Question

Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3

Answer 1

The given system of equations can be written as follows:
AX = B
Here,
\[ A = \begin{bmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{vmatrix}\]
\[ = 3\left( 0 - 5 \right) + 1\left( 0 + 3 \right) - 2(0 - 6)\]
\[ = - 15 + 3 + 12\]
\[ = 0\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 1 \\ - 5 & 0\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 1 \\ 3 & 0\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 3 & - 5\end{vmatrix} = - 6\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & - 2 \\ - 5 & 0\end{vmatrix} = 10, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 2 \\ 3 & 0\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 1 \\ 3 & - 5\end{vmatrix} = 12\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & - 2 \\ 2 & - 1\end{vmatrix} = 5, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 2 \\ 0 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 1 \\ 0 & 2\end{vmatrix} = 6\]
\[adj A = \begin{bmatrix}- 5 & - 3 & - 6 \\ 10 & 6 & 12 \\ 5 & 3 & 6\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}\begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18\end{bmatrix}\]
\[ = \begin{bmatrix}- 5 \\ - 3 \\ - 6\end{bmatrix} \neq 0\]
Hence, the given system of equations is consistent.