaluated as follows

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

Question:2 Evaluate the following determinant-

(i) $| \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} |$

(ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

Answer:

(i) The given two determinant is calculated as follows

$100 | \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} | = c o s θ (\cos θ) - (- \sin θ) \sin θ = \cos^{2} θ + \sin^{2} θ = 1$

(ii) We have determinant $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

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

$\Rightarrow$ $(x + 1) (x^{2} - x + 1 - x + 1) = (x + 1) (x^{2} - 2 x + 2)$

$\Rightarrow$ $x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2$

$\Rightarrow$ $x^{3} - x^{2} + 2$

Question:3 If $A = [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ , then show that $| 2 A | = 4 | A |$

Answer:

Given determinant $A = [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ then we have to show that $| 2 A | = 4 | A |$ ,

So, $A = [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ then, $2 A = 2 [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}] = [\begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix}]$

Hence we have $| 2 A | = | \begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix} | = 2 (4) - 4 (8) = - 24$

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

then calculating R.H.S. $4 | A |$

We have,

$| A | = | \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} | = 1 (2) - 2 (4) = - 6$

hence R.H.S becomes $4 | A | = 4 \times (- 6) = - 24$

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

Hence proved.

Question:4 If $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}]$ then show that $| 3 A | = 27 | A |$

Answer:

Given Matrix $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}]$

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

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

calculating $27 | A |$ ,

$\Rightarrow$ $| A | = | \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} | = 1 | \begin{matrix} 1 & 2 \\ 0 & 4 \end{matrix} | - 0 | \begin{matrix} 0 & 2 \\ 0 & 4 \end{matrix} | + 1 | \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} | = 4 - 0 + 0 = 4$

So, $27 | A | = 27 (4) = 108$

Therefore $| 3 A | = 27 | A |$ .

Hence proved.

Question:5 Evaluate the determinants.

(i) $| \begin{matrix} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

(ii) $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} |$

(iii) $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$

(iv) $| \begin{matrix} 2 & - 1 & 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

Answer:

(i) Given the determinant $| \begin{matrix} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$ ;

now, calculating its determinant value,

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

$= 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 0) = - 15 + 3 - 0 = - 12$ .

(ii) Given determinant $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} |$ ;

Now calculating the determinant value;

$\Rightarrow$ $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} | = 3 | \begin{matrix} 1 & - 2 \\ 3 & 1 \end{matrix} | - (- 4) | \begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix} | + 5 | \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} |$

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

(iii) Given determinant $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$ ;

Now calculating the determinant value;

$\Rightarrow$ $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} | = 0 | \begin{matrix} 0 & - 1 \\ 3 & 0 \end{matrix} | - 1 | \begin{matrix} - 1 & - 3 \\ - 2 & 0 \end{matrix} | + 2 | \begin{matrix} - 1 & 0 \\ - 2 & 3 \end{matrix} |$

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

(iv) Given determinant: $| \begin{matrix} 2 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$ ,

We now calculate the determinant value:

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

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

Question:6 If $A = [\begin{matrix} 1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9 \end{matrix}]$ , then find $| A |$ .

Answer:

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

Finding the determinant value of A;

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

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

Question:7 Find values of x, if

(i) $| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} | = | \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} |$

(ii) $| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | = | \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} |$

Answer:

(i) Given that $| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} | = | \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} |$

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

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

So, we have then,

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

(ii) Given $| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | = | \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} |$ ;

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

L.H.S. determinant value;

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

Similarly R.H.S. determinant value;

$| \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} | = 5 (x) - 3 (2 x) = 5 x - 6 x = - x$

So, we have then;

$- 2 = - x$ or $x = 2$ .

Question:8 If $| \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = | \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} |$ , then $x$ is equal to

(A) $6$ (B) $\pm 6$ (C) $- 6$ (D) $0$

Answer:

Solving the L.H.S. determinant ;

$100 | \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = x^{2} - 36$

and solving R.H.S determinant;

$| \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} | = 36 - 36 = 0$

So equating both sides;

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

Hence answer is (B).

Class 12 Maths chapter 4 solutions Exercise: 4.2
Page number: 83-84
Total questions: 5

Question:1 Find the area of the triangle with vertices at the point given in each of the following :

(i) $(1, 0), (6, 0), (4, 3)$

(ii) $(2, 7), (1, 1), (10, 8)$

(iii) $(- 2, - 3), (3, 2), (- 1, - 8)$

Answer:

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

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{matrix} |$

Expanding using the second column

$\Rightarrow$ $\frac{1}{2} (- 3) | \begin{matrix} 1 & 1 \\ 6 & 1 \end{matrix} |$

$\Rightarrow$ $\frac{15}{2} s q u a r e u n i t s .$

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

$\Rightarrow$ $△ = | \begin{matrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{matrix} |$

$\Rightarrow$ $\frac{1}{2} | \begin{matrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{matrix} | = \frac{1}{2} [2 (1 - 8) - 7 (1 - 10) + 1 (8 - 10)]$

$\Rightarrow$ $\frac{1}{2} [2 (- 7) - 7 (- 9) + 1 (- 2)] = \frac{1}{2} [- 14 + 63 - 2] = \frac{47}{2} s q u a r e u n i t s .$

(iii) Area of the triangle by the determinant method:

$\Rightarrow$ $A r e a △ = \frac{1}{2} | \begin{matrix} - 2 & - 3 & 1 \\ 3 & 2 & 1 \\ - 1 & - 8 & 1 \end{matrix} |$

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

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

Hence the area is equal to $| - 15 | = 15 s q u a r e u n i t s .$

Question:2 Show that points $A (a b + c), B (b, c + a), C (c, a + b)$ are collinear.

Answer:

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,

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{matrix} |$

calculating the area:

$\Rightarrow$ $\frac{1}{2} [a | \begin{matrix} c + a & 1 \\ a + b & 1 \end{matrix} | - (b + c) | \begin{matrix} b & 1 \\ c & 1 \end{matrix} | + 1 | \begin{matrix} b & c + a \\ c & a + b \end{matrix} |]$

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

$\Rightarrow$ $\frac{1}{2} [a c - a b - b^{2} + c^{2} + a b + b^{2} - c^{2} - a c] = \frac{1}{2} [0] = 0$

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

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

Question:3 Find values of k if the area of a triangle is 4 sq. units and vertices are

(i) $(k, 0), (4, 0), (0, 2)$

(ii) $(- 2, 0), (0, 4), (0, k)$

Answer:

(i) We can easily calculate the area by the formula :

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{matrix} | = 4 s q . u n i t s$

$\Rightarrow$ $\frac{1}{2} [k | \begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix} | - 0 | \begin{matrix} 4 & 1 \\ 0 & 1 \end{matrix} | + 1 | \begin{matrix} 4 & 0 \\ 0 & 2 \end{matrix} |] = 4 s q . u n i t s$

$\Rightarrow$ $\frac{1}{2} [k (0 - 2) - 0 + 1 (8 - 0)] = \frac{1}{2} [- 2 k + 8] = 4 s q . u n i t s$

$\Rightarrow$ $[- 2 k + 8] = 8 s q . u n i t s$ or $- 2 k + 8 = \pm 8 s q . u n i t s$

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

Hence two values are possible for k.

