Question

Use product \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]  to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

Answer 1

Suppose, A = \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\]

\[B = \begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]

\[A \times B = \begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]
\[ = \begin{bmatrix}- 2 - 9 + 12 & 0 - 2 + 2 & 1 + 3 - 4 \\ 0 + 18 - 18 & 0 + 4 - 3 & 0 - 6 + 6 \\ - 6 - 18 + 24 & 0 - 4 + 4 & 3 + 6 - 8\end{bmatrix}\]
\[ = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
Since, A × B = I,

\[\therefore\] B = A⁻¹                    .....(1)
Now, the given system of equations is
x + 3z = 9
−x + 2y − 2z = 4
2x − 3y + 4z = −3
This can also be represented as,

\[\begin{bmatrix}1 & 0 & 3 \\ - 1 & 2 & - 2 \\ 2 & - 3 & 4\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]

Here, we can observe that \[\begin{bmatrix}1 & 0 & 3 \\ - 1 & 2 & - 2 \\ 2 & - 3 & 4\end{bmatrix} = A^T\]
So,  \[A^T \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
Multiply the above expression by \[\left( A^T \right)^{- 1}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \left( A^T \right)^{- 1} \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \left( A^{- 1} \right)^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix} \]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = B^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix} \left[\text{ Using }(1) \right]\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & - 3 & - 2\end{bmatrix}\begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 18 + 36 - 18 \\ 0 + 8 - 3 \\ 9 - 12 + 6\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 5 \\ 3\end{bmatrix}\]

Hence, x = 0, y = 5 and z = 3.