NCERT solutions for Class 12 Maths Chapter 4 Determinants

NCERT solutions for Class 12 Maths Chapter 4 - Determinants

Edited By Komal Miglani | Updated on Jun 20, 2025 09:44 PM IST | #CBSE Class 12th

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In mathematics, Determinants are the values associated with a square matrix that give us some useful information about the matrix, such as whether the matrix is invertible or not. Also, determinants help us simplify and solve systems of equations. The determinants chapter from Class 12 Maths contains the concepts of Determinants and how to evaluate them, Area of a triangle using determinants, Minors and Cofactors, and also the adjoint and the inverse of a matrix. Understanding these concepts will help the students in many ways, from finding the value of determinants to using their properties in solving systems of equations. The solutions for various subjects and classes can be downloaded from the NCERT Solutions.

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Determinants Class 12 Questions And Answers PDF Free Download
NCERT Solutions for Class 12 Maths Chapter 4: Exercises
Determinants Class 12 Chapter 4: Topics
NCERT Class 12 Maths Chapter 4 Question Answer - Important Formulae
Approach to Solve Questions of Determinants Class 12
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NCERT solutions for class 12 Maths: Chapter-Wise

NCERT solutions for Class 12 Maths Chapter 4 - Determinants

This article on NCERT solutions for class 12 Maths Chapter 4 Determinants offers clear and step-by-step solutions for the exercise problems given in the ncert book. It covers all the important Class 12 Maths Chapter 4 question answers. Furthermore, these solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 12 maths and other subjects can be downloaded from the NCERT Solutions for Class 12.

Determinants Class 12 Questions And Answers PDF Free Download

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NCERT Solutions for Class 12 Maths Chapter 4: Exercises

Determinants class 12 questions and answers: 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(i): Evaluate the following determinant- $| \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} |$

Answer:

The given two by two determinant is calculated as follows

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

Question 2(ii): Evaluate the following determinant- $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

Answer:

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

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

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

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

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

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 $27 | A |$ ,

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

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

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

Hence proved.

Question 5(i): Evaluate the determinants.

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

Answer:

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

now, calculating its determinant value,

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

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

Question 5(ii): Evaluate the determinants.

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

Answer:

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

Now calculating the determinant value;

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

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

Question 5(iii): Evaluate the determinants.

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

Answer:

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

Now calculating the determinant value;

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

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

Question 5(iv): Evaluate the determinants.

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

Answer:

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

We now calculate determinant value:

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

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

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(i): Find values of x, if

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

Answer:

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.,

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

So, we have then,

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

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

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

Answer:

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;

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

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 ;

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

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).

Determinants class 12 questions and answers: Exercise: 4.2
Page number: 83
Total Questions: 5

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

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

Answer:

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

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

Expanding using second column

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

Area of the triangle by the determinant method:

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

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

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

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,

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

calculating the area:

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

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

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

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

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

Question 3(i): Find values of k if area of triangle is 4 sq. units and vertices are

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

Answer:

We can easily calculate the area by the formula :

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

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

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

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

Question 3(ii): Find values of k if area of triangle is 4 sq. units and vertices are

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

Answer:

The area of the triangle is given by the formula:

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

Now, calculating the area:

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

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

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

Question 4(i): Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.

Answer:

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.

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

So, we have:

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

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

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

Question 4(ii): Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.

Answer:

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

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

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

Calculating the determinant:

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

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

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

Hence we have the line equation:

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

Question 5: If the area of 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:

Area of triangle is given by:

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

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

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

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

$50 - 10 k = \pm 70$

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

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

Therefore option (D) is correct.

Determinants class 12 questions and answers: Exercise: 4.3
Page number: 87
Total Questions: 5

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

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

Answer:

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:

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

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

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

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

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

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

Answer:

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(i): Write Minors and Cofactors of the elements of following determinants:

$| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} |$

Answer:

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

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

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

Answer:

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

By the definition itself, $Δ$ is equal to the sum of the products of the elements of any row or column with their corresponding cofactors.