(ii) The area of the triangle is given by the formula:

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} - 2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{matrix} | = 4 s q . u n i t s .$

Now, calculating the area:

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

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

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

Question:4 Find the equation of the line joining

(i) $(1, 2)$ and $(3, 6)$ using determinants.

(ii) $(3, 1)$ and $(9, 3)$ using determinants.

Answer:

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

Therefore area formed by them will be equal to zero.

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{matrix} | = 0$

So, we have:

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

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

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

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

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

$△ = \frac{1}{2} | \begin{matrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{matrix} | = 0$

Calculating the determinant:

$\Rightarrow$ $\frac{1}{2} [3 | \begin{matrix} 3 & 1 \\ y & 1 \end{matrix} | - 1 | \begin{matrix} 9 & 1 \\ x & 1 \end{matrix} | + 1 | \begin{matrix} 9 & 3 \\ x & y \end{matrix} |]$

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

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

Hence we have the line equation:

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

Question:5 If the area of a triangle is 35 sq units with vertices $(2, - 6), (5, 4)$ and $(k, 4)$ . Then k is

(A) $12$ (B) $- 2$ (C) $- 12, - 2$ (D) $12, - 2$

Answer:

The area of a triangle is given by:

$\Rightarrow$ $△ = \frac{1}{2} | \begin{matrix} 2 & - 6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} | = 35 s q . u n i t s .$

or $| \begin{matrix} 2 & - 6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} | = 70 s q . u n i t s .$

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

$\Rightarrow$ $2 (4 - 4) + 6 (5 - k) + (20 - 4 k) = \pm 70$

$\Rightarrow$ $50 - 10 k = \pm 70$

$\Rightarrow$ $k = 12$ or $k = - 2$

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

Therefore option (D) is correct.

Class 12 Maths chapter 4 solutions Exercise: 4.3
Page number: 87
Total questions: 5

Question:1 Write Minors and Cofactors of the elements of the following determinants:

(i) $| \begin{matrix} 2 & - 4 \\ 0 & 3 \end{matrix} |$

(ii) $| \begin{matrix} a & c \\ b & d \end{matrix} |$

Answer:

(i) GIven determinant: $| \begin{matrix} 2 & - 4 \\ 0 & 3 \end{matrix} |$

Minor of element $a_{i j}$ is $M_{i j}$ .

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 $a_{i j}$ is $A_{i j}$ = $(- 1)^{i + j} M_{i j}$ .

Therefore we have:

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

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

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

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

(ii) GIven determinant: $| \begin{matrix} a & c \\ b & d \end{matrix} |$

Minor of element $a_{i j}$ is $M_{i j}$ .

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 $a_{i j}$ is $A_{i j}$ = $(- 1)^{i + j} M_{i j}$ .

Therefore we have:

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

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

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

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

Question:2 Write Minors and Cofactors of the elements of the following determinants:

(i) $| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} |$

(ii) $| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & - 1 \\ 0 & 1 & 2 \end{matrix} |$

Answer:

(i) Given determinant : $| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} |$

Finding Minors: by the definition,

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

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

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

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

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

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

(ii) Given determinant : $| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & - 1 \\ 0 & 1 & 2 \end{matrix} |$

Finding Minors: by the definition,

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

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

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

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

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

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

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

Question:3 Using Cofactors of elements of the second row, evaluate. $Δ = | \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} |$

Answer:

Given determinant : $Δ = | \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} |$

First finding Minors of the second rows by the definition,

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

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

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

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 $△$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

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

Question:4 Using Cofactors of elements of third column, evaluate $Δ = | \begin{matrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{matrix} |$

Answer:

Given determinant : $Δ = | \begin{matrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{matrix} |$

First finding Minors of the third column by the definition,

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

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

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

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 $△$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$△ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$

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

$= y z^{2} - y^{2} z + z x^{2} - x z^{2} + x y^{2} - x^{2} y$

$= z (x^{2} - y^{2}) + z^{2} (y - x) + x y (y - x)$

$= (x - y) [z x + z y - z^{2} - x y]$

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

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

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

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

Question:5 If $Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$ and $A_{i j}$ is Cofactors of $a_{i j}$ , then the value of $Δ$ is given by

(A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$

(B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$

(D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$

Answer:

The answer is (D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$ by the definition itself, $Δ$ is equal to the product of the elements of the row/column with their corresponding cofactors.

Class 12 Maths chapter 4 solutions Exercise: 4.4
Page number: 92-93
Total questions: 18

Question:1 Find the adjoint of each of the matrices.

$[\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$

Answer:

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

Then we have,

$A_{11} = 4, A_{12} = - (1) 3, A_{21} = - (1) 2, a n d A_{22} = 1$

Hence we get:

$a d j A = {[\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}]}^{T} = [\begin{matrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{matrix}] = [\begin{matrix} 4 & - 2 \\ - 3 & 1 \end{matrix}]$

Question:2 Find the adjoint of each of the matrices

$[\begin{matrix} 1 & - 1 & 2 \\ 2 & 3 & 5 \\ - 2 & 0 & 1 \end{matrix}]$

Answer:

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

Then we have,

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

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

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

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

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

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

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

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

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

Hence we get:

$a d j A = [\begin{matrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{matrix}] = [\begin{matrix} 3 & 1 & - 11 \\ - 12 & 5 & - 1 \\ 6 & 2 & 5 \end{matrix}]$

Question:3 Verify $A (a d j A) = (a d j A) A = | A | I$ .

$[\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}]$

Answer:

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

Let $A = [\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}]$

Calculating the cofactors;

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

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

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

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

Hence, $a d j A = [\begin{matrix} - 6 & - 3 \\ 4 & 2 \end{matrix}]$

Now,

$A (a d j A) = [\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}] ([\begin{matrix} - 6 & - 3 \\ 4 & 2 \end{matrix}])$

$[\begin{matrix} - 12 + 12 & - 6 + 6 \\ 24 - 24 & 12 - 12 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

also,

$(a d j A) A = [\begin{matrix} - 6 & - 3 \\ 4 & 2 \end{matrix}] [\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}]$

$= [\begin{matrix} - 12 + 12 & - 18 + 18 \\ 8 - 8 & 12 - 12 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Now, calculating |A|;

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

So, $| A | I = 0 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Hence we get

$A (a d j A) = (a d j A) A = | A | I$

Question:4 Verify $A (a d j A) = (a d j A) A = | A | I$ .

