Show that the following system of linear equations is consistent …

Question

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

✓

Answer

This system can be written as: $\begin{bmatrix}2&2&-2\\ 4&4&-1\\ 6&6&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$ or
$\text{AX = B}$ $\text{|A|}=2{(14)}-2(14)-2{(0)}=0$
So, $A$ is singular and the system has either no solution or infinite solutions according as $\text{(Adj A)}\times\text{(B)}\neq0$ or $\text{(Adj A)}\times\text{(B)}=0$
Let $C_{ij}$ be the co$-$factors of $a_{ij}$ in $A$ $\text{C}_{11}=14\\ \text{C}_{21}=-16\\ \text{C}_{31}=6$
$\text{C}_{12}=-14\\ \text{C}_{22}=16\\ \text{C}_{32}=-6$ $\text{C}_{1}=0\\ \text{C}_{23}=0\\ \text{C}_{33}=0$ $\text{adj A}=\begin{bmatrix}14&-14&0\\ -16&16&0\\ 0&0&0\end{bmatrix}^\text{T}=\begin{bmatrix}14&-16&6\\ -14&16&-6\\ 0&0&0\end{bmatrix}$ $(\text{adj A})\times\text{B}=\begin{bmatrix}14&-16&0\\ -14&16&-6\\ 0&0&0\end{bmatrix}\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}=\begin{bmatrix}14-32+18\\ -14+32-18\\ 0+0+0\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
So, $\text{AX}=\text{B}$ has infinite solutions.
Now,
let $z = k$
So, $2x + 2y = 1 + 2k 4x + 4y = 2 + k$ which can be written as: $\begin{bmatrix}2&2\\ 4&4\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}1+2\text{k}\\ 2+\text{k}\end{bmatrix}$
or $\text{AX = B}|A| = 0, z = 0$
Again, $2\text{x}+2\text{y}=1$ $4\text{x}+4\text{y}=2$
Let $\text{y = k}$
$2\text{x}=1-2\text{k}$
$\text{x}=\frac{1}{2}-\text{k}$
Hence, $\text{x}=\frac{1}{2}-\text{k}$
$\text{y}=\text{k}$
$\text{z}=0$