#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 5 Subquestion (iii) Maths Textbook Solution.

Answer: $diag (37 -22 -14)$

Hint:  using this formula in$A, B, C$ and solve $2A + 3B - 5C.$

Given:$A = diag (2 -5 9), B = diag (1 1 -4) and\: C = diag (-6 3 4)$

Here, we have to compute:$2A + 3B - 5C.$

Solution:

$A=\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix},B=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix},C= \begin{bmatrix} -6 &0 & 0\\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}$

$2A+3B-5C=2\begin{bmatrix} 2 &0 &0 \\ 0& -5 &0 \\ 0 & 0 & 9 \end{bmatrix}+3\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &0 & -4 \end{bmatrix}-5 \begin{bmatrix} -6 &0 & 0\\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}$

$=\begin{bmatrix} 4 &0 &0 \\ 0& -10 &0 \\ 0 & 0 & 18 \end{bmatrix}+3\begin{bmatrix} 3 & 0 &0 \\ 0 & 3 &0 \\ 0 &0 & -12 \end{bmatrix}-5 \begin{bmatrix} -30 &0 & 0\\ 0 & 15 &0 \\ 0 & 0 & 20 \end{bmatrix}$

$=\begin{bmatrix} 4+3+30 &0+0-0 &0+0-0 \\ 0+0-0& -10+3-15 &0+0-0 \\ 0+0-0 & 0+0-0 & 18-12-20 \end{bmatrix}$

$=\begin{bmatrix} 37 &0 &0\\ 0& -22 &0\\ 0 & 0 & -14 \end{bmatrix}$

$= diag (37 -22 -14)$