Q: Consider the matrices A = bmatrix 2 & -2 \\ 4 & -2 bmatrix and B = bmatrix 3 & 9 \\ 1 & 3 bmatrix . If matrices P and Q are such that PA = B and AQ = B, then the absolute value of the sum of the diago

Given A = bmatrix 2 & -2 \\ 4 & -2 bmatrix , we find its determinant: |A| = (2)(-2) - (-2)(4) = 4 Since |A| 0, A is invertible. The inverse of A is: A^ -1 = 1 4 bmatrix -2 & 2 \\ -4 & 2 bmatrix = bmatrix -1/2 & 1/2 \\ -1 & 1/2 bmatrix We are given PA = B P = BA^ -1 and AQ = B Q = A^ -1 B. Using the cyclic property of trace Tr (XY) = Tr (YX), we have: Tr (P) = Tr (BA^ -1 ) = Tr (A^ -1 B) = Tr (Q) Let us calculate Tr (P) by finding the diagonal elements of P = BA^ -1 : P = bmatrix 3 & 9 \\ 1 & 3 b

Question 1

Let A = bmatrix & 1 & 2 \ 2 & 3 & 0 \ 0 & 4 & 5 bmatrix and B = bmatrix 1 & 0 & 0 \ 0 & -5 & 0 \ 0 & 4 & -2 bmatrix + adj (A). If (B)=66, then ( adj (A)) equals:

Accepted Answer

The cofactor matrix of A is calculated as follows: C_ 11 = 15, C_ 12 = -10, C_ 13 = 8 C_ 21 = 3, C_ 22 = 5 , C_ 23 = -4 C_ 31 = -6, C_ 32 = 4, C_ 33 = 3 - 2 The adjoint of A is the transpose of the cofactor matrix: adj (A) = bmatrix 15 & 3 & -6 \ -10 & 5 & 4 \ 8 & -4 & 3 - 2 bmatrix The matrix B is given by: B = bmatrix 1 & 0 & 0 \ 0 & -5 & 0 \ 0 & 4 & -2 bmatrix + bmatrix 15 & 3 & -6 \ -10 & 5 & 4 \ 8 & -4 & 3 - 2 bmatrix = bmatrix 16 & 3 & -6 \ -10 & 0 & 4 \ 8 & 0 & - 2 bmatrix Expandi

Question 2

Let A = bmatrix 1 & 0 & 0 \ 3 & 1 & 0 \ 9 & 3 & 1 bmatrix and B = [b_ ij ], 1 i, j 3. If B = A^ 99 - I, then the value of b_ 31 - b_ 21 b_ 32 is :

Accepted Answer

Let A = I + C, where I is the identity matrix and C = bmatrix 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 bmatrix . Calculating the powers of C, we get: C^2 = bmatrix 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 bmatrix bmatrix 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 bmatrix = bmatrix 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 bmatrix C^3 = bmatrix 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 bmatrix bmatrix 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 bmatrix = bmatrix 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 bmatrix Since C^3 = 0, all higher p

Question 3

Let A = bmatrix -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 bmatrix satisfy A^2 + (adj(adj(A))) + (adj(A)(adj(adj(A)))) = bmatrix 2 & -2 & 2 \ -2 & 0 & -1 \ 0 & 0 & -1 bmatrix for some , R . Then ( - )^2

Accepted Answer

The determinant of matrix A is: |A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1 Using the properties of the adjoint of a 3 3 matrix: adj(adj(A)) = |A|^ 3-2 A = -A adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I The given equation simplifies to: A^2 - A + I = bmatrix 2 & -2 & 2 \ -2 & 0 & -1 \ 0 & 0 & -1 bmatrix Calculating A^2: A^2 = bmatrix -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 bmatrix bmatrix -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 bmatrix = bmatrix 2 & -1 & 1 \ -1 & 1 & 0 \ 0 & 0 & 1 bmatr

Question 4

Let M be a 3 3 matrix such that M pmatrix 1 \ 0 \ 0 pmatrix = pmatrix 1 \ 2 \ 3 pmatrix , M pmatrix 0 \ 1 \ 0 pmatrix = pmatrix 0 \ 1 \ 2 pmatrix and M pmatrix 0 \ 0 \ 1 pmatrix = pmatrix -1

