#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 7

$\begin{bmatrix} 4 &-20 \\ 38& -10 \end{bmatrix}$

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: $A= \begin{bmatrix} 2 &-1 \\ 3 &2 \end{bmatrix},B = \begin{bmatrix} 0 &4 \\ -1& 7 \end{bmatrix}$

Consider,

$A^2=AA$

$A^{2}=\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 \times 2+(-1) \times 3 & 2 \times(-1)+(-1) \times 2 \\ 3 \times 2+2 \times 3 & 3 \times(-1)+2 \times 2 \end{array}\right]$

$=\left[\begin{array}{ll} 4-3 & -2-2 \\ 6+6 & -3+4 \end{array}\right]=\left[\begin{array}{cc} 1 & -4 \\ 12 & 1 \end{array}\right]$

Now, we have to find $3A^2-2B+I$

Where$I =\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$ identity matrix

$3 A^{2}-2 B+I=3\left[\begin{array}{cc} 1 & -4 \\ 12 & 1 \end{array}\right]-2\left[\begin{array}{cc} 0 & 4 \\ -1 & 7 \end{array}\right]+\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]$

\begin{aligned} &=\left[\begin{array}{cc} 3 & -12 \\ 36 & 3 \end{array}\right]-\left[\begin{array}{cc} 0 & 8 \\ -2 & 14 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{cc} 3-0+1 & -12-8+0 \\ 36+2+0 & 3-14+1 \end{array}\right] \end{aligned}

$=\begin{bmatrix} 4 &-20 \\ 38& -10 \end{bmatrix}$