#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 46

Answer:  Hence, prove $A^{2}-7 A+10 I_{3}=0$

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:

$A =\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]$

Prove: $A^{2}-7 A+10 I_{3}=0$

$I_3$ is identity matrix of size 3

$I_{3}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\\\\\ \mathrm{LHS}=A^{2}-7 A+10 I_{3}\\\\ =\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]-7\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right]+10\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\$

$\begin {array} {ll} =\left[\begin{array}{ccc} 9+2+0 & 6+8+0 & 0+0+0 \\ 3+4+0 & 2+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+25 \end{array}\right]-\left[\begin{array}{ccc} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{array}\right]+\left[\begin{array}{ccc} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{array}\right]-\left[\begin{array}{ccc} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{array}\right]+\left[\begin{array}{ccc} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 11-21+10 & 14-14+0 & 0-0+0 \\ 7-7+0 & 18-28+10 & 0-0+0 \\ 0-0+0 & 0-0+0 & 25-35+10 \end{array}\right]\\\\\\ =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=0\\ \end{}$

Hence, $A^{2}-7 A+10 I_{3}=0$