Q: Let R denote the set of all real numbers and let i = -1 . Consider the matrices S = bmatrix 0 & -1 \\ 1 & 0 bmatrix and T = bmatrix 1 & 1 \\ 0 & 1 bmatrix . Let a, b, c, d be real numbers such that ST

We are given the matrices S = bmatrix 0 & -1 \\ 1 & 0 bmatrix and T = bmatrix 1 & 1 \\ 0 & 1 bmatrix . First, we compute the product ST: ST = bmatrix 0 & -1 \\ 1 & 0 bmatrix bmatrix 1 & 1 \\ 0 & 1 bmatrix = bmatrix 0 & -1 \\ 1 & 1 bmatrix Comparing this with ST = bmatrix a & b \\ c & d bmatrix , we get a = 0, b = -1, c = 1, and d = 1. Evaluating Option (A): b + ia d + ic = -1 + i(0) 1 + i(1) = -1 1 + i = -1(1 - i) (1 + i)(1 - i) = -1 + i 2 This is not equal to i, so statement (A) is FALSE. Evalu

Q: Consider the matrix, P= ( array lll 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 array ) Let the transpose of a matrix X be denoted by X ^T. Then the number of 3 3 invertible matrices Q with integer entries, s

$ aligned & PQ = QP [ array ccc 2 a _1 & 2 ~b _1 & 2 c _1 \\ 2 a _2 & 2 ~b _2 & 2 c _2 \\ 3 a _3 & 3 ~b _3 & 3 c _3 array ]= [ array ccc 2 a _1 & 2 ~b _1 & 3 c _1 \\ 2 a _2 & 2 ~b _2 & 3 c _2 \\ 2 a _3 & 2 ~b _3 & 3 c _3 array ] \\ & c _1=0, c _2=0, a _3=0, ~b _3=0 \\ & Q = [ array ccc a _1 & ~b _1 & 0 \\ a _2 & ~b _2 & 0 \\ 0 & 0 & c _3 array ] \\ & a _1 a _2+ b _1 ~b _2=0 \\ & a _1^2+ b _1^2=1, a _2^2+ b _2^2=1, c _3^2=1 aligned $ \( array lllll a_1 & b_1 & a_2 & b_2 & c_3 \\ 1 & 0 & 0 & 1,-

Q: Let I= ( array ll 1 & 0 \\ 0 & 1 array ) and P= ( array ll 2 & 0 \\ 0 & 3 array ). Let Q= ( array ll x & y \\ z & 4 array ) for some non-zero real numbers x, y, and z, for which there is 2 2 matrix R

aligned & QR = RP \\ & P= ( array ll 2 & 0 \\ 0 & 3 array ) Q= ( array ll x & y \\ z & 4 array ) \\ & ( array ll x & y \\ z & 4 array ) ( array ll r_1 & r_2 \\ r_3 & r_4 array )= ( array ll r_1 & r_2 \\ r_3 & r_4 array ) ( array ll 2 & 0 \\ 0 & 3 array ) \\ & . array l xr _1+ yr _3=2 r _1 \\ zr _1+4 r _3=2 r _3 array \ x -2 - y = r _3 r _1 = z -2 aligned $2 x-4=y z$ . array l xr _2+ yr _4=3 r _2 \\ zr _2+4 r _4=3 r _4 array \ x -3 - y = r _4 r _2 =- z $ x -3= yz $ 2 x-4=x-3 \( x=1 ~\&~ y z=-

Q: Let and be the distinct roots of the equation x^2+x-1=0. Consider the set T=\ 1, , \ . For a 3 3 matrix M= (a_ i j )_ 3 3 , define R_i=a_ i 1 +a_ i 2 +a_ i 3 and C_j=a_ 1 j +a_ 2 j +a_ 3 j for i=1,2,3

aligned & , are roots of x ^2+ x -1=0 \\ & + =-1 1+ + =0 \\ & M= [ array lll a _ 11 & a _ 12 & a _ 13 \\ a _ 21 & a _ 22 & a _ 23 \\ a _ 31 & a _ 32 & a _ 33 array ] aligned (P) M= [ array lll 1 & & \\ & & 1 \\ & 1 & array ] 3! 2=12 For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3 ! ways 3! 2=12 ways (Q) M= [ array lll x & a & b \\ a & y & c \\ b & c & z array ] Consider one such arrangement with a= , b= , c=1 M= [ array lll 1 & & \\ &

