Visual SolutionPYQ 2024 · Jan (1 Feb) Shift 1Standard

Question

A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix}

B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}

C = ABA^T

and

X = A^TC^2A

, then

\det X

is equal to:

(A)

243

(B)

729

(C)

27

(D)

891

Solution Path

|A|=3

|B|=1

|C|=|ABA^T|=9

|X|=|A^T|\cdot|C|^2\cdot|A| = 3 \times 81 \times 3 = 729

01Question Setup

1/4

Given

A, B

matrices,

C = ABA^T

X = A^TC^2A

. Find

\det(X)

Find

\det(X)

02Base Determinants

2/4

|A| = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3

|B| = 1

. Then

|C| = |A| \cdot |B| \cdot |A^T| = 3 \times 1 \times 3 = 9

|A| = 3, \; |B| = 1, \; |C| = 9

03Determinant Product RuleKEY INSIGHT

3/4

|X| = |A^T C^2 A| = |A^T| \cdot |C|^2 \cdot |A| = 3 \times 81 \times 3 = 729

|X| = 3 \times 9^2 \times 3 = 729

04Final Answer

4/4

The determinant product rule avoids multiplying the actual matrices.

\det(X) = \boxed{729}

- Answer (B)

Concepts from this question1 concepts unlocked

★

EASY

det(AB) = det(A) × det(B). Also det(Aᵀ) = det(A) and det(kA) = kⁿ det(A) for n×n matrix.

|AB| = |A| \cdot |B|, \quad |A^T| = |A|, \quad |kA| = k^n|A|

JEE loves chained matrix products like det(ABA^T). This rule avoids explicit multiplication entirely.

Product of matricesTranspose determinantScalar multiplication

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