Question

Solve the following system of equations by matrix method:
3x + y = 19
3x − y = 23

Answer 1

The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix} \binom{x}{y} = \binom{19}{23}\]
\[AX=B\]
Here, 
\[A = \begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{19}{23}\]
Now,
\[\left| A \right| = \begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix} \]
\[ = - 3 - 3\]
\[ = - 6 \neq 0\]
\[\text{ So, the given system has a unique solution given by }X = A^{- 1} B . \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \left( - 1 \right) = - 1, C_{12} = \left( - 1 \right)^{1 + 2} \left( 3 \right) = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \left( 1 \right) = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \left( 3 \right) = 3\]
\[adj A = \begin{bmatrix}- 1 & - 3 \\ - 1 & 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ = \frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}\binom{19}{23}\]
\[ = \frac{1}{- 6}\binom{ - 19 - 23}{ - 57 + 69}\]
\[ = \binom{x}{y}\]
\[ = \binom{\frac{- 42}{- 6}}{\frac{12}{- 6}}\]
\[ \therefore x = 7\text{ and }y = - 2\]