Q: Let M = a i j , i , j ∈ 1 , 2 , 3 , be the 3 × 3 matrix such that a i j = 1 if j + 1 is divisible by i , otherwise a i j = 0 . Then which of the following statements is(

Given, M = a i j , i , j ∈ 1 , 2 , 3 a i j = 1 if j + 1 is divisible by i otherwise a i j = 0 So, a 11 = 1 , a 12 = 1 , a 13 = 1 , a 21 = 1 , a 22 = 0 . . . . . . , a 33 = 0 So, M = 1 1 1 1 0 1 0 1 0 ⇒ | M | = 1 ( − 1 ) &#87

Q: Let R = a 3 b c 2 d 0 5 0 : a , b , c , d ∈ 0 , 3 , 5 , 7 , 11 , 13 , 17 , 19 . Then the number of invertible matrices in R is

Given, R = a 3 b c 2 d 0 5 0 ⇒ | R | = - 5 a b c d We know that, for invertible matrices R ≠ 0 Now | R | can be zero in following cases: (i) Two of a , b , c , d are zeroes which can be ( a and b ),( b and d ),( d and c ) or ( c and a ) → 4 × 7 2 ways = 196 (ii) Any three of a , b , c , d are zeroes → C 3 4 × 7 = 28 (iii) All four of a , b , c

Q: Let β be a real number. Consider the matrix A = β 0 1 2 1 - 2 3 1 - 2 . If A 7 - β - 1 A 6 - β A 5 is a singular matrix, then the value of 9 β is _____ .

Given, A = β 0 1 2 1 - 2 3 1 - 2 So, det A = - 1 . . . 1 Now for A 7 - β - 1 A 6 - β A 5 to be singular A 7 - β - 1 A 6 - β A 5 = 0 ⇒ A 5 A 2 - ( β - 1 ) A - β = 0 . . . 2 ⇒ A 5 A + I A - β l = 0 ∴ A 5 A + I A - β = 0 As A ≠ 0 A + 1 or A - β = 0 ⇒ β + 1 0 1 2 2 - 2 3 1 - 1 = 0 ⇒ A + I = - 4 , here A ≠ 0

Question 1

Which one of the following matrices can be obtained by performing elementary row transformations on the 3 3 identity matrix?

Accepted Answer

Any matrix obtained by performing elementary row transformations on the identity matrix is equivalent to the identity matrix, which means it must be non-singular (invertible). Therefore, its determinant must be non-zero. Evaluating the determinant of the matrix in option (A): vmatrix 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 vmatrix = 0 Evaluating the determinant of the matrix in option (B): vmatrix 1 & 1 & 1 \ 2 & 3 & 4 \ 1 & 2 & 1 vmatrix = 1(3 - 8) - 1(2 - 4) + 1(4 - 3) = -5 + 2 + 1 = -2 0 Evalua

Question 2

Consider the matrix M = bmatrix 2 & -1 \ 1 & 0 bmatrix . Let p, q, r, s, a, b, c and d be integers such that M^ 26 = bmatrix p & q \ r & s bmatrix and _ k=1 ^ 26 M^k = bmatrix a & b \ c & d bmatrix

Accepted Answer

The characteristic equation of M is given by (M - I) = 0: vmatrix 2 - & -1 \ 1 & - vmatrix = ^2 - 2 + 1 = ( - 1)^2 = 0 The only eigenvalue is = 1 with algebraic multiplicity 2. For the eigenvector, (M - I)X = 0 bmatrix 1 & -1 \ 1 & -1 bmatrix bmatrix x \ y bmatrix = bmatrix 0 \ 0 bmatrix x = y. Since there is only one linearly independent eigenvector, M is not diagonalizable but can be reduced to its Jordan canonical form J = bmatrix 1 & 1 \ 0 & 1 bmatrix . Thus, there exists an invertible

Question 3

For real numbers , , , and , consider the matrix M = bmatrix & 1 2 & - 1 2 \ 1 3 & & 1 3 \ & & bmatrix . Suppose that MM^T = I, where M^T is the transpose of the matrix M, and I is the 3 3 identity