$[\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}]$

Answer:

Given matrix: $[\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}]$

Let $A = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}]$

Calculating the cofactors;

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

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

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

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

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

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

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

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

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

Hence, $a d j A = [\begin{matrix} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{matrix}]$

Now,

$A (a d j A) = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}] [\begin{matrix} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{matrix}]$

$= [\begin{matrix} 0 + 11 + 0 & 3 - 1 - 2 & 2 - 8 + 6 \\ 0 + 0 + 0 & 9 + 0 + 2 & 6 + 0 - 6 \\ 0 + 0 + 0 & 3 + 0 - 3 & 2 + 0 + 9 \end{matrix}] = [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}]$

also,

$A (a d j A) = [\begin{matrix} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}]$

$= [\begin{matrix} 0 + 9 + 2 & 0 + 0 + 0 & 0 - 6 + 6 \\ - 11 + 3 + 8 & 11 + 0 + 0 & - 22 - 2 + 24 \\ 0 - 3 + 3 & 0 + 0 + 0 & 0 + 2 + 9 \end{matrix}] = [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}]$

Now, calculating |A|;

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

So, $| A | I = 11 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}]$

Hence we get,

$A (a d j A) = (a d j A) A = | A | I$ .

Question:5 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 2 & - 2 \\ 4 & 3 \end{matrix}]$

Answer:

Given matrix : $[\begin{matrix} 2 & - 2 \\ 4 & 3 \end{matrix}]$

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

|A| = (6+8) = 14

Now, calculate the cofactors terms and then adj A.

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

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

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

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

So, we have $a d j A = [\begin{matrix} 3 & 2 \\ - 4 & 2 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

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

Question:6 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} - 1 & 5 \\ - 3 & 2 \end{matrix}]$

Answer:

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

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

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

Now, calculate 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 $a d j A = [\begin{matrix} 2 & - 5 \\ 3 & - 1 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

$= \frac{1}{13} [\begin{matrix} 2 & - 5 \\ 3 & - 1 \end{matrix}] = [\begin{matrix} \frac{2}{13} & \frac{- 5}{13} \\ \frac{3}{13} & \frac{- 1}{13} \end{matrix}]$

Question:7 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix}]$

Answer:

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

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

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

Now, calculate 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 $a d j A = [\begin{matrix} 10 & - 10 & 2 \\ 0 & 5 & - 4 \\ 0 & 0 & 2 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

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

Question:8 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{matrix}]$

Answer:

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

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

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

Now, calculate 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 $a d j A = [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

$= \frac{- 1}{3} [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$

Question 9 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 2 & 1 & 3 \\ 4 & - 1 & 0 \\ - 7 & 2 & 1 \end{matrix}]$

Answer:

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

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

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

Now, calculate 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 $a d j A = [\begin{matrix} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

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

Question:10 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}]$

Answer:

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

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

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

Now, calculate 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 $a d j A = [\begin{matrix} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

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

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

Question:11 Find the inverse of each of the matrices (if it exists).

$[\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

Answer:

Given the matrix : $[\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}] = A$

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

$A^{- 1} = \frac{1}{| A |} a d j A$

So, calculating |A| :

$| A | = 1 (- \cos^{2} α - \sin^{2} α) + 0 (0 - 0) + 0 (0 - 0)$

$= - (\cos^{2} α + \sin^{2} α) = - 1$

Now, calculate the cofactors terms and then adjA.

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

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

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

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

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

So, we have $a d j A = [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}]$

Therefore inverse of A will be:

$A^{- 1} = \frac{1}{| A |} a d j A$

$A^{- 1} = \frac{1}{- 1} [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

Question:12 Let $A = [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}]$ and $B = [\begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix}]$ . Verify that $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Answer:

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

then calculating;

$A B = [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}] [\begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix}]$

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

Finding the inverse of AB.

Calculating the cofactors fo AB:

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

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

Then we have adj(AB):

$a d j (A B) = [\begin{matrix} 61 & - 87 \\ - 47 & 67 \end{matrix}]$

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

Therefore we have the inverse:

$(A B)^{- 1} = \frac{1}{| A B |} a d j (A B) = - \frac{1}{2} [\begin{matrix} 61 & - 87 \\ - 47 & 67 \end{matrix}]$

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

Now, calculate the inverses of A and B.

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

$a d j A = [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}]$ and $a d j B = [\begin{matrix} 9 & - 8 \\ - 7 & 6 \end{matrix}]$

therefore we have

$A^{- 1} = \frac{1}{| A |} a d j A = \frac{1}{1} [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}]$ and $B^{- 1} = \frac{1}{| B |} a d j B = \frac{1}{- 2} [\begin{matrix} 9 & - 8 \\ - 7 & 6 \end{matrix}] = [\begin{matrix} \frac{- 9}{2} & 4 \\ \frac{7}{2} & - 3 \end{matrix}]$

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

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

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

From (1) and (2) we get

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Hence proved.

Question:13 If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$ ? , show that $A^{2} - 5 A + 7 I = O$ . Hence find $A^{- 1}$

Answer:

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

So, calculating each term;

$\Rightarrow$ $A^{2} = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] = [\begin{matrix} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{matrix}] = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}]$

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

$\Rightarrow$ $[\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - 5 [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] + 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - [\begin{matrix} 15 & 5 \\ - 5 & 10 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

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

$∴ A . A - 5 A = - 7 I$

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

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

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

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

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

$∴ A^{- 1} = \frac{1}{7} (5 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]) = \frac{1}{7} [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}]$

Question:14 For the matrix $A = [\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}]$ , find the numbers $a$ and $b$ such that $A^{2} + a A + b I = 0$ .

Answer:

Given $A = [\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}]$ then we have the relation $A^{2} + a A + b I = O$

So, calculating each term;

$\Rightarrow$ $A^{2} = [\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}] [\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}] = [\begin{matrix} 9 + 2 & 6 + 2 \\ 3 + 1 & 2 + 1 \end{matrix}] = [\begin{matrix} 11 & 8 \\ 4 & 3 \end{matrix}]$

Therefore $A^{2} + a A + b I = O$ ;

$\Rightarrow$ $[\begin{matrix} 11 & 8 \\ 4 & 3 \end{matrix}] + a [\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}] + b [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 11 + 3 a + b & 8 + 2 a \\ 4 + a & 3 + a + b \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

So, we have equations;

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

We get $a = - 4 a n d b = 1$ .

Question:15 For the matrix $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$ Show that $A^{3} - 6 A^{2} + 5 A + 11 I = O$ Hence, find $A^{- 1}$ .

