Question

Show that the following systems of linear equations is consistent and also find their solutions:
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1

Answer 1

 Here,
\[x - y + z = 3 . . . (1)\]
\[2x + y - z = 2 . . . (2)\]
\[ - x - 2y + 2z = 1 . . . (3)\]
or, AX = B
where, 
\[ A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}\]
\[\begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{vmatrix}\]
\[ = 1\left( 2 - 2 \right) + 1\left( 4 - 1 \right) + 1( - 4 + 1)\]
\[ = 0 + 3 - 3\]
\[ = 0\]
So, A is singular . Thus, the given system of equations is either inconsistent or it is consistent with 
\[\text{ infinitely many solutions because } \left( adj A \right)B \neq 0\text{ or }\left( adj A \right)B = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co-factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & - 1 \\ - 2 & 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & - 1 \\ - 1 & 2\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ - 1 & - 2\end{vmatrix} = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ - 2 & 2\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ - 1 & - 2\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 1\end{vmatrix} = 0, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = 3 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = 3\]
\[adj A = \begin{bmatrix}0 & - 3 & - 3 \\ 0 & 3 & 3 \\ 0 & 3 & 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ - 9 + 6 + 3 \\ - 9 + 6 + 3\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[\text{ If }\left| A \right|=0\text{ and }\left( adjA \right)B=0, \text{ then the system is consistent and has infinitely many solutions. }\]
Thus, AX=B has infinitely many solutions.
\[\text{ Substituting z=k in eq.}\left( 1 \right)\text{ and eq.}\left( 2 \right),\text{ we get }\]
\[x - y = 3 - k\text{ and }2x + y = 2 + k\]
\[\begin{bmatrix}1 & - 1 \\ 2 & 1\end{bmatrix}\binom{x}{y} = \binom{3 - k}{2 + k}\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix}\]
\[ = 1 + 2 = 3 \neq 0\]
\[adj A = \begin{vmatrix}1 & 2 \\ - 1 & 1\end{vmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{3}\begin{bmatrix}1 & 1 \\ - 2 & 1\end{bmatrix}\]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{3}\begin{bmatrix}1 & 1 \\ - 2 & 1\end{bmatrix}\binom{3 - k}{2 + k}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{3}\binom{3 - k + 2 + k}{ - 6 + 2k + 2 + k}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{5}{3}}{\frac{3k - 4}{3}}\]
\[ \therefore x = \frac{5}{3}, y = \frac{3k - 4}{3}\text{ and }z = k\]
These values of x, y and z also satisfy the third equation .
\[\text{ Thus, }x = \frac{5}{3}, y = \frac{3k - 4}{3}\text{ and }z = k \left(\text{ where k is a real number} \right)\text{ satisfy the given system of equations }.\]