Accepted Answer

Since MM^T = I, M is an orthogonal matrix. The columns of an orthogonal matrix form an orthonormal basis for R ^3. The given vectors u , v , w are exactly the columns of M. Therefore, u , v , w are mutually orthogonal unit vectors. For (P): The rows of M are also orthonormal. The third row is bmatrix & & bmatrix , so ^2 + ^2 + ^2 = 1. The third column w is a unit vector, so | w |^2 = (- 1 2 )^2 + ( 1 3 )^2 + ^2 = 1. 1 2 + 1 3 + ^2 = 1 ^2 = 1 6 Substituting ^2 into the row equation gives ^2 + ^2

Question 4

Let R denote the set of all real numbers and let i = -1 . Consider the matrices S = bmatrix 0 & -1 \ 1 & 0 bmatrix and T = bmatrix 1 & 1 \ 0 & 1 bmatrix . Let a, b, c, d be real numbers such that ST

Accepted Answer

We are given the matrices S = bmatrix 0 & -1 \ 1 & 0 bmatrix and T = bmatrix 1 & 1 \ 0 & 1 bmatrix . First, we compute the product ST: ST = bmatrix 0 & -1 \ 1 & 0 bmatrix bmatrix 1 & 1 \ 0 & 1 bmatrix = bmatrix 0 & -1 \ 1 & 1 bmatrix Comparing this with ST = bmatrix a & b \ c & d bmatrix , we get a = 0, b = -1, c = 1, and d = 1. Evaluating Option (A): b + ia d + ic = -1 + i(0) 1 + i(1) = -1 1 + i = -1(1 - i) (1 + i)(1 - i) = -1 + i 2 This is not equal to i, so statement (A) is FALSE. Evalu

Question 5

Consider the matrix, P= ( array lll 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 array ) Let the transpose of a matrix X be denoted by X ^T. Then the number of 3 3 invertible matrices Q with integer entries, s

Accepted Answer

$ aligned & PQ = QP [ array ccc 2 a _1 & 2 ~b _1 & 2 c _1 \ 2 a _2 & 2 ~b _2 & 2 c _2 \ 3 a _3 & 3 ~b _3 & 3 c _3 array ]= [ array ccc 2 a _1 & 2 ~b _1 & 3 c _1 \ 2 a _2 & 2 ~b _2 & 3 c _2 \ 2 a _3 & 2 ~b _3 & 3 c _3 array ] \ & c _1=0, c _2=0, a _3=0, ~b _3=0 \ & Q = [ array ccc a _1 & ~b _1 & 0 \ a _2 & ~b _2 & 0 \ 0 & 0 & c _3 array ] \ & a _1 a _2+ b _1 ~b _2=0 \ & a _1^2+ b _1^2=1, a _2^2+ b _2^2=1, c _3^2=1 aligned $ \( array lllll a_1 & b_1 & a_2 & b_2 & c_3 \ 1 & 0 & 0 & 1,-

Question 6

Let I= ( array ll 1 & 0 \ 0 & 1 array ) and P= ( array ll 2 & 0 \ 0 & 3 array ). Let Q= ( array ll x & y \ z & 4 array ) for some non-zero real numbers x, y, and z, for which there is 2 2 matrix R

Accepted Answer

aligned & QR = RP \ & P= ( array ll 2 & 0 \ 0 & 3 array ) Q= ( array ll x & y \ z & 4 array ) \ & ( array ll x & y \ z & 4 array ) ( array ll r_1 & r_2 \ r_3 & r_4 array )= ( array ll r_1 & r_2 \ r_3 & r_4 array ) ( array ll 2 & 0 \ 0 & 3 array ) \ & . array l xr _1+ yr _3=2 r _1 \ zr _1+4 r _3=2 r _3 array \ x -2 - y = r _3 r _1 = z -2 aligned $2 x-4=y z$ . array l xr _2+ yr _4=3 r _2 \ zr _2+4 r _4=3 r _4 array \ x -3 - y = r _4 r _2 =- z $ x -3= yz $ 2 x-4=x-3 \( x=1 ~\&~ y z=-

Question 7

Let and be the distinct roots of the equation x^2+x-1=0. Consider the set T=\ 1, , \ . For a 3 3 matrix M= (a_ i j )_ 3 3 , define R_i=a_ i 1 +a_ i 2 +a_ i 3 and C_j=a_ 1 j +a_ 2 j +a_ 3 j for i=1,2,3