Accepted Answer

Let M be the matrix whose columns are the given vectors. We have: M = pmatrix 1 & 0 & -1 \ 2 & 1 & 1 \ 3 & 2 & 1 pmatrix Given M pmatrix x \ y \ z pmatrix = pmatrix 1 \ 7 \ 11 pmatrix , we can write this as a system of linear equations: x - z = 1 2x + y + z = 7 3x + 2y + z = 11 Subtracting the second equation from the third equation gives: x + y = 4 Adding the first equation and the second equation gives: 3x + y = 8 Subtracting the equation x + y = 4 from 3x + y = 8 gives: 2x = 4 x = 2 Sub

Question 5

Let A be a 3 3 matrix such that A^T bmatrix 1\0\1 bmatrix = bmatrix 5\2\2 bmatrix , A^T bmatrix 0\0\1 bmatrix = bmatrix 3\1\1 bmatrix , A bmatrix 1\0\1 bmatrix = bmatrix 3\4\4 bmatrix and

Accepted Answer

Let C_1, C_2, C_3 be the columns of A and R_1, R_2, R_3 be the rows of A. From the given equations, we have: A bmatrix 0\0\1 bmatrix = C_3 = bmatrix 1\3\1 bmatrix A bmatrix 1\0\1 bmatrix = C_1 + C_3 = bmatrix 3\4\4 bmatrix C_1 = bmatrix 2\1\3 bmatrix Similarly, for the transpose A^T, the columns correspond to the rows of A: A^T bmatrix 0\0\1 bmatrix = R_3^T = bmatrix 3\1\1 bmatrix R_3 = bmatrix 3 & 1 & 1 bmatrix A^T bmatrix 1\0\1 bmatrix = R_1^T + R_3^T = bmatrix 5\2\2 bmatrix

Question 6

Let A= bmatrix 1 & 2 & 7 \ 4 & -2 & 8 \ 3 & 8 & -7 bmatrix and (A- I)=0, where is a real number. If the largest possible value of is p, then the circle (x-p)^2+(y-2p)^2=320, intersects the co-ordina

Accepted Answer

The characteristic equation is given by (A- I) = 0. bmatrix 1- & 2 & 7 \ 4 & -2- & 8 \ 3 & 8 & -7- bmatrix = 0 Expanding the determinant along the first row: (1- )[(-2- )(-7- ) - 64] - 2[4(-7- ) - 24] + 7[32 - 3(-2- )] = 0 (1- )( ^2 + 9 - 50) - 2(-4 - 52) + 7(3 + 38) = 0 - ^3 - 8 ^2 + 59 - 50 + 8 + 104 + 21 + 266 = 0 ^3 + 8 ^2 - 88 - 320 = 0 By inspection, = 8 is a root since 8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0. Factoring out ( - 8), we get: ( - 8)( ^2 + 16 + 40) = 0 The roo

Question 7

Let S = \ A = bmatrix a & b \ c & d bmatrix : a, b, c, d \ 0, 1, 2, 3, 4\ and A^2 - 4A + 3I = 0 \ be a set of 2 2 matrices. Then the number of matrices in S, for which the sum of the diagonal element

Accepted Answer

The characteristic equation of a 2 2 matrix A = bmatrix a & b \ c & d bmatrix is given by: A^2 - Tr (A)A + (A)I = 0 We are given that the sum of the diagonal elements is 4, so Tr (A) = a + d = 4. Substituting this into the characteristic equation, we get: A^2 - 4A + (ad - bc)I = 0 We are also given that the matrix satisfies: A^2 - 4A + 3I = 0 Comparing the two equations, we must have: ad - bc = 3 We need to find the number of quadruples (a, b, c, d) with elements from the set \ 0, 1, 2, 3, 4\ s

Question 8

Let A = bmatrix 1 & 1 & 2 \ -2 & 0 & 1 \ 1 & 3 & 5 bmatrix . Then the sum of all elements of the matrix adj ( adj (2( adj A)^ -1 )) is equal to:

