Question

Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0

Answer 1

 The given system of equations can be written in matrix form as folllows:
\[\begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix} \binom{x}{y} = \binom{ - 2}{ - 3}\]
\[AX=B\]
Here,
\[A = \begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix}, X = \binom{x}{y}\text{ and }B = \binom{ - 2}{ - 3}\]
Now,
\[\left| A \right| = \begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix} \]
\[ = 30 - 28\]
\[ = 2 \neq 0 \]
\[\text{ The given system has a unique solution given by }X = A^{- 1} B . \]
\[{ \text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \left( 6 \right) = 6 , C_{12} = \left( - 1 \right)^{1 + 2} \left( 4 \right) = - 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \left( 7 \right) = - 7, C_{22} = \left( - 1 \right)^{2 + 2} \left( 5 \right)\]
\[ = 5\]
\[adj A = \begin{bmatrix}6 & - 4 \\ - 7 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ \Rightarrow A^{- 1} = \frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ = \frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}\binom{ - 2}{ - 3}\]
\[ = \frac{1}{2}\binom{ - 12 + 21}{8 - 15}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{9}{2}}{\frac{- 7}{2}}\]
\[ \therefore x = \frac{9}{2}\text{ and } y = \frac{- 7}{2}\]