Accepted Answer

aligned & , are roots of x ^2+ x -1=0 \ & + =-1 1+ + =0 \ & M= [ array lll a _ 11 & a _ 12 & a _ 13 \ a _ 21 & a _ 22 & a _ 23 \ a _ 31 & a _ 32 & a _ 33 array ] aligned (P) M= [ array lll 1 & & \ & & 1 \ & 1 & array ] 3! 2=12 For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3 ! ways 3! 2=12 ways (Q) M= [ array lll x & a & b \ a & y & c \ b & c & z array ] Consider one such arrangement with a= , b= , c=1 M= [ array lll 1 & & \ &

Question 8

Let M = a i j , &#160; i , &#160; j &#8712; 1 , 2 , 3 , be the 3 &#215; 3 matrix such that a i j = 1 if j + 1  is divisible by i , otherwise a i j = 0 . Then which of the following statements is(

Accepted Answer

Given, M = a i j , &#160; i , &#160; j &#8712; 1 , 2 , 3 a i j = 1 &#160;if&#160; j + 1 &#160;is divisible by&#160; i otherwise&#160; a i j = 0 So,  a 11 = 1 , &#160; a 12 = 1 , &#160; a 13 = 1 , &#160; a 21 = 1 , &#160; a 22 = 0 . . . . . . , a 33 = 0 So,  M = 1 &#8197;&#8197;&#8197;&#8197; 1 &#8197;&#8197;&#8197;&#8197; 1 1 &#8197;&#8197;&#8197;&#8197; 0 &#8197;&#8197;&#8197;&#8197; 1 0 &#8197;&#8197;&#8197;&#8197; 1 &#8197;&#8197;&#8197;&#8197; 0 &#8658; | M | = 1 ( &#8722; 1 ) &#87

Question 9

Let R = a 3 b c 2 d 0 5 0 : a , b , c , d &#8712; 0 , 3 , 5 , 7 , 11 , 13 , 17 , 19 . Then the number of invertible matrices in R  is

Accepted Answer

Given, R = a 3 b c 2 d 0 5 0 &#8658; | R | = - 5 a b c d We know that, for invertible matrices  R &#8800; 0 Now  | R | can be zero in following cases: (i) Two of a , &#160; b , &#160; c , &#160; d  are zeroes which can be ( a  and b ),( b  and d ),( d  and c ) or ( c  and a ) &#8594; 4 &#215; 7 2 &#160;ways&#160; = 196 (ii) Any three of a , &#160; b , &#160; c , &#160; d  are zeroes &#8594; C 3 4 &#215; 7 = 28 (iii) All four of a , &#160; b , &#160; c

Question 10

Let &#946;  be a real number. Consider the matrix A = &#946; 0 1 2 1 - 2 3 1 - 2 . If A 7 - &#946; - 1 A 6 - &#946; A 5 is a singular matrix, then the value of 9 &#946; is _____ .

Accepted Answer

Given, A = &#946; 0 1 2 1 - 2 3 1 - 2 So,  det A = - 1 &#160; &#160; &#160; &#160; . . . 1 Now for A 7 - &#946; - 1 A 6 - &#946; A 5  to be singular A 7 - &#946; - 1 A 6 - &#946; A 5 = 0 &#8658; A 5 A 2 - ( &#946; - 1 ) A - &#946; = 0 &#160; &#160; &#160; &#160; &#160; &#160; . . . 2 &#8658; A 5 A + I A - &#946; l = 0 &#8756; A 5 A + I A - &#946; = 0 As A &#8800; 0 A + 1 or A - &#946; = 0 &#8658; &#946; + 1 0 1 2 2 - 2 3 1 - 1 = 0 &#8658; A + I = - 4 , &#160; here&#160; A &#8800; 0

Question 11

If M = 5 2 3 2 - 3 2 - 1 2 , then which of the following matrices is equal to M 2022 ?

