3 and 4 .Determinants and Matrices
hard

यदि $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\top}$ तथा $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$ हैं, तो $\operatorname{det} \mathrm{X}$ बराबर है :

A

$243$

B

$729$

C

$27$

D

$891$

(JEE MAIN-2024)

Solution

$\begin{aligned} & A=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right] \Rightarrow \operatorname{det}(A)=3 \\ & B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right] \Rightarrow \operatorname{det}(B)=1\end{aligned}$

Now $C=A B A^T \Rightarrow \operatorname{det}(C)=(\operatorname{dct}(A))^2 x \operatorname{det}(B)$

$|C|=9$

$\text { Now }|X|=\left|A^T C^2 A\right|$

$=\left|A^T\right||C|^2|A|$

$=|A|^2|C|^2$

$=9 \times 81$

$=729$

Standard 12
Mathematics

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