Answer:

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

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

Finding each term:

$\Rightarrow$ $A^{2} = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 1 + 1 + 2 & 1 + 2 - 1 & 1 - 3 + 3 \\ 1 + 2 - 6 & 1 + 4 + 3 & 1 - 6 - 9 \\ 2 - 1 + 6 & 2 - 2 - 3 & 2 + 3 + 9 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}]$

$\Rightarrow$ $A^{3} = [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3 \\ - 3 + 8 - 28 & - 3 + 16 + 14 & - 3 - 24 - 42 \\ 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{matrix}]$

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

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

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

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

$\Rightarrow$ $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0$

Now finding the inverse of A;

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

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

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

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

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

Now,

From equation (1) we get;

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

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

$\Rightarrow$ $A^{- 1} = \frac{- 1}{11} ([\begin{matrix} 3 & - 4 & - 5 \\ - 9 & 1 & 4 \\ - 5 & 3 & 1 \end{matrix}]$

Question:16 If $A = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]$ , verify that $A^{3} - 6 A^{2} + 9 A - 4 I = O$ . Hence find $A^{- 1}$ .

Answer:

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

To show: $A^{3} - 6 A^{2} + 9 A - 4 I$

Finding each term:

$\Rightarrow$ $A^{2} = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 4 + 1 + 1 & - 2 - 2 - 1 & 2 + 1 + 2 \\ - 2 - 2 - 1 & 1 + 4 + 1 & - 1 - 2 - 2 \\ 2 + 1 + 2 & - 1 - 2 - 2 & 1 + 1 + 4 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}]$

$\Rightarrow$ $A^{3} = [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 12 + 5 + 5 & - 6 - 10 - 5 & 6 + 5 + 10 \\ - 10 - 6 - 5 & 5 + 12 + 5 & - 5 - 6 - 10 \\ 10 + 5 + 6 & - 5 - 10 - 6 & 5 + 5 + 12 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}]$

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

$\Rightarrow$ $[\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - 6 [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] + 9 [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] - 4 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - [\begin{matrix} 36 & - 30 & 30 \\ - 30 & 36 & - 30 \\ 30 & - 30 & 36 \end{matrix}] + [\begin{matrix} 18 & - 9 & 9 \\ - 9 & 18 & - 9 \\ 9 & - 9 & 18 \end{matrix}] - [\begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 22 - 36 + 18 - 4 & - 21 + 30 - 9 & 21 - 30 + 9 \\ - 21 + 30 - 9 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 \\ 21 - 30 + 9 & - 21 + 30 - 9 & 22 - 36 + 18 - 4 \end{matrix}]$

$\Rightarrow$ $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = O$

Now finding the inverse of A;

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

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

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

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

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

Now,

From equation (1) we get;

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

$\Rightarrow$ $A^{- 1} = \frac{1}{4} [\begin{matrix} 6 - 12 + 9 & - 5 + 6 & 5 - 6 \\ - 5 + 6 & 6 - 12 + 9 & - 5 + 6 \\ 5 - 6 & - 5 + 6 & 6 - 12 + 9 \end{matrix}]$

Hence inverse of A is :

$\Rightarrow$ $A^{- 1} = \frac{1}{4} [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}]$

Question:17 Let A be a nonsingular square matrix of order $3 \times 3$ . Then $| a d j A |$ is equal to

(A) $| A |$ (B) $| A |^{2}$ (C) $| A |^{3}$ (D) $3 | A |$

Answer:

We know the identity $(a d j A) A = | A | I$

Hence we can determine the value of $| (a d j A) |$ .

Taking both sides determinant value we get,

$\Rightarrow$ $| (a d j A) A | = | | A | I |$ or $| (a d j A) | | A | = | | A | | | I |$

or taking R.H.S.,

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

$\Rightarrow$ $| A | (| A |^{2}) = | A |^{3}$

or, we have then $| (a d j A) | | A | = | A |^{3}$

Therefore $| (a d j A) | = | A |^{2}$

Hence the correct answer is B.

Question:18 If A is an invertible matrix of order 2, then det $(A^{- 1})$ is equal to

(A) $d e t (A)$ (B) $\frac{1}{d e t (A)}$ (C) $1$ (D) $0$

Answer:

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

Let us assume a matrix of the order of 2;

$\Rightarrow$ $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ .

Then $| A | = a d - b c$ .

$a d j A = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ and $| a d j A | = a d - b c$

Now,

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} a d j A$

Taking determinant both sides;

$\Rightarrow$ $| A^{- 1} | = | \frac{1}{| A |} a d j A | = [\begin{matrix} \frac{d}{| A |} & \frac{- b}{| A |} \\ \frac{- c}{| A |} & \frac{a}{| A |} \end{matrix}]$

$∴ | A^{- 1} | = | \begin{matrix} \frac{d}{| A |} & \frac{- b}{| A |} \\ \frac{- c}{| A |} & \frac{a}{| A |} \end{matrix} | = \frac{1}{| A |^{2}} | \begin{matrix} d & - b \\ - c & a \end{matrix} | = \frac{1}{| A |^{2}} (a d - b c) = \frac{1}{| A |^{2}} . | A | = \frac{1}{| A |}$

Therefore we get;

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

Hence the correct answer is B.

Class 12 Maths chapter 4 solutions Exercise: 4.5
Page number: 97-98
Total questions: 16

Question:1 Examine the consistency of the system of equations.

$x + 2 y = 2$

$2 x + 3 y = 3$

Answer:

We have given the system of equations:18967

$x + 2 y = 2$

$2 x + 3 y = 3$

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

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

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

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

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

Hence, the given system of equations is consistent.

Question:2 Examine the consistency of the system of equations

$2 x - y = 5$

$x + y = 4$

Answer:

We have given the system of equations:

$2 x - y = 5$

$x + y = 4$

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

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

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

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

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

Hence, the given system of equations is consistent.

Question:3 Examine the consistency of the system of equations.

$x + 3 y = 5$

$2 x + 6 y = 8$

Answer:

We have given the system of equations:

$x + 3 y = 5$

$2 x + 6 y = 8$

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

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

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

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

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

$\Rightarrow$ $a d j A = [\begin{matrix} 6 & - 3 \\ - 2 & 1 \end{matrix}]$

So, $(a d j A) B = [\begin{matrix} 6 & - 3 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 5 \\ 8 \end{matrix}] = [\begin{matrix} 30 - 24 \\ - 10 + 8 \end{matrix}] = [\begin{matrix} 6 \\ - 2 \end{matrix}] \neq 0$

As, $(a d j A) B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question:4 Examine the consistency of the system of equations.

$x + y + z = 1$

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

$a x + a y + 2 a z = 4$

Answer:

We have given the system of equations:

$x + y + z = 1$

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

$a x + a y + 2 a z = 4$

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

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

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

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

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

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

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

Hence, the given system of equations is consistent.

Question:5 Examine the consistency of the system of equations.