Accepted Answer

Given  M = 5 2 3 2 - 3 2 - 1 2 M 2 = 5 2 3 2 - 3 2 - 1 2 5 2 3 2 - 3 2 - 1 2 = 4 3 - 3 - 2 M 3 = 5 2 3 2 - 3 2 - 1 2 4 3 - 3 - 2 = 11 2 9 2 - 9 2 - 7 2 M 4 = 11 2 9 2 - 9 2 - 7 2 11 2 9 2 - 9 2 - 7 2 = 7 6 - 6 - 5 and so on i.e.  M n = 3 n 2 + 1 3 n 2 - 3 n 2 - 3 n 2 + 1 Hence, M 2022 = 3 &#215; 2022 2 + 1 3 &#215; 2022 2 - 3 &#215; 2022 2 - 3 &#215; 2022 2 + 1 = 3034 3033 - 3033 - 3032

Question 12

For any 3 &#215; 3 matrix M , let | M | denote the determinant of M . Let E = 1 2 3 2 3 4 8 13 18 , &#160;&#160; P = 1 0 0 0 0 1 0 1 0 &#160; &#38; &#160; F = 1 3 2 8 18 13 2 4 3 If Q is a nonsingular

Accepted Answer

(1)  P 2 = 1 0 0 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 = 1 0 0 0 1 0 0 0 1 = I PEP = 1 0 0 0 0 1 0 1 0 1 2 3 2 3 4 8 13 18 1 0 0 0 0 1 0 1 0 = 1 2 3 8 13 18 2 3 4 1 0 0 0 0 1 0 1 0 = 1 3 2 8 18 13 2 4 3 = F (2) Here,  F = P E P &#8658; P F = P 2 E P &#8658; P F = E P ,  E = 0 , &#160; F = 0 E Q + P F Q - 1 = E Q + E P Q - 1 = &#8739; E Q + P Q - 1 | = | E | | Q + P Q - 1 &#8739; = 0 | E Q | + P F Q - 1 = 0 + 0 = 0 (3) Since,  | E | = 0 &#8658; | E F | = 0 Thus,  E F 3 &#62; E F

Question 13

For any 3 &#215; 3 matrix M , let | M | denote the determinant of M . Let I be the 3 &#215; 3 identity matrix. Let  E and F  be two 3 &#215; 3 matrices such that ( I - E F ) is invertible. I

Accepted Answer

(A)  G = ( I - E F ) - 1 &#8658; F G E = F ( I - E F ) - 1 E = E - 1 ( I - E F ) F - 1 - 1 &#8658; F G E = E - 1 &#160; F - 1 - I - 1 &#8658; | F G E | &#183; E - 1 &#160; F - 1 - I = 1 &#8658; | F G E | &#183; E - 1 &#160; F - 1 - I | F E | = | F E | &#8658; | F G E | | I - F E | = | F E | (B) & (D)  ( I - F E ) &#183; ( I + F G E ) = I + F G E - F E - F E &#183; F G E = 0 = I + F G E - F E - F ( G - I ) E = I + F G E - F E - F G E + F E = I (C)  ( I - E F ) G = I = G ( I - E

Question 14

The trace of a square matrix is defined to be the sum of its diagonal entries. If A  is a  2 &#215; 2  matrix such that the trace of A is  3  and the trace of   A 3

Accepted Answer

Let  A = a b c 3 - a &#160; &#160; &#8658; &#160; A = 3 a - a 2 - b c A 3 = a b c 3 - a a b c 3 - a a b c 3 - a = a 2 + b c a b + 3 b - a b a c + 3 c - a c c b + ( 3 - a ) 2 a b c 3 - a = a 2 + b c 3 b 3 c c b + ( 3 - a ) 2 a b c 3 - a tr A 3 = a 3 + a b c + 3 b c + 3 b c + 3 b c - a b c + ( 3 - a ) 3 = - 18 a 3 + 9 b c + 3 - a 3 = - 18 9 b c + 27 - 9 a ( 3 - a ) = - 18 b c + 3 - 3 a + a 2 = - 2 3 a - a 2 - b c = 5 &#160;&#160; &#8658; &#160;&#160; | A | = 5

Question 15

Let M be a 3 &#215; 3 invertible matrix with real entries and let I denote the 3 &#215; 3 identity matrix. M &#8722; 1 = a d j a d j &#8201; M , then which of the following statements is/are ALWAYS TR

