# Solve this

Question:

If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then find $\lambda, \mu$ so that $A^{2}=\lambda A+\mu l$

Solution:

Given : $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]$

$A^{2}=\lambda A+\mu I$

$\Rightarrow\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\lambda\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]+\mu\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{cc}2 \lambda & 3 \lambda \\ \lambda & 2 \lambda\end{array}\right]+\left[\begin{array}{ll}\mu & 0 \\ 0 & \mu\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{cc}2 \lambda+\mu & 3 \lambda+0 \\ \lambda+0 & 2 \lambda+\mu\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]=\left[\begin{array}{cc}2 \lambda+\mu & 3 \lambda \\ \lambda & 2 \lambda+\mu\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\therefore 7=2 \lambda+\mu$                 $\ldots(1)$

$12=3 \lambda$

$\Rightarrow \lambda=\frac{12}{3}=4$

Putting the value of $\lambda$ in eq. (1), we get

$7=2(4)+\mu$

$\Rightarrow 7-8=\mu$

$\therefore \mu=-1$