Question

If A = diag (2 − 59), B = diag (11 − 4) and C = diag (−6 3 4), find

B + C − 2A

Answer 1

LaTeX

\[Here, \]
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} 

⇒ 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}
\[ \Rightarrow B + C - 2A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4\end{bmatrix} + \begin{bmatrix}- 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{bmatrix} - \begin{bmatrix}4 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & 0 & 18\end{bmatrix}\]
\[ \Rightarrow B + C - 2A = \begin{bmatrix}1 - 6 - 4 & 0 + 0 - 0 & 0 + 0 - 0 \\ 0 + 0 - 0 & 1 + 3 + 10 & 0 + 0 - 0 \\ 0 + 0 - 0 & 0 + 0 - 0 & - 4 + 4 - 18\end{bmatrix}\]
\[ \Rightarrow B + C - 2A = \begin{bmatrix}- 9 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & - 18\end{bmatrix} = diag\left( - 9 14 - 18 \right)\]