Question

The system $\left(\begin{array}{ccc}1& -1& 2\\ 3& 5& -3\\ 2& 6& a\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3\\ b\\ 2\end{array}\right)$ has no solution, if

• A. $a=-5,b\ne 5$
• B. $a=-5,b=5$
• C. $a\ne -5,b=5$
• D. $a\ne -5,b\ne 5$

Answer 1

The correct option is A $a=-5,b\ne 5$

Given
$\left(\begin{array}{ccc}1& -1& 2\\ 3& 5& -3\\ 2& 6& a\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3\\ b\\ 2\end{array}\right)$
The augment matrix of given system is
$\left[A|B\right]=\left[\begin{array}{ccccc}1& -1& 2& |& 3\\ 3& 5& -3& |& b\\ 2& 6& a& |& 2\end{array}\right]$
$R_2\to R_2-3R_1$ gives
$\left[A|B\right]=\left[\begin{array}{ccccc}1& -1& 2& |& 3\\ 0& 8& -9& |& b-9\\ 2& 6& a& |& 2\end{array}\right]$
$R_3\to R_3-2R_1$ gives
$\left[A|B\right]=\left[\begin{array}{ccccc}1& -1& 2& |& 3\\ 0& 8& -9& |& b-9\\ 0& 8& a-4& |& -4\end{array}\right]$
$R_3\to R_3-R_2$ gives
$\left[A|B\right]=\left[\begin{array}{ccccc}1& -1& 2& |& 3\\ 0& 8& -9& |& b-9\\ 0& 0& a+5& |& 5-b\end{array}\right]$
We know that
For no solution
Rank A < Rank $\left[A|B\right]$
$\therefore$ $a+5=0$ and $5-b\ne 0$
$a=-5,b\ne 5$