JEE Advanced Maths Matrices Previous Year Questions with Solutions

Matrices JEE Advanced previous year questions with solutions are given on this page. These are prepared by our subject experts. Matrix is a set of numbers arranged in rows and columns so as to form a rectangular array. Students can expect questions from this topic. Aspirants are recommended to revise these solutions so that they can understand the difficulty level and pattern of the JEE Advanced exam.

Question 1:

(a) 52

(b) 103

(c) 201

(d) 205

Solution:

Given,

= I + A

Where

And

So Aⁿ = 0 for all n greater than or equal to 3.

Now P⁵⁰ = (I+A)⁵⁰ = ⁵⁰C₀ I⁵⁰ + ⁵⁰C₁ I⁴⁹ + ⁵⁰C₂ I⁴⁸ A² + 0

= I + 50A + 25 × 49 A²

⇒ q₂₁ = 50 × 4 = 200

q₃₁ = 50 × 16 + 25 × 49 × 16

= 20400

q₃₂ = 50 × 4 = 200

So (q₃₁+q₃₂)/q₂₁ = 20600/200 = 103

Hence option b is the answer.

Question 2: If A is a square matrix, then adj A^T– (adj A)^T is equal to

(a) 2|A|

(b) 2|A| I

(c) null matrix

(d) unit matrix

Solution:

We know A^-1 = adj A/det A

(A^T)^-1 = adj A^T/det A^T

⇒ adj A^T = det A^T.(A^T)^-1

⇒ adj A^T = det A.(A^-1)^T (since det A^T = det A, and (A^T)^-1 = (A^-1)^T

⇒ adj A^T = det A.(adj A/det A)^T

⇒ adj A^T = (det A/det A)(adj A)^T

⇒ adj A^T = (adj A)^T

⇒ adj A^T – (adj A)^T = 0

⇒ null matrix

Hence option c is the answer.

Question 3:

(a) A² + I = A(A² – I)

(b) A⁴ – I = A² + I

(c) A³ + I = A(A³ – I)

(d) A³ – I = A(A – I)

Solution:

Given

A²+I = A³-A

-I+I = A³-A

A³ ≠ A

So option a is the wrong statement.

Hence option a is the answer.

Question 4: If A is a 3 × 3 non-singular matrix such that AA^T = A^TA and B = A^-1A^T, then BB^T equals

(a) I+B

(b) I

(c) B^-1

(d) (B^-1)^T

Solution:

Given that AA^T = A^TA and B = A^-1A^T

BB^T = (A^-1A^T) (A^-1A^T)^T

= A^-1A^T A(A^-1)^T (since (A^T)^T = A)

= A^-1A A^T(A^T)^-1

= I.I

= I

Hence option b is the answer.

Question 5:

Solution:

Given,

Question 6: Let ω be a complex cube root of unity with ω ≠ 1 and P = [p_ij] be an n×n matrix with p_ij = ω^i+j. Then p² ≠ 0, when n =

(a) 55

(b) 56

(c) 57

(d) 58

Solution:

For n = 3,

It shows P² = 0 when n is a multiple of 3.

So for P² ≠ 0, n should not be a multiple of 3. So n can take the values 55, 56, and 58.

Hence, option a, b and d are correct.

Question 7:

(a) A – 4I

(b) 6I – A

(c) A – 6I

(d) 4I – A

Solution:

Given,

Characteristic equation of matrix A is |A – λI| = 0

⇒ λ² – 6λ – 10 = 0

So A² – 6A – 10I = 0

Multiply by A^-1

⇒ A^-1(A²) – 6A A^-1– 10 I A^-1 = 0

⇒ A^-1A A – 6A A^-1– 10 I A^-1 = 0

⇒ 10A^-1 = A – 6I

Hence, option c is the answer.

Question 8: Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible if

(a) The first column of M is the transpose of the second row of M

(b) The second row of M is the transpose of the first column of M

(c) M is a diagonal matrix with non-zero entries in the main diagonal

(d) The product of entries in the main diagonal of M is not the square of an integer

Solution:

If a matrix is invertible, determinant must not be equal to zero.

⇒ The matrix should be non-singular.

Let

Det M = ac – b² ≠ 0

⇒ ac ≠ b²

So M is a diagonal matrix with non-zero entries in the main diagonal.

The product of entries in the main diagonal of M is not the square of an integer.

Hence, options c and d are correct.

Question 9: Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

(a) Y³Z⁴ – Z⁴Y³

(b) X⁴Z³ – Z³X⁴

(c) X²³ + Y²³

(d) X⁴⁴ + Y⁴⁴

Solution:

We use the properties (A + B)^T = A^T + B^T

(AB)^T = B^TA^T

X^T = -X

Y^T = -Y

Z^T = Z

(Y³Z⁴ – Z⁴Y³)^T = (Z⁴)^T(Y³)^T – (Y³)^T(Z⁴)^T

= (Z^T)⁴(Y^T)³ – (Y^T)³(Z^T)⁴

= -Z⁴Y³ + Y³Z⁴

= Y³Z⁴ – Z⁴Y³

So (Y³Z⁴ – Z⁴Y³) is a symmetric matrix.

Similarly, X⁴⁴ + Y⁴⁴ is a symmetric matrix.

(X⁴Z³ – Z³X⁴)^T = (X⁴Z³)^T – (Z³X⁴)^T

= (Z^T)³(X^T)⁴ – (X^T)⁴(Z^T)³

= Z³X⁴ – X⁴Z³

= -(X⁴Z³ – Z³X⁴)

⇒ skew-symmetric.

(X²³ + Y²³)^T = (X^T)²³ + (Y^T)²³

= -X²³ – Y²³

= -(X²³ + Y²³)

⇒ skew-symmetric.

X⁴Z³ – Z³X⁴ and X²³ + Y²³ are skew-symmetric matrices.

Hence, options b and c are the answer.

Question 10: How many 3×3 matrices M with entries from {0, 1, 2} are there for which the sum of the diagonal entries of M^TM is 5?

(a) 198

(b) 126

(c) 162

(d) 135

Solution:

Given that the sum of the diagonal elements in M^TM = 5

⇒ (a₁² + a₄² + a₇²) + (a₂² + a₅² + a₈²) + (a₃² + a₆² + a₉²) = 5, which is possible when

Case 1: 5 a_i’s are 1 and 4 a_i’s are zero, which can be done in ⁹C₄ ways = 126

Case 2: 1 a_i is 1, 1 a_i is 2 and the rest 7a_is are zero.

It can be done in ⁹C₁ × ⁸C₁ = 9×8 = 72 ways

So total number of ways = 126 + 72

= 198

Hence option a is the answer.

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