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#### Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 38 math

Answer:  $\lambda=4$ and  $\mu=-1$

Hint: Iis an identity matrix. Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and$I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

So,

$\mu I=\mu\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}\mu & 0 \\ 0 & \mu\end{array}\right]$

Now, we will find the matrix for $A^2$, we get

$A^{2}=A A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right] ....(1)$

Now, we will find the matrix for , $\lambda A$we get

$\lambda A=\lambda\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \\\\ \lambda A=\left[\begin{array}{ll}2 \lambda & 3 \lambda \\ 1 \lambda & 2 \lambda\end{array}\right] ....(ii)$

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

So, substituting corresponding values from equation i & ii we get

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

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal.
Hence,

$\lambda+0=4 \\\\ \lambda=4\\\\ 2 \lambda+\mu=7\\\\ 8+\mu=7\\\\ \mu=7-8=-1$

Therefore, the value of $\lambda = 4$ and $\mu = -1$