Accepted Answer

Given  M &#8722; 1 = adj adjM &#8756;&#160;adj&#160;M &#8901; M &#8722; 1 = adj&#160;M&#160; .&#160;adj&#160; adjM &#8658;adj&#160;M . M &#8722; 1 = AdjM I &#8658;adj&#160;M = M 2 M &#160; . . . . . . . . . 1 &#8658; a d j &#8201; M = M 2 M = M 6 M &#8658; M 2 = M 7 &#8658; M &#8800; 0 , &#160; M = 1 &#160; . . . . . . . . . . . . . 2 By equations  1 &#38; &#160; 2 , we get adj&#160;M = M &#160; . . . . . . . . . . . 3 &#8658;M . adjM = M 2 &#8658; M I = M 2 &#8658; M 2 = I Now, by equ

Question 16

Let P 1 = I = 1 0 0 0 1 0 0 0 1 , &#160; P 2 = 1 0 0 0 0 1 0 1 0 , &#160; P 3 = 0 1 0 1 0 0 0 0 1 ,&#160; P 4 = 0 1 0 0 0 1 1 0 0 , &#160; P 5 = 0 0 1 1 0 0 0 1 0 , P 6 = 0 0 1 0 1 0 1 0 0 , and X = &

Accepted Answer

&#8757; &#160; P 1 = P 1 T = P 1 - 1 , P 3 = P 3 T = P 3 - 1 , P 5 = P 5 T = P 5 T , P 2 = P 2 T = P 2 - 1 , P 4 = P 4 T = P 4 T , P 6 = P 6 T = P 6 - 1 Now Let Q = 2 1 3 1 0 2 3 2 1 X = &#8721; K = 1 6 P K Q P K T &#8658; T r a c e X = &#8721; K = 1 6 T r P K Q P K T &#8757; T r A B = T r B A = &#8721; K = 1 6 T r P K T P K Q &#8757; P K T &#160; P K = I = &#8721; K = 1 6 T r Q = 6 &#160; T r a c e Q = 6 &#215; 3 = 18 &#8658; X = &#8721; K = 1 6 P K Q P K T X T = &#8721; K = 1 6 P K Q P K T T X

Question 17

Let M = sin 4 &#8289; &#952; - 1 - sin 2 &#8289; &#952; 1 + cos 2 &#8289; &#952; cos 4 &#8289; &#952; = &#945; I + &#946; M - 1 , where &#945; = &#945; &#952; and &#946; = &#946; &#952; are real numbe

Accepted Answer

Method I As given, M = s i n 4 &#952; - 1 - s i n 2 &#952; 1 + c o s 2 &#952; c o s 4 &#952; = &#945; I + &#946; M - 1 a d j &#8201; M = c o s 4 &#952; 1 + s i n 2 &#952; - 1 - c o s 2 &#952; s i n 4 &#952; M - 1 = 1 | M | c o s 4 &#952; 1 + s i n 2 &#952; - 1 - c o s 2 &#952; s i n 4 &#952; s i n 4 &#952; - 1 - s i n 2 &#952; 1 + c o s 2 &#952; c o s 4 &#952; = &#945; 1 0 0 1 + &#946; M c o s 4 &#952; 1 + s i n 2 &#952; - 1 - c o s 4 &#952; s i n 4 &#952; Compare element a 12 in L.H.S. and R.H.

Question 18

Let M = 0 1 a 1 2 3 3 b 1 and a d j &#8201; M = - 1 1 - 1 8 - 6 2 - 5 3 - 1 where a and b are real numbers. Which of the following options is/are correct?

Accepted Answer

Given, M = 0 1 a 1 2 3 3 b 1 ...(i) a d j M = - 1 1 - 1 8 - 6 2 - 5 3 - 1 ...(ii) Option ( A ) : Now from matrix M a d j &#160; M 11 = 2 - 3 b As given 2 - 3 b = - 1 &#8658; b = 1 Similarly,  a = 2 a + b = 3 Using the values of a and b in matrix M &#8658; | M | = - 2 | a d j M 2 | &#8201; = &#8201; | M 2 | 2 = | M | 4 = 16 Now, M - 1 = a d j &#8201; M M a d j &#160; M = M M - 1 a d j &#160; ( M ) = &#160; - 2 M - 1 a d j M - 1 = - 1 2 M - 1 - 1 K A - 1 = 1 K A - 1 ( a d j ( M ) ) - 1 = - M

Matrices — Previous Year Questions & Solutions