Determinants class 12 questions and answers: Exercise: 4.4
Page number: 92-93
Total Questions: 18

Question 1: Find 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 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,

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

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

aslo,

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

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

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 adjA then as we know the relation:

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

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

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

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

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

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

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

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, calculating the cofactors terms and then adjA.

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

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

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

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

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

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, calculating the cofactors terms and then adjA.

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

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

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

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

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

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

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, calculating the cofactors terms and then adjA.

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

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

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

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

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

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

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, calculating the cofactors terms and then adjA.

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

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

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

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

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

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

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, calculating the cofactors terms and then adjA.

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

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

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

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

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

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

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, calculating 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 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, calculating 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;

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

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

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

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

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

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;

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

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

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

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

So, we have equations;

$11 + 3 a + b = 0, 8 + 2 a = 0$ 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:

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

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

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

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

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

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

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

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

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

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

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

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;

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

Now finding the inverse of A;

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

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

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

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

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

Now,

From equation (1) we get;

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

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

Hence inverse of A is :

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

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,

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

or taking R.H.S.,

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

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

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 $\frac{1}{det (A)}$ .

(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;

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

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,

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

Taking determinant both sides;

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

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

Therefore we get;

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

Hence the correct answer is B.

Determinants class 12 questions and answers: 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:

$x + 2 y = 2$

$2 x + 3 y = 3$

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;

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

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 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;

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

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;

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

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

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

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;

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

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

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

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

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 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;

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

$= - 15 + 3 + 12 = 0$

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;

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

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

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

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using 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

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

we have,

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

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 system of linear equations, using 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

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

we have,

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

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

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

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

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 = -5/11 and y = 12/11.

Question:9 Solve system of linear equations, using 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,

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

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

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

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

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 = -6/11 and y = -19/11.

Question:10 Solve system of linear equations, using 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

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

we have,

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

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

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

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

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 and y = 4

Question:11 Solve system of linear equations, using 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

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

we have,

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

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;

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

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

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

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

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

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

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

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 = 1/2, and z = -3/2.

Question:12 Solve system of linear equations, using 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

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

we have,

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

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;

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

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

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

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

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

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

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

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, and z = 1.

Question:13 Solve system of linear equations, using 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

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

we have,

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

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;

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

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

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

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

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

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

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

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, and z = -1.

Question:14 Solve system of linear equations, using 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

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

we have,

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

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;

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

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

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

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

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

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

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

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, and 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

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

we have,

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

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;

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

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

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

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

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

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

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

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, and 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 cost of each item per kg by matrix method.

Answer:

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

Then we have the equations for the given situation :

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

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

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

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;

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

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

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;

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

$A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{50} [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}]$ 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, and z = 8

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

Determinants class 12 questions and answers: 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 $| \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ ;

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

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

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

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

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

Clearly, the determinant is independent of $Θ$ .

Question:2 Without expanding the determinant, prove that
$| \begin{matrix} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{matrix} | = | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} |$

Answer:

We have the

$L . H . S . = | \begin{matrix} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{matrix} |$

Multiplying rows with a, b, and c respectively.

$R_{1} \to a R_{1}, R_{2} \to b R_{2}, a n d R_{3} \to c R_{3}$

we get;

$= \frac{1}{a b c} | \begin{matrix} a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c \end{matrix} |$

$= \frac{1}{a b c} . a b c | \begin{matrix} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{matrix} |$ $[a f t e r t a k i n g o u t a b c f r o m c o l u m n 3] .$

$= | \begin{matrix} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{matrix} | = | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} |$ $[A p p l y i n g C_{1} \leftrightarrow C_{3} a n d C_{2} \leftrightarrow C_{3}]$

= R.H.S.

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

Question:3 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 $| \begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix} |$ ;

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

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

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

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

Question 4: 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

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

Then we have to basically 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 5(i): Let $A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ . Verify that $100 [adj A]^{- 1} = adj (A^{- 1})$ .
Answer:

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.

Question 5(ii):Let $A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ , verify that $(A^{- 1})^{- 1} = A$ .