$3 x - y - 2 z = 2$

$2 y - z = - 1$

$3 x - 5 y = 3$

Answer:

We have given the system of equations:

$3 x - y - 2 z = 2$

$2 y - z = - 1$

$3 x - 5 y = 3$

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

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

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

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

$= - 15 + 3 + 12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(a d j A) B,$

$(a d j A) = [\begin{matrix} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{matrix}]$

$∴ (a d j A) B = [\begin{matrix} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{matrix}] [\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}] = [\begin{matrix} - 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18 \end{matrix}] = [\begin{matrix} - 5 \\ - 3 \\ - 6 \end{matrix}] \neq 0$

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

Question:6 Examine the consistency of the system of equations.

$5 x - y + 4 z = 5$

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

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

Answer:

We have given the system of equations:

$5 x - y + 4 z = 5$

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

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

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

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

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

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

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

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

Hence, the given system of equations is consistent.

Question:7 Solve the system of linear equations, using the matrix method.

$5 x + 2 y = 4$

$7 x + 3 y = 5$

Answer:

The given system of equations

$5 x + 2 y = 4$

$7 x + 3 y = 5$

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

$\Rightarrow$ $A = [\begin{matrix} 5 & 4 \\ 7 & 3 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \end{matrix}]$ and $B = [\begin{matrix} 4 \\ 5 \end{matrix}]$

we have,

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

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

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

$A^{- 1} = \frac{1}{| A |} (a d j A) = (a d j A) = [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

x = 2 and y = -3 .

Question:8 Solve the system of linear equations, using the matrix method.

$2 x - y = - 2$

$3 x + 4 y = 3$

Answer:

The given system of equations

$2 x - y = - 2$

$3 x + 4 y = 3$

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

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

we have,

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

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

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

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{11} [\begin{matrix} 4 & 1 \\ - 3 & 2 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = \frac{- 5}{11} a n d y = \frac{12}{11} .$

Question:9 Solve the system of linear equations, using the matrix method.

$4 x - 3 y = 3$

$3 x - 5 y = 7$

Answer:

The given system of equations

$4 x - 3 y = 3$

$3 x - 5 y = 7$

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

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

we have,

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

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

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

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{- 1}{11} [\begin{matrix} - 5 & 3 \\ - 3 & 4 \end{matrix}] = \frac{1}{11} [\begin{matrix} 5 & - 3 \\ 3 & - 4 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = \frac{- 6}{11} a n d y = \frac{- 19}{11} .$

Question:10 Solve the system of linear equations, using the matrix method.

$5 x + 2 y = 3$

$3 x + 2 y = 5$

Answer:

The given system of equations

$5 x + 2 y = 3$

$3 x + 2 y = 5$

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

$\Rightarrow$ $A = [\begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \end{matrix}]$ and $B = [\begin{matrix} 3 \\ 5 \end{matrix}]$

we have,

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

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

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

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = - 1 a n d y = 4.$

Question:11 Solve a system of linear equations, using the matrix method.

$2 x + y + z = 1$

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

$3 y - 5 z = 9$

Answer:

The given system of equations

$2 x + y + z = 1$

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

$3 y - 5 z = 9$

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

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

we have,

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

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

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

Now, we will find the cofactors;

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

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

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

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

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

$\Rightarrow$ $(a d j A) = [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}]$

$A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{34} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = 1, y = \frac{1}{2}, a n d z = - \frac{3}{2} .$

Question:12 Solve the system of linear equations, using the matrix method.

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

Answer:

The given system of equations

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

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

$\Rightarrow$ $A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ $a n d B = [\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}] .$

we have,

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

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

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

Now, we will find the cofactors;

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

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

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

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

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

$\Rightarrow$ $(a d j A) = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = 2, y = - 1, a n d z = 1.$

Question:13 Solve the system of linear equations, using the matrix method.

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

$x - 2 y + z = - 4$

$3 x - y - 2 z = 3$

Answer:

The given system of equations

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

$x - 2 y + z = - 4$

$3 x - y - 2 z = 3$

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

$\Rightarrow$ $A = [\begin{matrix} 2 & 3 & 3 \\ 1 & - 2 & 1 \\ 3 & - 1 & - 2 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ $a n d B = [\begin{matrix} 5 \\ - 4 \\ 3 \end{matrix}] .$

we have,

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

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

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

Now, we will find the cofactors;

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

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

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

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

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

$\Rightarrow$ $(a d j A) = [\begin{matrix} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{matrix}]$

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{40} [\begin{matrix} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = 1, y = 2, a n d z = - 1.$

Question:14 Solve the system of linear equations, using the matrix method.

$x - y + 2 z = 7$

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

$2 x - y + 3 z = 12$

Answer:

The given system of equations

$x - y + 2 z = 7$

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

$2 x - y + 3 z = 12$

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

$\Rightarrow$ $A = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ $a n d B = [\begin{matrix} 7 \\ - 5 \\ 12 \end{matrix}] .$

we have,

$\Rightarrow$ $| A | = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) = 7 + 19 - 22 = 4 \neq 0$ .

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

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

Now, we will find the cofactors;

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

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

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

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

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

$\Rightarrow$ $(a d j A) = [\begin{matrix} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{matrix}]$

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{4} [\begin{matrix} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = 2, y = 1, a n d z = 3.$

Question:15 If $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$ , find $A^{- 1}$ . Using $A^{- 1}$ solve the system of equations

$2 x - 3 y + 5 z = 11$

$3 x + 2 y - 4 z = - 5$

$x + y - 2 z = - 3$

Answer:

The given system of equations

$2 x - 3 y + 5 z = 11$

$3 x + 2 y - 4 z = - 5$

$x + y - 2 z = - 3$

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

$\Rightarrow$ $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ $a n d B = [\begin{matrix} 11 \\ - 5 \\ - 3 \end{matrix}] .$

we have,

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

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

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

Now, we will find the cofactors;

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

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

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

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

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

$\Rightarrow$ $(a d j A) = [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}]$

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = - 1 [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}] = [\begin{matrix} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}]$

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

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

Hence the solutions of the given system of equations;

$x = 1, y = 2, a n d z = 3.$

Question:16 The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat, and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find the cost of each item per kg by matrix method.

Answer:

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

Then we have the equations for the given situation :

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

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

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

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;

$\Rightarrow$ $A = [\begin{matrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{matrix}],$ $X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ $a n d B = [\begin{matrix} 60 \\ 90 \\ 70 \end{matrix}] .$

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

Now, we will find the cofactors of A;

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

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

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

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

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

Now we have adjA;

$\Rightarrow$ $a d j A = [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}]$

$\Rightarrow$ $A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{50} [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}]$ s

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

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

Hence the solutions of the given system of equations;

$x = 5, y = 8, a n d z = 8.$

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.

Class 12 Maths chapter 4 solutions - Miscellaneous Exercise
Page number: 99-100
Total questions: 9

Question:1 Prove that the determinant $| \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ is independent of $θ$ .

