Visual SolutionPYQ 2024 · Jan 29 Shift 1Standard

Question

Let

A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}

and

|2A|^3 = 2^{21}

where

\alpha, \beta \in \mathbb{Z}

. Then a value of

\alpha

is:

(A)

3

(B)

5

(C)

17

(D)

9

Solution Path

|kA| = k^n|A|

gives

|2A| = 8(\alpha^2 - \beta^2) = 128

, so

\alpha^2 - \beta^2 = 16

. With

(\alpha,\beta) = (5,3)

\alpha = 5

01Question Setup

1/4

Given

A = \begin{bmatrix}1&0&0\\0&\alpha&\beta\\0&\beta&\alpha\end{bmatrix}

with

|2A|^3 = 2^{21}

. Find

\alpha

|2A|^3 = 2^{21}

02Determinant of Scalar Multiple

2/4

Key property:

|kA| = k^n|A|

for

n \times n

matrix. So

|2A| = 8|A|

. Also

|A| = \alpha^2 - \beta^2

|2A| = 8|A|

03Solve for AlphaKEY INSIGHT

3/4

|2A|^3 = 2^{21} \Rightarrow |2A| = 128

. Then

8|A| = 128

, so

|A| = 16

. Thus

\alpha^2 - \beta^2 = 16

, and

(5,3)

works.

\alpha^2 - \beta^2 = 16

04Final Answer

4/4

\alpha = 5

satisfies all conditions.

\boxed{5}

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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