Question

Let $a_{r}=r$ $^{4}C_{r}$ , $b_{r}=\left(4 - r\right)\left( \, \right)^{4}C_{r}$ , $A_{r}=\begin{bmatrix} a_{r} & 2 \\ 3 & b_{r} \end{bmatrix}$ and $ \, A=\displaystyle \sum _{r = 0}^{4}A_{r}$ , then the value of $\left|A\right|$ is equal to

Answer 1

$A=\displaystyle \sum _{r = 0}^{4}A_{r}=\begin{bmatrix} \displaystyle \sum _{r = 0}^{4}r^{4}C_{r} & \displaystyle \sum _{r = 0}^{4}2 \\ \displaystyle \sum _{r = 0}^{4}3 & \displaystyle \sum _{r = 0}^{4}\left(4 - r\right)^{4}C_{r} \end{bmatrix}$
Now, $\displaystyle \sum _{r = 0}^{4}r^{4}C_{r}=\displaystyle \sum _{r = 0}^{4}\left(4 - r\right)^{4}C_{r}=4\times 2^{3}=2^{5}$
$A=\begin{bmatrix} 32 & 10 \\ 15 & 32 \end{bmatrix}$
$\left|A\right|=1024-150=874$