Answer:

Calculating the determinant value of $| \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ ;

$\Rightarrow$ $x [\begin{matrix} - x & 1 \\ 1 & x \end{matrix}] - \sin Θ [\begin{matrix} - \sin Θ & 1 \\ \cos Θ & x \end{matrix}] + \cos Θ [\begin{matrix} - \sin Θ & - x \\ \cos Θ & 1 \end{matrix}]$

$\Rightarrow$ $x (- x^{2} - 1) - \sin Θ (- x \sin Θ - \cos Θ) + \cos Θ (- \sin Θ + x \cos Θ)$

$\Rightarrow$ $- x^{3} - x + x \sin^{2} Θ + \sin Θ \cos Θ - \cos Θ \sin Θ + x \cos^{2} Θ$

$\Rightarrow$ $- x^{3} - x + x (\sin^{2} Θ + \cos^{2} Θ)$

$\Rightarrow$ $- x^{3} - x + x = - x^{3}$

Clearly, the determinant is independent of $θ$ .

Question:2 Evaluate $| \begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix} |$ .

Answer:

Given determinant $| \begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix} |$ ;

$\Rightarrow$ $\cos α \cos β | \begin{matrix} \cos β & 0 \\ \sin α \sin β & \cos α \end{matrix} | - \cos α \sin β | \begin{matrix} - \sin β & 0 \\ \sin α \cos β & \cos α \end{matrix} | - \sin α | \begin{matrix} - \sin β & \cos β \\ \sin α \cos β & \sin α \sin β \end{matrix} |$ $= \cos α \cos β (\cos β \cos α - 0) - \cos α \sin β (- \cos α \sin β - 0) - \sin α (- \sin α \sin^{2} β - \sin α \cos^{2} β)$

$\Rightarrow$ $\cos^{2} α \cos^{2} β + \cos^{2} α \sin^{2} β + \sin^{2} α \sin^{2} β + \sin^{2} α \cos^{2} β$

$\Rightarrow$ $\cos^{2} α (\cos^{2} β + \sin^{2} β) + \sin^{2} α (\sin^{2} β + \cos^{2} β)$

$\Rightarrow$ $\cos^{2} α (1) + \sin^{2} α (1) = 1$ .

Question:3 If $A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ and $B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$ , find $(A B)^{- 1}$ .

Answer:

We know from the identity that;

$(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Then we can find easily,

Given $A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ and $B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$

Then we have to find the $B^{- 1}$ matrix.

So, Given matrix $B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$

$| B | = 1 (3 - 0) - 2 (- 1 - 0) - 2 (2 - 0) = 3 + 2 - 4 = 1 \neq 0$

Hence its inverse $B^{- 1}$ exists;

Now, as we know that

$B^{- 1} = \frac{1}{| B |} a d j B$

So, calculating cofactors of B,

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

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

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

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

$B_{33} = (- 1)^{3 + 3} (3 + 2) = 5$

$a d j B = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}]$

$B^{- 1} = \frac{1}{| B |} a d j B = \frac{1}{1} [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}]$

Now, We have both $A^{- 1}$ as well as $B^{- 1}$ ;

Putting in the relation we know; $(A B)^{- 1} = B^{- 1} A^{- 1}$

$(A B)^{- 1} = \frac{1}{1} [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$

$= [\begin{matrix} 9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10 \end{matrix}]$

$= [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}]$

Question:4 Let $A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ . Verify that,

(i) $100 [a d j A]^{- 1} = a d j (A^{- 1})$

(ii) $(A^{- 1})^{- 1} = A$

Answer:

(i) Given that $A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ ;

So, let us assume that $A^{- 1} = B$ matrix and $a d j A = C$ then;

$| A | = 1 (15 - 1) - 2 (10 - 1) + 1 (2 - 3) = 14 - 18 - 1 = - 5 \neq 0$

Hence its inverse exists;

$A^{- 1} = \frac{1}{| A |} a d j A$ or $B = \frac{1}{| A |} C$ ;

so, we now calculate the value of $a d j A$

Cofactors of A;

$A_{11} = (- 1)^{1 + 1} (15 - 1) = 14$ $A_{12} = (- 1)^{1 + 2} (10 - 1) = - 9$

$A_{13} = (- 1)^{1 + 3} (2 - 3) = - 1$ $A_{21} = (- 1)^{2 + 1} (10 - 1) = - 9$

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

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

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

$\Rightarrow a d j A = C = [\begin{matrix} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{matrix}]$

$A^{- 1} = B = \frac{1}{| A |} a d j A = \frac{1}{- 5} [\begin{matrix} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{matrix}]$

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;

$C^{- 1} = \frac{1}{| C |} a d j C$

Now, finding the $a d j C$ ;

$C_{11} = (- 1)^{1 + 1} (- 4 - 1) = - 5$ $C_{12} = (- 1)^{1 + 2} (9 + 1) = - 10$

$C_{13} = (- 1)^{1 + 3} (- 9 + 4) = - 5$ $C_{21} = (- 1)^{2 + 1} (9 + 1) = - 10$

$C_{22} = (- 1)^{2 + 2} (- 14 - 1) = - 15$ $C_{23} = (- 1)^{2 + 3} (14 - 9) = - 5$

$C_{31} = (- 1)^{3 + 1} (- 9 + 4) = - 5$ $C_{32} = (- 1)^{3 + 2} (14 - 9) = - 5$

$C_{33} = (- 1)^{3 + 3} (56 - 81) = - 25$

$a d j C = [\begin{matrix} - 5 & - 10 & - 5 \\ - 10 & - 15 & - 5 \\ - 5 & - 5 & - 25 \end{matrix}]$

$C^{- 1} = \frac{1}{| C |} a d j C = \frac{1}{25} [\begin{matrix} - 5 & - 10 & - 5 \\ - 10 & - 15 & - 5 \\ - 5 & - 5 & - 25 \end{matrix}] = [\begin{matrix} - \frac{1}{5} & - \frac{2}{5} & - \frac{1}{5} \\ - \frac{2}{5} & - \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & - \frac{1}{5} & - 1 \end{matrix}]$

or $L . H . S . = C^{- 1} = [a d j A]^{- 1} = [\begin{matrix} - \frac{1}{5} & - \frac{2}{5} & - \frac{1}{5} \\ - \frac{2}{5} & - \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & - \frac{1}{5} & - 1 \end{matrix}]$

Now, finding the R.H.S.