Accepted Answer

The determinant of matrix A is given by: |A| = 1(0 - 3) - 1(-10 - 1) + 2(-6 - 0) |A| = -3 + 11 - 12 = -4 We know the property A( adj A) = |A|I, which gives adj A = |A|A^ -1 . Taking the inverse on both sides: ( adj A)^ -1 = A |A| = A -4 Let B = 2( adj A)^ -1 . Substituting the above expression, we get: B = 2 ( A -4 ) = - 1 2 A We need to find the matrix adj ( adj B). For a square matrix of order n, we have adj ( adj B) = |B|^ n-2 B. Since the order is n = 3: adj ( adj B) = |B|^ 3-2 B = |B|B Now,

Question 9

Consider the matrices A = bmatrix 2 & -2 \ 4 & -2 bmatrix and B = bmatrix 3 & 9 \ 1 & 3 bmatrix . If matrices P and Q are such that PA = B and AQ = B, then the absolute value of the sum of the diago

Accepted Answer

Given A = bmatrix 2 & -2 \ 4 & -2 bmatrix , we find its determinant: |A| = (2)(-2) - (-2)(4) = 4 Since |A| 0, A is invertible. The inverse of A is: A^ -1 = 1 4 bmatrix -2 & 2 \ -4 & 2 bmatrix = bmatrix -1/2 & 1/2 \ -1 & 1/2 bmatrix We are given PA = B P = BA^ -1 and AQ = B Q = A^ -1 B. Using the cyclic property of trace Tr (XY) = Tr (YX), we have: Tr (P) = Tr (BA^ -1 ) = Tr (A^ -1 B) = Tr (Q) Let us calculate Tr (P) by finding the diagonal elements of P = BA^ -1 : P = bmatrix 3 & 9 \ 1 & 3 b

Question 10

Let A = bmatrix 1 & 2 \ 1 & bmatrix and B = bmatrix 3 & 3 \ & 2 bmatrix . If A^2 - 4A + I = O and B^2 - 5B - 6I = O, then among the two statements : (S1): [(B-A)(B+A)]^T = bmatrix 13 & 15 \ 7 & 10

Accepted Answer

For a 2 2 matrix M, the characteristic equation is given by M^2 - Tr (M)M + (M)I = O. For matrix A = bmatrix 1 & 2 \ 1 & bmatrix , we are given A^2 - 4A + I = O. Comparing the trace, we get Tr (A) = 1 + = 4 = 3. Thus, A = bmatrix 1 & 2 \ 1 & 3 bmatrix . For matrix B = bmatrix 3 & 3 \ & 2 bmatrix , we are given B^2 - 5B - 6I = O. Comparing the determinant, we get (B) = 6 - 3 = -6 3 = 12 = 4. Thus, B = bmatrix 3 & 3 \ 4 & 2 bmatrix . Evaluating Statement (S1): B - A = bmatrix 3 & 3 \ 4 & 2 bm

Question 11

Let A= [ array ll 3 & -4 \ 1 & -1 array ] and B be two matrices such that A^ 100 =100 B+I. Then the sum of all the elements of B ^ 100 is \_\_\_\_

Accepted Answer

A = I + bmatrix 2 & -4 \ 1 & -2 bmatrix , let M = bmatrix 2 & -4 \ 1 & -2 bmatrix M^2 = bmatrix 0 & 0 \ 0 & 0 bmatrix = M^3 = M^4 = = M^ 100 A^ 100 = (I + M)^ 100 = _ r=0 ^ 100 100 r M^r I A^ 100 = I + 100M = I + 100B M = B M^ 100 = B^ 100 = bmatrix 0 & 0 \ 0 & 0 bmatrix

Question 12

Let A, B and C be three 2 2 matrices with real entries such that B=(I+A)^ -1 and A + C = I . If BC = [ array cc 1 & -5 \ -1 & 2 array ] and CB [ array l x_ 1 \ x_ 2 array ]= [ array c 12 \ -6 array

