Solve this

Question:

If $A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$, then $A^{4}=$_______

Solution:

The given matrix is $A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$.

Now,

$A^{2}=A \cdot A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]=\left[\begin{array}{lll}3 \times 3+0 \times 0+0 \times 0 & 3 \times 0+0 \times 3+0 \times 0 & 3 \times 0+0 \times 0+0 \times 3 \\ 0 \times 3+3 \times 0+0 \times 0 & 0 \times 0+3 \times 3+0 \times 0 & 0 \times 0+3 \times 0+0 \times 3 \\ 0 \times 3+0 \times 0+3 \times 0 & 0 \times 0+0 \times 3+3 \times 0 & 0 \times 0+0 \times 0+3 \times 3\end{array}\right]=\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]$

$\therefore A^{4}=A^{2} . A^{2}=\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]=\left[\begin{array}{cc}9 \times 9+0 \times 0+0 \times 0 & 9 \times 0+0 \times 9+0 \times 0 & 9 \times 0+0 \times 0+0 \times 9 \\ 0 \times 9+9 \times 0+0 \times 0 & 0 \times 0+9 \times 9+0 \times 0 & 0 \times 0+9 \times 0+0 \times 9 \\ 0 \times 9+0 \times 0+9 \times 0 & 0 \times 0+0 \times 9+9 \times 0 & 0 \times 0+0 \times 0+9 \times 9\end{array}\right]\left[\begin{array}{cc}81 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81\end{array}\right]$

If $A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$, then $A^{4}=\left[\begin{array}{ccc}81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81\end{array}\right]$

Leave a comment