$a d j (A^{- 1}) = a d j B$

$A^{- 1} = B = [\begin{matrix} \frac{- 14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & \frac{- 4}{5} & \frac{- 1}{5} \\ \frac{1}{5} & \frac{- 1}{5} & \frac{1}{5} \end{matrix}]$

Cofactors of B;

$B_{11} = (- 1)^{1 + 1} (\frac{- 4}{25} - \frac{1}{25}) = \frac{- 1}{5}$

$B_{12} = (- 1)^{1 + 2} (\frac{9}{25} + \frac{1}{25}) = - \frac{2}{5}$

$B_{13} = (- 1)^{1 + 3} (\frac{- 9}{25} + \frac{4}{25}) = \frac{- 1}{5}$

$B_{21} = (- 1)^{2 + 1} (\frac{9}{25} + \frac{1}{25}) = - \frac{2}{5}$

$B_{22} = (- 1)^{2 + 2} (\frac{- 14}{25} - \frac{1}{25}) = \frac{- 3}{5}$

$B_{23} = (- 1)^{2 + 3} (\frac{14}{25} - \frac{9}{25}) = \frac{- 1}{5}$

$B_{31} = (- 1)^{3 + 1} (\frac{- 9}{25} + \frac{4}{25}) = \frac{- 1}{5}$

$B_{32} = (- 1)^{3 + 2} (\frac{14}{25} - \frac{9}{25}) = \frac{- 1}{5}$

$B_{33} = (- 1)^{3 + 3} (\frac{56}{25} - \frac{81}{25}) = - 1$

$R . H . S . = a d j B = a d j (A^{- 1}) = [\begin{matrix} - \frac{1}{5} & - \frac{2}{5} & - \frac{1}{5} \\ - \frac{2}{5} & - \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & - \frac{1}{5} & - 1 \end{matrix}]$

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

(ii) Given that $A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ ;

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;

$A^{- 1} = \frac{1}{| A |} a d j A$ or $B = \frac{1}{| A |} C$ ;

so, we now calculate the value of $a d j A$

Cofactors of A;

$A_{11} = (- 1)^{1 + 1} (15 - 1) = 14$ $A_{12} = (- 1)^{1 + 2} (10 - 1) = - 9$

$A_{13} = (- 1)^{1 + 3} (2 - 3) = - 1$ $A_{21} = (- 1)^{2 + 1} (10 - 1) = - 9$

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

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

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

$\Rightarrow a d j A = C = [\begin{matrix} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{matrix}]$

$A^{- 1} = B = \frac{1}{| A |} a d j A = \frac{1}{- 5} [\begin{matrix} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{matrix}] = [\begin{matrix} \frac{- 14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & \frac{- 4}{5} & \frac{- 1}{5} \\ \frac{1}{5} & \frac{- 1}{5} & \frac{1}{5} \end{matrix}]$

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;

$B^{- 1} = \frac{1}{| B |} a d j B$

Now, finding the $a d j B$ ;

$B_{11} = (- 1)^{1 + 1} (\frac{- 4}{25} - \frac{1}{25}) = \frac{- 1}{5}$ $B_{12} = (- 1)^{1 + 2} (\frac{9}{25} + \frac{1}{25}) = \frac{- 2}{5}$

$B_{13} = (- 1)^{1 + 3} (\frac{- 9}{25} + \frac{4}{25}) = \frac{- 1}{5}$ $B_{21} = (- 1)^{2 + 1} (\frac{9}{25} + \frac{1}{25}) = - \frac{2}{5}$

$B_{22} = (- 1)^{2 + 2} (\frac{- 14}{25} - \frac{1}{25}) = \frac{- 3}{5}$ $B_{23} = (- 1)^{2 + 3} (\frac{14}{25} - \frac{9}{25}) = \frac{- 1}{5}$

$B_{31} = (- 1)^{3 + 1} (\frac{- 9}{25} + \frac{4}{25}) = \frac{- 1}{5}$

$B_{32} = (- 1)^{3 + 2} (\frac{14}{25} - \frac{9}{25}) = \frac{- 1}{5}$

$B_{33} = (- 1)^{3 + 3} (\frac{56}{25} - \frac{81}{25}) = \frac{- 25}{25} = - 1$

$a d j B = [\begin{matrix} \frac{- 1}{5} & \frac{- 2}{5} & \frac{- 1}{5} \\ \frac{- 2}{5} & \frac{- 3}{5} & \frac{- 1}{5} \\ \frac{- 1}{5} & \frac{- 1}{5} & - 1 \end{matrix}]$

$B^{- 1} = \frac{1}{| B |} a d j B = \frac{- 5}{1} [\begin{matrix} \frac{- 1}{5} & \frac{- 2}{5} & \frac{- 1}{5} \\ \frac{- 2}{5} & \frac{- 3}{5} & \frac{- 1}{5} \\ \frac{- 1}{5} & \frac{- 1}{5} & - 1 \end{matrix}] = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$

$L . H . S . = B^{- 1} = (A^{- 1})^{- 1} = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$

$R . H . S . = A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$

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

Question:5 Evaluate $| \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} |$

Answer:

We have determinant $△ = | \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} |$

Applying row transformations; $R_{1} \to R_{1} + R_{2} + R_{3}$ , we have then;

$△ = | \begin{matrix} 2 (x + y) & 2 (x + y) & 2 (x + y) \\ y & x + y & x \\ x + y & x & y \end{matrix} |$

Take out the common factor 2(x+y) from the row first.

$= 2 (x + y) | \begin{matrix} 1 & 1 & 1 \\ y & x + y & x \\ x + y & x & y \end{matrix} |$

Now, applying the column transformation; $C_{1} \to C_{1} - C_{2}$ and $C_{2} \to C_{2} - C_{1}$ we have ;

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

Expanding the remaining determinant;

$= 2 (x + y) (- x (x - y) - y^{2}) = 2 (x + y) [- x^{2} + x y - y^{2}]$

$= - 2 (x + y) [x^{2} - x y + y^{2}] = - 2 (x^{3} + y^{3})$ .

Question:6 Evaluate $| \begin{matrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{matrix} |$

Answer:

We have determinant $△ = | \begin{matrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{matrix} |$

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

$△ = | \begin{matrix} 0 & - y & 0 \\ 0 & y & - x \\ 1 & x & x + y \end{matrix} |$

Taking out the common factor-y from the row first.

$△ = - y | \begin{matrix} 0 & 1 & 0 \\ 0 & y & - x \\ 1 & x & x + y \end{matrix} |$

Expanding the remaining determinant;

$- y [1 (- x - o)] = x y$

Question:7 Solve the system of equations

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

Answer:

We have a system of equations;

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

So, we will convert the given system of equations into a simple form to solve the problem by the matrix method;

Let us take, $\frac{1}{x} = a$ , $\frac{1}{y} = b a n d \frac{1}{z} = c$

Then we have the equations;

$2 a + 3 b + 10 c = 4$

$4 a - 6 b + 5 c = 1$

$6 a + 9 b - 20 c = 2$

We can write it in the matrix form as $A X = B$ , where

$A = [\begin{matrix} 2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20 \end{matrix}], X = [\begin{matrix} a \\ b \\ c \end{matrix}] a n d B = [\begin{matrix} 4 \\ 1 \\ 2 \end{matrix}] .$

Now, Finding the determinant value of A;

$| A | = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36)$

$= 150 + 330 + 720$

$= 1200 \neq 0$

Hence we can say that A is non-singular $∴$ its inverse exists;

Finding cofactors of A;

$A_{11} = 75$ , $A_{12} = 110$ , $A_{13} = 72$

$A_{21} = 150$ , $A_{22} = - 100$ , $A_{23} = 0$

$A_{31} = 75$ , $A_{31} = 30$ , $A_{33} = - 24$

$∴$ as we know $A^{- 1} = \frac{1}{| A |} a d j A$

$= \frac{1}{1200} [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}]$

Now we will find the solutions by relation $X = A^{- 1} B$ .