Accepted Answer

From B = (I + A)^ -1 we get B(I + A) = I, so BA = I - B. From A + C = I: C = I - A. Calculating CB = C(I+A)^ -1 = (I-A)(I+A)^ -1 . Since B(I+A) = I, we have (I+A) = B^ -1 , so C = (I-A)B^ -1 . Multiplying by B: CB = (I-A). However, using the given BC = bmatrix 1 & -5 \ -1 & 2 bmatrix and B(I+A) = I implies B - BA = B - (I-B) = 2B - I = BC, i.e. BC = 2B - I. Therefore B = 1 2 (BC + I) = 1 2 bmatrix 2 & -5 \ -1 & 3 bmatrix = bmatrix 1 & -5/2 \ -1/2 & 3/2 bmatrix . Also CB = 2B - I = bmatrix 1 &

Question 13

Let f(x)= 7 x^ 10 +9 x^ 8 (1+x^ 2 +2 x^ 9 )^ 2 d x, x>0, _ x 0 f(x)=0 and f(1)= 1 4 . If A = [ array ccc 0 & 0 & 1 \ 1 4 & f^ (1) & 1 \ ^ 2 & 4 & 1 array ] and B = adj ( adj A ) be such that | B |=8

Accepted Answer

Divide numerator and denominator by x^ 18 , let t = x^ -9 + x^ -7 + 2, then dt = -(9x^ -10 +7x^ -8 )dx. f(x) = 1 t + C = x^9 1+x^2+2x^9 + C. _ x 0 f(x) = 0 C = 0. f(1) = 1 4 (verified). f'(x) = 9x^8(1+x^2+2x^9) - x^9(2x+18x^8) (1+x^2+2x^9)^2 . f'(1) = 36-20 16 = 1. |A| = vmatrix 0 & 0 & 1 \ 1/4 & 1 & 1 \ ^2 & 4 & 1 vmatrix = 1 - ^2. |B| = | adj ( adj A)| = |A|^ (3-1)^2 = (1- ^2)^4 = 81 = 3^4. |1- ^2| = 3 ^2 = 4.

Question 14

Let P= [p_ i j ] and Q= [q_ i j ] be two square matrices of order 3 such that q_ ij =2^ ( i + j -1) p _ ij and det ( Q )=2^ 10 . Then the value of det ( adj ( adj P )) is:

Accepted Answer

vmatrix 2p_ 11 & 2^2p_ 12 & 2^3p_ 13 \ 2^2p_ 21 & 2^3p_ 22 & 2^4p_ 23 \ 2^3p_ 31 & 2^4p_ 32 & 2^5p_ 33 vmatrix = 2^ 10 2^2 2 2^3 vmatrix p_ 11 & p_ 12 & p_ 13 \ 2p_ 21 & 2p_ 22 & 2p_ 23 \ 2^2p_ 31 & 2^2p_ 32 & 2^2p_ 33 vmatrix = 2^ 10 2^9 vmatrix p_ 11 & p_ 12 & p_ 13 \ p_ 21 & p_ 22 & p_ 23 \ p_ 31 & p_ 32 & p_ 33 vmatrix = 2^ 10 |P| = 2 | adj ( adj (P))| = |P|^ (n-1)^2 = |P|^4 = 2^4 = 16

Question 15

The number of 3 2 matrices A, which can be formed using the elements of the set \ -2,-1,0,1,2\ such that the sum of all the diagonal elements of A ^ T A is 5, is

Accepted Answer

For a 3×2 matrix A, the diagonal sum of A^T A equals the sum of squares of all 6 elements of A. We need: a_ 11 ^2 + a_ 21 ^2 + a_ 31 ^2 + a_ 12 ^2 + a_ 22 ^2 + a_ 32 ^2 = 5 where each element is from \ -2, -1, 0, 1, 2\ . Possible squares: 0^2=0, ( 1)^2=1, ( 2)^2=4. Case 1: Distribution 4+1+0+0+0+0 (one ±2, one ±1, four 0's). Arrangements: 6 1 5 1 = 30. Sign choices: 2 2 = 4. Total: 30 4 = 120. Case 2: Distribution 1+1+1+1+1+0 (five ±1, one 0). Arrangements: 6 1 = 6 ways to place the 0. Sign choi

Question 16

Let A= [ array ccc 0 & 2 & -3 \ -2 & 0 & 1 \ 3 & -1 & 0 array ] and B be a matrix such that B(I-A)=I+A. Then the sum of the diagonal elements of B ^ T B is equal to \_\_\_\_.

