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

Answer 1

$$\begin{gathered} \mathrm{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]\mathrm{and}C=\left[\begin{array}{ccc}-6& 0& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right] \\ \left(\mathrm{i}\right) \\ A-2B \\ \Rightarrow A-2B=\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-2B=\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-2B=\left[\begin{array}{ccc}0& 0& 0\\ 0& -7& 0\\ 0& 0& 17\end{array}\right]=\mathrm{diag}\left(0-717\right) \\ \left(\mathrm{ii}\right) \\ B+C-2A \\ \Rightarrow B+C-2A=\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-2A=\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-2A=\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-2A=\left[\begin{array}{ccc}-9& 0& 0\\ 0& 14& 0\\ 0& 0& -18\end{array}\right]=\mathrm{diag}\left(-914-18\right) \\ \left(\mathrm{iii}\right) \\ 2A+3B-5C \\ \Rightarrow 2A+3B-5C=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 2A+3B-5C=\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 2A+3B-5C=\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 2A+3B-5C=\left[\begin{array}{ccc}37& 0& 0\\ 0& -22& 0\\ 0& 0& -14\end{array}\right]=\mathrm{diag}\left(37-22-14\right) \end{gathered}$$