Question

Let $X=\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$ be a matrix. If $X^{6}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then the number of divisors of $\left(a + b + 2020 c + d\right)$ is equal to

• A. $7$
• B. $14$
• C. $21$
• D. $28$

Answer 1

Answer: (B)

$X^{2}=\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}=\begin{bmatrix} 4 & 5 \\ 0 & 9 \end{bmatrix}$
$X^{3}=X^{2}X=\begin{bmatrix} 4 & 5 \\ 0 & 9 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}=\begin{bmatrix} 8 & 19 \\ 0 & 27 \end{bmatrix}$
$X^{6} = X^{3} \cdot X^{3} = \begin{bmatrix} 8 & 19 \\ 0 & 27 \end{bmatrix} \begin{bmatrix} 8 & 19 \\ 0 & 27 \end{bmatrix} = \begin{bmatrix} 64 & 665 \\ 0 & 729 \end{bmatrix}$
$\Rightarrow a=64,b=665,c=0,d=729$
$\Rightarrow a+b+2020c+d=1458=2\times 3^{6}$
Number of divisors $=\left(1 + 1\right)\left(6 + 1\right)=14$