Matrices Question 1
Question 1 - 2024 (01 Feb Shift 1)
If $A=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], C=ABA^{T}$ and $X=A^{T} C^{2} A$, then $\operatorname{det} X$ is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Show Answer
Answer (2)
Solution
$A=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right] \Rightarrow \operatorname{det}(A)=3$
$B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right] \Rightarrow \operatorname{det}(B)=1$
Now $C=ABA^{T} \Rightarrow \operatorname{det}(C)=(\operatorname{dct}(A))^{2} x \operatorname{det}(B)$
$|C|=9$
Now $|X|=\left|A^{T} C^{2} A\right|$
$=\left|A^{T}\right||C|^{2}|A|$
$=|A|^{2}|C|^{2}$
$=9 \times 81$
$=729$
