Question

Solve the following system of equations by matrix method:
 5x + 2y = 3
 3x + 2y = 5

Answer 1

The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix} \binom{x}{y} = \binom{3}{5}\]
\[AX=B\]
Here,
\[A = \begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{3}{5}\]
Now, 
\[\left| A \right| = \begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix} \]
\[ = 10 - 6\]
\[ = 4 \neq 0\]
\[\text{ The given system has a unique solution given by} X = A^{- 1} B . \]
\[ {\text{Let }C}_{ij} {\text{ be the cofactors of the elements a}}_{ij}\text{ in }A=\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \left( 2 \right) = 2 , C_{12} = \left( - 1 \right)^{1 + 2} \left( 3 \right) = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \left( 2 \right) = - 2, C_{22} = \left( - 1 \right)^{2 + 2} \left( 5 \right) = 5\]
\[ \therefore adj A = \begin{bmatrix}2 & - 3 \\ - 2 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ = \frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}\binom{3}{5}\]
\[ = \frac{1}{4}\binom{6 - 10}{ - 9 + 25}\]
\[\binom{x}{y} = \binom{\frac{- 4}{4}}{\frac{16}{4}}\]
\[ \therefore x = - 1\text{ and }y = 4\]