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Question 1:

Let M be any 3 × 3 matrix with entries from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements of M^TM is seven, is

Solution:

Let

Then

Sum of the diagonal elements of M^TM = a² + b² + c² + d² + e² + f² + g² + h² + i² = 7

Case I∶ Seven (1’s) and two (0’s)

⁹C₂ = 36

Case II∶ One (2) and three (1’s) and five (0’s)

9!/5!3! = 504

The sum of diagonal elements = 36 + 504 = 540.

Question 2:

Let A and B be 3 × 3 real matrices such that A is a symmetric matrix and B is a skew-symmetric matrix. Then the system of linear equations (A²B² − B²A²)X = O, where X is a 3×1 column matrix of unknown variables, and O is a 3×1 null matrix, has:

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

Solution:

Since A is a symmetric matrix, A^T = A.

B is a skew-symmetric matrix, B^T = -B.

Let A²B² − B²A²= P

P^T = (A²B² – B²A²)^T

= (A²B²)^T – (B²A²)^T

= (B²)^T (A²)^T– (A²)^T (B²)^T

= B²A² – A²B²

So P is a skew-symmetric matrix.

So ay + bz = 0 …(1)

–ax + cz = 0 …(2)

–bx – cy = 0 …(3)

From equation (1), (2), (3)

Δ = 0 and Δ₁ = Δ₂= Δ₃ = 0

Therefore, the system of equations has an infinite number of solutions.

Hence, option (c) is the answer.

Question 3:

If for the matrix,

(a) 1

(b) 3

(c) 2

(d) 4

Solution:

AA^T = I₂

1 + α² = 1

α² = 0

α² + β² = 1

β² = 1

So, α⁴ = 0 and β⁴ = 1

Hence, α⁴ + β⁴ = 1

Hence, option (a) is the answer.

Question 4:

Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A² is 1, then the possible number of such matrices is

(a) 6

(b) 1

(c) 4

(d) 12

Solution:

Let

a² + 2b² + c² = 1

a = 1, b = 0, c = 0

a = 0, b = 0, c = 1

a = –1, b = 0, c = 0

c = –1, b = 0, a = 0

The possible number of matrices = 4.

Hence, option (c) is the answer.

Question 5:

The value of

(a) -2

(b) (a + 1) (a + 2) (a + 3)

(c) 0

(d) (a + 2) (a + 3) (a + 4)

Solution:

Given matrix is

C₁ → C₁ – C₂

∆ =

C₂ → C₂ – C₃

R₂ → R₂ – R₁ and R₃ → R₃ – R₁

= 6 – 8

= -2

Hence, option (a) is the answer.

Question 6:

The total number of 3 × 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AA^T is 9, is equal to

Solution:

Tr (AA^T) = x² + y² + z² + a² + b² + c² + d² + e² + f² = 9

all⇒1 = 1

one 3, rest = 0 ⇒ (9! / 8!) = 9

two 2 , one 1 and rest 0 ⇒ (9!) / (2! 6!) = 63 × 4 = 252

one 2 , five 1, rest 0 ⇒ (9!) / (5! 3!) = 63 × 8 = 504

Total number of 3 × 3 matrices = 766

Question 7:

Let

Solution:

Given that A = XB

b₁ – b₂ = √3a₁

3a₁² = b₁² + b₂² – 2b₁b₂ ..(i)

b₁ + kb₂ = √3a₂

3a₂² = b₁² + k²b₂² + 2kb₁b₂ ..(ii)

Adding both equations, 3 (a₁² + a₂²) = 2b₁² + (k² + 1) b₂² + 2b₁b₂ (k – 1)

(k² – 1) b₂² + 2b₁b₂ (k – 1) = 0

(k – 1) [(k + 1) b₂² + 2b₁b₂] = 0

k = 1

Question 8:

Let A be a 2 × 2 real matrix with entries from {0,1} and A ≠ 0. Consider the following two statements:

(P) If A ≠I₂ , then |A| = -1

(Q) If |A| =1 , then tr(A) = 2,

where I₂ denotes 2 × 2 identity matrix, and tr(A) denotes the sum of the diagonal entries of A. Then:

(a) Both (P) and (Q) are false

(b) (P) is true, and (Q) is false

(c) Both (P) and (Q) are true

(d) (P) is false, and (Q) is true

Solution:

P:

And A ≠0 and A = 1

(P) is false.

Q:

then Tr(A) = 2

(Q) is true.

Hence, option (d) is the answer.

Question 9:

Let

(a) 0

(b) 1

(c) 3

(d) 2

Solution:

Given that

tr(A + 2B) ≡ tr(A) + 2 tr(B) = –1 …..(1)

And tr (2A – B) ≡ 2tr (A) – tr (B) = 3 …..(2)

On solving (1) and (2), we get

tr(A) = 1

tr(B) = -1

tr(A) – tr(B) = 1 + 1

= 2

Hence, option (d) is the answer.

Question 10:

If element of matrix A = [a_ij]_(3×3) where

(a) 72

(b) 36

(c) 108

(d) 48

Solution:

|A| = 2(4-1)+1(-2+1)+1(1-2)

= 2(3)+1(-1)+1(-1)

= 4

|3 adj(2A^-1)| = 3³|adj(2A^-1)|

= 3³×|2A^-1|²

= 3³×2⁶×|A⁻¹|²

= 3³×2⁶×1/|A|²

= 27 × 64/16

= 108

Hence, option (c) is the answer.

Question 11:

Let α be a root of the equation x² + x+ 1= 0 and the matrix

(a) A

(b) A²

(c) A³

(d) I₃

Solution:

Given x² + x+ 1 = 0

The roots of equation x² + x+ 1 = 0 are complex cube roots of unity.

= ω or ω²

Substituting α = ω, we get;

A⁴ = I

A²⁸ = I

Therefore, we get

A³¹ = A²⁸ A³

= IA³

= A³

Hence, option (c) is the answer.

Question 12:

If

(a) 6I − A

(b) A − 6I

(c) 4I − A

(d) A − 4I

Solution:

Given that

Then,

10A^-1= A − 6I

Hence, option (b) is the answer.

Question 13:

If the matrices

(a) 16

(b) 2

(c) 8

(d) 72

Solution:

= 13 + 1 – 8

= 6

B = adj(A)

ǀadj Bǀ = ǀadj (adj A)ǀ = ǀAǀ⁴ = 6⁴

ǀCǀ = ǀ3Aǀ = 3³ǀAǀ = 3³ × 6

= 8

Hence, option (c) is the answer.

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