Question

Consider the matrices $A, \, B, \, C$ and $D$ with order $2\times 4, \, 4\times 3, \, 3\times 3$ and $3\times 2$ respectively. Let $M=\left(2 A B C^{2020} D\right)^{6}$ and $\left|M\right|=\lambda \left|A B C^{2020} D\right|^{6}$ , then $\lambda $ is equal to

• A. $1024$
• B. $4096$
• C. $64$
• D. $4$

Answer 1

Answer: (B)

$M=\left(2 A B C^{2020} D\right)^{6}$ (Given)
$\left|M\right|=\left|2 A B C^{2020} D\right|^{6}$
$\because ABC^{2020}D$ is a $2\times 2$ matrix
Hence, $\left|2 A B C^{2020} D\right|=2^{2}\left|A B C^{2020} D\right|$
Therefore,
$\left|M\right|=4^{6}\left|A B C^{2020} D\right|^{6}$
$\Rightarrow \lambda =4096$