$\Rightarrow [\begin{matrix} a \\ b \\ c \end{matrix}] = \frac{1}{1200} [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}] [\begin{matrix} 4 \\ 1 \\ 2 \end{matrix}]$

$= \frac{1}{1200} [\begin{matrix} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{matrix}]$

$= \frac{1}{1200} [\begin{matrix} 600 \\ 400 \\ 240 \end{matrix}] = [\begin{matrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{matrix}]$

Therefore we have the solutions $a = \frac{1}{2}, b = \frac{1}{3}, a n d c = \frac{1}{5} .$

Or in terms of x, y, and z;

$x = 2, y = 3, a n d z = 5$

Question:8 Choose the correct answer.

If x, y, z are nonzero real numbers, then the inverse of matrix $A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ is

$(A) [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$ $(B) x y z [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

$(C) \frac{1}{x y z} [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ $(D) \frac{1}{x y z} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Answer:

Given Matrix $A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ ,

$| A | = x (y z - 0) = x y z$

As we know,

$A^{- 1} = \frac{1}{| A |} a d j A$

So, we will find the $a d j A$ ,

Determining its cofactor first,

$A_{11} = y z$ $A_{12} = 0$ $A_{13} = 0$

$A_{21} = 0$ $A_{22} = x z$ $A_{23} = 0$

$A_{31} = 0$ $A_{32} = 0$ $A_{33} = x y$

Hence $A^{- 1} = \frac{1}{| A |} a d j A = \frac{1}{x y z} [\begin{matrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{matrix}]$

$A^{- 1} = [\begin{matrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{matrix}]$

Therefore the correct answer is (A)

Question:9 Choose the correct answer.

Let $A = | \begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix} |,$ where $0 \leq θ \leq 2 π$ . Then

(A) $D e t (A) = 0$ ; (B) $D e t (A) \in (2, \infty)$

Answer:

Given determinant $A = | \begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix} |$

$| A | = 1 (1 + \sin^{2} Θ) - \sin Θ (- \sin Θ + \sin Θ) + 1 (\sin^{2} Θ + 1)$

$= 1 + \sin^{2} Θ + \sin^{2} Θ + 1$

$= 2 + 2 \sin^{2} Θ = 2 (1 + \sin^{2} Θ)$

Now, given the range of $Θ$ from $0 \leq Θ \leq 2 π$

$\Rightarrow 0 \leq \sin Θ \leq 1$

$\Rightarrow 0 \leq \sin^{2} Θ \leq 1$

$\Rightarrow 1 \leq 1 + \sin^{2} Θ \leq 2$

$\Rightarrow 2 \leq 2 (1 + \sin^{2} Θ) \leq 4$

Therefore the $| A | ϵ [2, 4]$ .

Hence the correct answer is D.

If you are interested in Determinants Class 12 NCERT Solutions exercises then these are listed below.

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

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Importance of solving NCERT questions for class 12 Math Chapter 4

Understanding determinants is not complete without practicing the problems in Chapter 4 of Class 12 Maths. Determinants help in solving linear equations, finding the area of triangles, and understanding matrices. Before solving these questions, students should first build a strong foundation in the properties of determinants, cofactors, and applications.

class 12 chapter 4 ncert solutions will help the students with the exam preparation in a strategic way.
these solutions are prepared by the experts. therefore, students can depend upon the same without any second thought.
Determinants class 12 questions and answers provides the detailed solution for all the questions. This will help the students in analyzing and understanding the questions in a better way.

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NCERT solutions for class 12 - subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Solutions - Class Wise

Given below are the class-wise solutions of class 12 NCERT:

NCERT Exemplar solutions for class 12 - subject-wise

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Frequently Asked Questions (FAQs)

1. How to find the determinant of a 3 x 3 matrix?

To find the determinant of a 3*3 matrix, we use the method of cofactor expansion. This involves multiplying each element of a row (or column) by its cofactor, which is the determinant of the 2*2 matrix obtained by deleting the row and column of that element. You calculate the 2*2 determinants inside each term, and then multiply them by the respective elements. The signs alternate (positive, negative, positive) based on the position of the element. This process gives you the determinant of the matrix.

2. What are the properties of determinants in NCERT Class 12?

Determinants possess several key properties that simplify their manipulation. Some of the important properties are:

The determinant of the identity matrix is always 1.
If two rows or columns of a matrix are interchanged, the sign of the determinant changes.
If a row or column is multiplied by a constant, the determinant is also multiplied by that constant.
Additionally, if any row or column of a matrix is entirely zero, the determinant of that matrix is zero.
For triangular matrices (both upper and lower), the determinant is simply the product of the diagonal elements.
Another important property is that the determinant of the product of two matrices equals the product of their individual determinants.
These properties make it easier to compute and analyze determinants in different scenarios.

3. What is the difference between minors and cofactors in determinants?

Minors and cofactors are closely related concepts in the calculation of determinants. The minor of an element in a matrix is the determinant of the submatrix that remains after removing the row and column containing that element. It is denoted as M_ij for an element a_ij. The cofactor, on the other hand, is the signed version of the minor. The cofactor of an element a_ij is given by C_ij=(-1)^i+jM_ij, where the factor (-1)^i+j introduces a sign change depending on the position of the element in the matrix. Minors are used to calculate the determinant, while cofactors are used to expand the determinant and in the adjoint of a matrix.

4. What are the important formulas in Chapter 4 Determinants?

There are many formulas in this chapter such as:
1. Formula to determine the value of the determinant of a matrix.
2. To find the inverse of a matrix using the adjoint of a matrix.

5. What is the inverse of a matrix using determinants?

The inverse of a matrix can be found using determinants if the matrix is square and its determinant is non-zero. The formula for the inverse of a matrix A is:

A^(-1) = (1 / det(A)) * adj(A)
To find the inverse, you first calculate the determinant of the matrix A. If the determinant is non-zero, you then find the adjoint (or adjugate) of the matrix. Finally, multiply the adjoint by (1 / det(A)) to obtain the inverse

If the determinant is zero, the matrix does not have an inverse. This method allows you to find the inverse of any invertible matrix using determinants and the adjoint.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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