Question

Examine the consistency of the system of equations.
$x+y+z=1$
$2x+3y+2z=2$
$ax+ay+2az=4$

Answer 1

Given: system of equation is
$x+y+z=1$
$2x+3y+2z=2$
$ax+ay+2az=4\Rightarrow x+y+2z=\frac{4}{a}$
Writing equation as $AX=B$
$\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 1& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ \frac{4}{a}\end{array}\right]$
Hence, $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 1& 1& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}1\\ 2\\ \frac{4}{a}\end{array}\right]$
Calculating $\left|\begin{array}{c}A\end{array}\right|$
$\left|\begin{array}{c}A\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 1& 1& 2\end{array}\right|$
$=1\left|\begin{array}{cc}3& 2\\ 1& 2\end{array}\right|-1\left|\begin{array}{cc}2& 2\\ 1& 2\end{array}\right|+1\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|$
$=1(6-2)-1(4-2)+1(2-3)$
$=4-2-1=1\ne 0$
Since, $\left|\begin{array}{c}A\end{array}\right|\ne 0$, the system of equations has solutions.
Hence, system of equations is consistent.