# if

Question:

If A = diag (2 − 59), B = diag (11 − 4) and C = diag (−6 3 4), find
(i) A − 2B
(ii) B + C − 2A
(iii) 2A + 3B − 5C

Solution:

Here,

$A=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9\end{array}\right], B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{array}\right]$ and $C=\left[\begin{array}{ccc}-6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$

(i)

$A-2 B$

$\Rightarrow A-2 B=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9\end{array}\right]-2\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{array}\right]$

$\Rightarrow A-2 B=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9\end{array}\right]-\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -8\end{array}\right]$

$\Rightarrow A-2 B=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & 17\end{array}\right]=\operatorname{diag}(0-717)$

(ii)

$B+C-2 A$

$\Rightarrow B+C-2 A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{array}\right]+\left[\begin{array}{ccc}-6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]-2\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9\end{array}\right]$

$\Rightarrow B+C-2 A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{array}\right]+\left[\begin{array}{ccc}-6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]-\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 18\end{array}\right]$

$\Rightarrow B+C-2 A=\left[\begin{array}{ccc}1-6-4 & 0+0-0 & 0+0-0 \\ 0+0-0 & 1+3+10 & 0+0-0 \\ 0+0-0 & 0+0-0 & -4+4-18\end{array}\right]$

$\Rightarrow B+C-2 A=\left[\begin{array}{ccc}-9 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & -18\end{array}\right]=\operatorname{diag}(-914-18)$

(iii)

$2 A+3 B-5 C$

$\Rightarrow 2 A+3 B-5 C=2\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9\end{array}\right]+3\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{array}\right]-5\left[\begin{array}{ccc}-6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$

$\Rightarrow 2 A+3 B-5 C=\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 18\end{array}\right]+\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -12\end{array}\right]-\left[\begin{array}{ccc}-30 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 20\end{array}\right]$

$\Rightarrow 2 A+3 B-5 C=\left[\begin{array}{ccc}4+3+30 & 0+0-0 & 0+0-0 \\ 0+0-0 & -10+3-15 & 0+0-0 \\ 0+0-0 & 0+0-0 & 18-12-20\end{array}\right]$

$\Rightarrow 2 A+3 B-5 C=\left[\begin{array}{ccc}37 & 0 & 0 \\ 0 & -22 & 0 \\ 0 & 0 & -14\end{array}\right]=\operatorname{diag}(37-22-14)$