#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 43

$f(A)=\left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]$

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:

$A=\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\ \ \ and \ \ f(x)=x^{3}+4 x^{2}-x$

Substitute x=A

Then, $f(A)=A^{3}+4 A^{2}-A ... ( i)$

Now we will find the matrix $A^ 2$

$\begin {array}{ll}A^{2}=\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{lll}0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9+0 & 4+0+0 \\ 0-2+0 & 1+3+0 & 2+0+0\end{array}\right] \\\\ \\A ^{2}=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right] \end{}$

Now we will find matrix $A^ 3$

$\begin{array} {ll} A^{3}=A^{2} A=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{3}=\left[\begin{array}{ccc}0-10+0 & 4+15+0 & 8+0+0 \\ 0+22+4 & -6-33-4 & -12+0+0 \\ 0+8+2 & -2-12-2 & -4+0+0\end{array}\right] \\\\ \\A^{3}=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right] \end{array}$

Put the value of A, $A^2$, $A^3$ in equation i

$f(A)=A^{3}+4 A^{2}-A\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+4\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+\left[\begin{array}{ccc}16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10+16-0 & 19-20-1 & 8+0-2 \\ 26-24-2 & -43+44+3 & -12+16-0 \\ 10-8-1 & -16+16+1 & -4+8-0\end{array}\right] \\\\\\ = \left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]$