Accepted Answer

A is skew-symmetric (A^T = -A). From B(I-A) = I+A: B = (I+A)(I-A)^ -1 . B^T = ((I-A)^T)^ -1 (I+A)^T = (I+A)^ -1 (I-A). Since A is skew-symmetric, (I-A) and (I+A) commute: (I-A)(I+A) = I - A^2 = (I+A)(I-A). B^TB = (I+A)^ -1 (I-A)(I+A)(I-A)^ -1 = (I+A)^ -1 (I+A)(I-A)(I-A)^ -1 = I. Trace of B^TB = tr (I) = 3.

Question 17

Let | A |=6, where A is a 3 3 matrix. If | adj (3 adj ( ~A ^ 2 adj (2 ~A ) ) ) |=2^ m 3^ n , m , n N , then m + n is equal to \_\_\_\_.

Accepted Answer

For 3 3 matrix: | adj (M)| = |M|^2, |kM| = k^3|M|. |2A| = 8 6 = 48, | adj (2A)| = 48^2 = 2^8 3^2. |A^2 adj (2A)| = |A|^2 | adj (2A)| = 2^2 3^2 2^8 3^2 = 2^ 10 3^4. Let B = A^2 adj (2A). | adj (B)| = |B|^2 = 2^ 20 3^8. |3\, adj (B)| = 3^3 2^ 20 3^8 = 2^ 20 3^ 11 . | adj (3\, adj (B))| = (2^ 20 3^ 11 )^2 = 2^ 40 3^ 22 . m = 40, n = 22 m + n = 62.

Question 18

If X= [ array l x \ y \ z array ] is a solution of the system of equations A X=B, where adj A= [ array ccc 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 array ] and B = [ array l 4 \ 0 \ 2 array ], then |

Accepted Answer

X = 1 |A| adj (A) B. | adj (A)| = 4(10)-2(-20)+2(10) = 100, so |A|^2=100 |A|= 10. adj (A) B = pmatrix 16+0+4\-20+0+10\4+0+6 pmatrix = pmatrix 20\-10\10 pmatrix x+y+z = 20-10+10 |A| = 20 10 = 2 |x+y+z|=2

Question 19

If A = [ array ll 2 & 3 \ 3 & 5 array ], then the determinant of the matrix ( A ^ 2025 -3 ~A ^ 2024 + A ^ 2023 ) is

Accepted Answer

A = bmatrix 2 & 3 \ 3 & 5 bmatrix A^2 = bmatrix 13 & 21 \ 21 & 34 bmatrix |A^ 2025 - 3A^ 2024 + A^ 2023 | = | A^ 2023 ( A^2 - 3A + I ) | = |A|^ 2023 |A^2 - 3A + I| = 1 vmatrix 8 & 12 \ 12 & 20 vmatrix = 160 - 144 = 16

Question 20

Let A be a 3 3 matrix such that A + A ^ T = O . If A [ array c 1 \ -1 \ 0 array ]= [ array l 3 \ 3 \ 2 array ], A ^ 2 [ array c 1 \ -1 \ 0 array ]= [ array c -3 \ 19 \ -24 array ] and det ( ad

Accepted Answer

A is skew-symmetric: A = bmatrix 0 & -c & b\ & 0 & -a\-b & a & 0 bmatrix . From A bmatrix 1\-1\0 bmatrix = bmatrix 3\3\2 bmatrix : c = 3, a+b = -2. From A^2 bmatrix 1\-1\0 bmatrix = bmatrix -3\19\-24 bmatrix : b = 3, a = -5. A+I = bmatrix 1&-3&3\3&1&5\-3&-5&1 bmatrix , (A+I) = 44. For 3 3 matrix: ( adj (M)) = ( M)^2. (2\, adj (A+I)) = 8 44^2 = 8 1936 = 15488 = 2^7 11^2. ( adj (2\, adj (A+I))) = 15488^2 = 2^ 14 11^4. = 14,\ = 0,\ = 4. + + = 18.

Matrices — Previous Year Questions & Solutions