#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 27 math

Answer: Hence prove, $A^{2}-A+2 I=0$

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: $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] and \ \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

Consider,

$A^{2}=A A \\\\ A^{2}=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2)\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}9-8 & -6+4 \\ 12-8 & -8+4\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]$

Now, put the value of $A^2$ in the given equation $A^2 - A + 2l$, we get

$=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]-\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]+2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ =\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]-\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]\\\\\\ =\left[\begin{array}{ll}1-3+2 & -2+2+0 \\ 4-4+0 & -4+2+2\end{array}\right]$

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

Hence, $A^{2}-A+2 I=0$