Question

Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0

Answer 1

 The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix} \binom{x}{y} = \binom{5}{ - 3}\]
\[AX=B\]
Here,
\[A = \begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{5}{ - 3}\]
Now, 
\[\left| A \right| = \begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix} \]
\[ = - 3 - 4\]
\[ = - 7 \neq 0\]
\[\text{ So, the given system has a unique solution given by }X = A^{- 1} B . \]
\[ {\text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \left( - 1 \right) = - 1, C_{12} = \left( - 1 \right)^{1 + 2} \left( 1 \right) = - 1\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \left( 4 \right) = - 4, C_{22} = \left( - 1 \right)^{2 + 2} \left( 3 \right) = 3\]
\[adj A = \begin{bmatrix}- 1 & - 1 \\ - 4 & 3\end{bmatrix}^T = \begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ = \frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}\binom{5}{ - 3}\]
\[ = \frac{1}{- 7}\binom{ - 5 + 12}{ - 5 - 9}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{7}{- 7}}{\frac{- 14}{- 7}}\]
\[ \therefore x = - 1\text{ and }y = 2\]