Question

Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3

Answer 1

 Here,
\[2x + 2y - 2z = 1 . . . (1) \]
\[4x + 4y - z = 2 . . . (2) \]
\[6x + 6y + 2z = 3 . . . (3) \]
\[or, AX = B\]
\[\text{ where, }A = \begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\]
\[\begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{vmatrix}\]
\[ = 2\left( 8 + 6 \right) - 2\left( 8 + 6 \right) - 2(24 - 24)\]
\[ = 28 - 28 - 0\]
\[ = 0\]
So, A is singular . Thus, the system of equations is either inconsistent or it is consistent with
\[\text{ infinitely many solutions because }\left( adj A \right)B \neq 0\text{ or }\left( adj A \right)B = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co-factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right]\text{. Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = 14, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = - 14, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 4 \\ 6 & 6\end{vmatrix} = 0\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = - 16, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = 16, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 2 \\ 6 & 6\end{vmatrix} = 0\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = - 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 2 \\ 4 & 4\end{vmatrix} = 0\]
\[adj A = \begin{bmatrix}14 & - 14 & 0 \\ - 16 & 16 & 0 \\ 6 & - 6 & 0\end{bmatrix}^T \]
\[ = \begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\]
\[ = \begin{bmatrix}14 - 32 + 18 \\ - 14 + 32 - 18 \\ 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[\text{ If }\left| A \right|=0\text{ and }\left( adjA \right)B=0, \text{ then the system is consistent and has infinitely many solutions}.\]
Thus, AX=B has infinitely many solutions.
 Substituting y=k in eq. (1) and eq. (2), we get
\[2x - 2z = 1 - 2k\text{ and }4x - z = 2 - 4k\]
\[\begin{bmatrix}2 & - 2 \\ 4 & - 1\end{bmatrix}\binom{x}{z} = \binom{1 - 2k}{2 - 4k}\]
Now, 
\[\left| A \right| = \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix}\]
\[ = - 2 + 8 = 6 \neq 0\]
\[adj A = \begin{vmatrix}- 1 & 2 \\ - 4 & 2\end{vmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{6}\begin{bmatrix}- 1 & 2 \\ - 4 & 2\end{bmatrix}\]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{z} = \frac{1}{6}\begin{bmatrix}- 1 & 2 \\ - 4 & 2\end{bmatrix}\binom{1 - 2k}{2 - 4k}\]
\[ \Rightarrow \binom{x}{z} = \frac{1}{6}\binom{ - 1 + 2k + 4 - 8k}{ - 4 + 8k + 4 - 8k}\]
\[ \Rightarrow \binom{x}{z} = \binom{\frac{3 - 6k}{6}}{0}\]
\[ \therefore x = \frac{1 - 2k}{2}, y = k \hspace{0.167em}\text{ and } z = 0 \]
These values of x, y and z satisfy the third equation .
\[\text{ Thus, } x = \frac{1 - 2k}{2}, y = k\text{ and }z = 0 \left(\text{ where k is real number } \right)\text{ satisfy the given system of equations } .\]