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

Answer: $diag (-9 14 -18)$

Hint:

$diag= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1 \end{bmatrix}$  using this formula in $A, B, C$ and solve $B + C -2A.$

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

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

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

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

$=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0 & 0 & -4 \end{bmatrix}+\begin{bmatrix} -6 & 0 &0 \\ 0 & 3 &0 \\ 0 &0 & 4 \end{bmatrix}-\begin{bmatrix} 4 &0 & 0\\ 0 & -10 &0 \\ 0 & 0 & 18 \end{bmatrix}$

$=\begin{bmatrix} 1-6-4 &0+0-0 &0+0-0 \\ 0+0-0& 1+3+10 &0+0-0 \\ 0 +0-0& 0+0-0 & -4+4-18 \end{bmatrix}$

$=\begin{bmatrix} -9 &0 &0\\ 0& 14 &0 \\ 0 & 0 & -18 \end{bmatrix}$
$= diag (-9\: \ 14 -18)$