Matrices Question 1
Question 1 - 2024 (01 Feb Shift 1)
If $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ and $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^{2} \mathrm{~A}$, then $\operatorname{det} \mathrm{X}$ 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 $\mathrm{C}=\mathrm{ABA}^{\mathrm{T}} \Rightarrow \operatorname{det}(\mathrm{C})=(\operatorname{dct}(\mathrm{A}))^{2} \mathrm{x} \operatorname{det}(\mathrm{B})$
$|C|=9$
Now $|\mathrm{X}|=\left|\mathrm{A}^{\mathrm{T}} \mathrm{C}^{2} \mathrm{~A}\right|$
$=\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{C}|^{2}|\mathrm{~A}|$
$=|\mathrm{A}|^{2}|\mathrm{C}|^{2}$
$=9 \times 81$
$=729$
