Question

Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = −1

Answer 1

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