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UPSC NDA II 2023 Mathematics Solved Paper

Question : 17 of 120

Marks: +1, -0

A=\begin{bmatrix} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{bmatrix}

, then what is the value of det[adj(adjA)] ?

Solution:

Concept:

Properties of adjugate (adjoint) and determinant of a square matrix.

For an

n \times n

matrix

A

with

\det A \neq 0

A(\operatorname{adj} A) = (\det A) I

and

\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2} A

Also,

\det(kA) = k^n \det A

Explanation:

First, compute

\det A

for

A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{bmatrix}

Expanding along the third column:

\det A = 0 \cdot (\cdots) + 0 \cdot (\cdots) + 1 \cdot \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} = 1 \cdot (6 - 1) = 5

For a

3 \times 3

matrix (

n=3

\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A = |A|^{1} A

Take determinant on both sides:

|\operatorname{adj}(\operatorname{adj} A)| = \det( |A| A ) = (|A|)^n \cdot |A| = |A|^{n+1}

Wait: Using the general formula

\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A

, we get

|\operatorname{adj}(\operatorname{adj} A)| = \left( |A|^{n-2} \right)^n \cdot |A| = |A|^{n^2 - 2n + 1}

For

n=3

, exponent

= 9 - 6 + 1 = 4

Thus

|\operatorname{adj}(\operatorname{adj} A)| = |A|^4 = 5^4 = 625

Answer:

625

(Option D).

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