Question

If $A$ is a non-null diagonal matrix of order $3$ such that $A^{4}=A^{2},$ then the possible number of matrices $A$ are

• A. $27$
• B. $26$
• C. $8$
• D. $7$

Answer 1

Answer: (B)

Let, $A=\begin{bmatrix} d_{1} & 0 & 0 \\ 0 & d_{2} & 0 \\ 0 & 0 & d_{3} \end{bmatrix}$
$A^{2}=\begin{bmatrix} d_{1}^{2} & 0 & 0 \\ 0 & d_{2}^{2} & 0 \\ 0 & 0 & d_{3}^{2} \end{bmatrix}\&A^{4}=\begin{bmatrix} d_{1}^{4} & 0 & 0 \\ 0 & d_{2}^{4} & 0 \\ 0 & 0 & d_{3}^{4} \end{bmatrix}$
$\because A^{4}=A^{2}\Rightarrow d_{1}^{4}=d_{1}^{2}\Rightarrow d_{1}=0,1,-1$
$d_{2}^{4}=d_{2}^{2}\Rightarrow d_{2}=0,1,-1$
$d_{3}^{4}=d_{3}^{2}\Rightarrow d_{3}=0,1,-1$
$\because $ $A$ can't be a null matrix so the total number of possible matrices is $=3^{3}-1=26$