Answer:

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:6 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;

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

Taking 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:7 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:8 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 in 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 invers 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 9: Choose the correct answer.

If $x$ , $y$ , $z$ are nonzero real numbers, then the inverse of the matrix

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

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

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

(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:10 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) $D e t (A) = 0$ nbsp; (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.

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Determinants Class 12 Chapter 4: Topics

Here is the list of important topics that are covered in Class 12 Chapter 4, Determinants:

4.1 Introduction
4.2 Determinant
4.3 Area of a Triangle
4.4 Minors and Cofactors
4.5 Adjoint and Inverse of a Matrix
4.6 Applications of Determinants and Matrices

JEE Main Highest Scoring Chapters & Topics

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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 ⋅ 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 ⋅ a 22 − a 12 ⋅ a 21

For a 3x3 matrix A: | A | = a 11 ⋅ | A 11 | − a 12 ⋅ | A 12 | + a 13 ⋅ | 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 | ⋅ | 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 = [a_ij]_n×n is a matrix [A_ji]_n×n, where A_ji is a cofactor of element a_ji.

Properties of Adjoint of a Matrix:

A ⋅ adj ( A ) = adj ( A ) ⋅ 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 | ≠ 0 ), the inverse A − 1 is defined as A − 1 = 1 | A | ⋅ 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 | ≠ 0 , the system is consistent, and X = A^{− 1} B has a unique solution.

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

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

Approach to Solve Questions of Determinants Class 12

Determinants play a quite significant role in Class 12 mathematics, so here are some key steps on how to approach determinants-related questions effectively:

Understanding the basics: First of all, learn what a determinant is and how it is calculated. Then study all the different types of determinants to have a clear idea about them.

Learn and apply the properties: Study all the important properties of determinants, like swapping two rows changes the sign or the value of the determinant becomes zero if two rows or two columns are equal. Applying these properties will make solving determinant problems quite easy to handle.

Use the row and column operations: You can use the elementary operations, like making zeros in the rows or columns, this will simplify the problem before expanding the determinants.

Avoid common mistakes: Be careful about the sign changes in the cofactor expansion. Always try to double-check the values after transformations.

Tips and tricks: To improve your speed and accuracy, you have to practice many different types of questions from the ncert book, the exemplar book, and the previous year papers. Also, you need to revise the key concepts and formulas often in between to memorise them.

What Extra Should Students Study Beyond the NCERT for JEE?

Here is a comparison list of the concepts in Inverse Trigonometric Functions that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

Concepts Name	JEE	NCERT
Properties of Determinants	✅	✅
Multiplication of Two Determinants	✅	✅
Minors And Cofactors	✅	✅
Adjoint and Inverse of a Matrix	✅	✅
Inverse Matrix	✅	✅
System of Linear Equations	✅	✅
Cramer’s Rule	✅	❌
Homogeneous System of Linear Equations	✅	❌

NCERT solutions for class 12 Maths: Chapter-Wise

Given below is the chapter-wise list of the NCERT Class 12 Maths solutions with their respective links:

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

Also Read,

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 Books and NCERT Syllabus

Here are some useful links for the NCERT books and the NCERT syllabus for class 12

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.

Also Read

NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

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Aldehydes, Ketones and Carboxylic Acids Class 12th Notes- Free NCERT Class 12 Chemistry Chapter 12 Notes - Download PDF

Biomolecules class 12th notes- Free ncert class 12 Chemistry Chapter 14 Notes- Download PDF

Dual nature of radiation and matter Class 12th Notes - Free NCERT Class 12 Physics Chapter 11 Notes - Download PDF

Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF

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NCERT Book Class 12 Maths 2025-26 PDF Free Download

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NCERT Book Class 12 English 2025-26 Free PDF Download

NCERT Books for Class 12 Biology 2025-26: Download Free PDF

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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.

Read Complete Answer

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!

Read Complete Answer

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.

Read Complete Answer

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!

Read Complete Answer

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.

Read Complete Answer

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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