If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ satisfies $A^{4}=\lambda A$, then write the value of $\lambda$.
$A^{2}=A \cdot A$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1+1 & 1+1 \\ 1+1 & 1+1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
Now.
$A^{4}=A^{2} A^{2}$
$\Rightarrow A^{4}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
$\Rightarrow A^{4}=\left[\begin{array}{ll}4+4 & 4+4 \\ 4+4 & 4+4\end{array}\right]$
$\Rightarrow A^{4}=\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]$
Also,
$A^{4}=\lambda A$
$\Rightarrow\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]=\lambda\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]=\left[\begin{array}{ll}\lambda & \lambda \\ \lambda & \lambda\end{array}\right]$
$\therefore \lambda=8$