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#### Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 34 math

$A^{2}-5 A-7 I=0, A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$

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

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

Prove:$A^{2}-5 A+7 I=0$

Solution: I is identity matrix so

$7 I=7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$

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

$A^{2}=A A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] ...(i)$

Now, we will find the matrix for  5A, we get

$5 A=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ 5 A=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right] ... (ii)$

So, substituting corresponding values from equation  i &  ii in
$A^{2}-5 A+7 I$

we get

$\begin{array}{l} =\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \\\\ =\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{array}$

$\therefore A^{2}-5 A+7 I=0$

Hence, proved.

We will find $A^4$

$A^{2}-5 A+7 I=0$

Multiply both sides by $A^2$, we get

$A^{2}\left(A^{2}-5 A+7 I\right)=A^{2}(0) \\\\ A^{4}-5 A^{2} A+7 I A^{2}=0 \\\\ A^{4}=5 A^{2} A-7 I A^{2}\\\\ A^{4}=5 A^{2} A-7 A^{2}$

Now we will substitute the corresponding values we get

$A^{4}=5\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\ \\ A^{4}=5\left[\begin{array}{cc}24-5 & 8+10 \\ -15-3 & -5+6\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\\\\\ A^{4}=5\left[\begin{array}{cc}19 & 18 \\ -18 & 1\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\\ A^{4}=\left[\begin{array}{cc}5 \times 19 & 5 \times 18 \\ -18 \times 5 & 1 \times 5\end{array}\right]-\left[\begin{array}{cc}7 \times 8 & 7 \times 5 \\ -5 \times 7 & 3 \times 7\end{array}\right] \\\\\\ A^{4}=\left[\begin{array}{cc}95 & 90 \\ -90 & 5\end{array}\right]-\left[\begin{array}{cc}56 & 35 \\ -35 & 21\end{array}\right]\\\\\\ A^{4}=\left[\begin{array}{cc}95-56 & 90-35 \\ -90+35 & 5-21\end{array}\right] \\\\\\A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$