Question

Solve the following system of equations by matrix method:
 x + y + z = 6
x + 2z = 7
3x + y + z = 12

Answer 1

Here, 
\[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{vmatrix}\]
\[ = 1\left( 0 - 2 \right) - 1\left( 1 - 6 \right) + 1(1 - 0)\]
\[ = - 2 + 5 + 1\]
\[ = 4\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 2 \\ 1 & 1\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 0 \\ 3 & 1\end{vmatrix} = 1\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = - 2, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 & 2\end{vmatrix} = 2, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = - 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix} = - 1\]
\[adj A = \begin{bmatrix}- 2 & 5 & 1 \\ 0 & - 2 & 2 \\ 2 & - 1 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}\begin{bmatrix}6 \\ 7 \\ 12\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 - 14 - 12\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}12 \\ 4 \\ - 20\end{bmatrix}\]
\[ \Rightarrow x = \frac{12}{4}, y = \frac{4}{4}\text{ and }z = \frac{- 20}{4}\]
\[ \therefore x = 3, y = 1\text{ and }z = - 5\]