Question

\[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\], find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

Answer 1

Here,
\[ A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[AB = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix} \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}7 + 4 + 0 & 2 - 2 + 0 & - 6 + 6 + 0 \\ 14 - 2 - 12 & 4 + 1 + 6 & - 12 - 3 + 15 \\ 0 + 4 - 4 & 0 - 2 + 2 & 0 + 6 + 5\end{bmatrix}\]
\[ = \begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}\]
\[ \Rightarrow AB = 11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[AB = 11 I_3 \]
\[ \Rightarrow \frac{1}{11}AB = I_3 \]
\[ \Rightarrow \left( \frac{1}{11}B \right)A = I_3 \]
\[ \Rightarrow A^{- 1} = \frac{1}{11}B\]
\[ \Rightarrow A^{- 1} = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[X = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 9 - 21 \\ - 40 + 16 + 35\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}44 \\ - 33 \\ 11\end{bmatrix}\]
\[ \therefore x = 4, y = - 3\text{ and }z = 1\]