If A = 2 1 − 1 2 , B 1 0 1 1 , C = A B A T and X = A T C 2 A , then det X is equal to:

If

A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{matrix}], B [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], C = A B A^{T}

and

X = A^{T} C^{2} A

, then det

X

is equal to:

$243$
$729$
$27$
$891$

Solution

Given: $A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{matrix}]$

$\Rightarrow |A| = 2 + 1 = 3$

$B = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}] \Rightarrow |B| = 1$

And $C = A B A^{T}$

$\Rightarrow |C| = |A B A^{T}|$

$\Rightarrow |C| = {|A|}^{2} |B| \{as |A| = |A^{T}|\}$

$\Rightarrow |C| = 9$

Now, solving $X = A C^{2} A^{T}$

. $\Rightarrow |X| = |A| |C^{2}| |A^{T}|$

$\Rightarrow |X| = 3 \times 9^{2} \times 3$

$\Rightarrow |X| = 729$

Asked in: JEE Main 2024 (01 Feb Shift 1)

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