Question

Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12

Answer 1

The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} \binom{x}{y} = \binom{7}{12}\]
\[AX=B\]
Here,
\[A = \begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{7}{12}\]
Now,
\[\left| A \right| = \begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} \]
\[ = 9 - 5\]
\[ = 4 \neq 0\]
\[\text{ So, the given system has a unique solution given by }X = A^{- 1} B . \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \left( 3 \right) = 3, C_{12} = \left( - 1 \right)^{1 + 2} \left( 5 \right) = - 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \left( 1 \right) = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \left( 3 \right) = 3\]
\[adj A = \begin{bmatrix}3 & - 5 \\ - 1 & 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ = \frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}\binom{7}{12}\]
\[ = \frac{1}{4}\binom{21 - 12}{ - 35 + 36}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{9}{4}}{\frac{1}{4}}\]
\[ \therefore x = \frac{9}{4}\text{ and }y = \frac{1}{4}\]