# Solve the following equations

Question:

If $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ such that $A^{5}=\lambda A$, then $\lambda=$

Solution:

The given matrix is $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$.

$\therefore A^{2}=A \cdot A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}4+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+4+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+4\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$

$A^{4}=A^{2} . A^{2}=\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]=\left[\begin{array}{ccc}16+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+16\end{array}\right]=\left[\begin{array}{ccc}16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16\end{array}\right]$

$A^{5}=A^{4} . A=\left[\begin{array}{ccc}16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]=\left[\begin{array}{ccc}32+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+32+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+32\end{array}\right]=\left[\begin{array}{ccc}32 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 32\end{array}\right]$

Now,

$A^{5}=\left[\begin{array}{ccc}32 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 32\end{array}\right]=16\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

$\Rightarrow A^{5}=16 A$

It is given that, $A^{5}=\lambda A$.

$\therefore \lambda=16$

If $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ such that $A^{5}=\lambda A$, then $\lambda=$ 16