Show that the following system of linear equations is consistent … - Vidyadip

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 $a_{ij}i$n 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, $\text{x}=\frac{1}{2}-\text{k}$
$\text{y}=\text{k}$
$\text{z}=0$

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