#### Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 15 math textbook solution.

Answer: $\left[\begin{array}{ccc}2 & 9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34\end{array}\right]$

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: $A=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right], B=\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]$

Find:$A^2 = B^2$

Consider, $A^2$

$A^2 = AA$

$A^{2}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]$

$A^{2}=\left[\begin{array}{ccc}(-1)(-1)+(1)(3)+(-1)(5) & (-1)(1)+(1)(-3)+(-1)(5) & (-1)(-1)+1(3)+(-1) 5 \\ 3(-1)+(-3)(3)+3(5) & 3(1)+(-3)(-3)+(3)(5) & (3)(-1)+(-3)(3)+3(5) \\ 5(-1)+5(3)+5(5) & 5(1)+(-3)(5)+5(5) & 5(-1)+5(3)+5(5)\end{array}\right] \\\\ \\\ A^{2}=\left[\begin{array}{ccc}1+3-5 & -1-3-5 & 1+3-5 \\ -3-9+15 & 3+9+15 & -3-9+15 \\ -5+15+25 & 5-15+25 & -5+15+25\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{ccc}-1 & 9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35\end{array}\right] .... ( i)$

Now, again consider,$B^{2}$

$B^{2}=B B \\\\ B^{2}=\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right] \\\\\\ B^{2}=\left[\begin{array}{ccc}(0)(0)+(4)(1)+3(-1) & 0(4)+4(-3)+3(4) & 0(3)+4(-3)+3(4) \\ 1(0)+(-3)(1)+(-3)(-1) & 1(4)+(-3)(-3)+(-3)(4) & (1)(3)+(-3)(-3)+(-3)(4) \\ (-1)(0)+4(1)+4(-1) & (-1)(4)+4(-3)+4(4) & (-1)(3)+4(-3)+4(4)\end{array}\right] \\\\ \\ B^{2}=\left[\begin{array}{ccc}0+4-3 & 0-12+12 & 0-12+12 \\ 0-3+3 & 4+9-12 & 3+9-12 \\ 0+4-4 & -4-12+16 & -3-12+16\end{array}\right] \\\\ B^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]-----(ii)$

Now by subtracting equation (ii) from (i) we, get

$\begin{array}{l} A^{2}-B^{2}=\left[\begin{array}{ccc} -1 & 9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35 \end{array}\right]-\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ A^{2}-B^{2}=\left[\begin{array}{ccc} -1-1 & 9-0 & -1-0 \\ 3-0 & 27-1 & 3-0 \\ 35-0 & 15-0 & 35-1 \end{array}\right] \\\\ A^{2}-B^{2}=\left[\begin{array}{ccc} -2 & 9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34 \end{array}\right] \end{array}$