Question

If the following system of linear equations
$2x+y+z=5x-y+z=3x+y+az=b$
has no solution, then :

• A. $a=-\frac{1}{3},b\ne \frac{7}{3}$
• B. $a\ne -\frac{1}{3},b=\frac{7}{3}$
• C. $a\ne \frac{1}{3},b=\frac{7}{3}$
• D. $a=\frac{1}{3},b\ne \frac{7}{3}$

Answer 1

The correct option is D $a=\frac{1}{3},b\ne \frac{7}{3}$

Given: $2x+y+z=5x-y+z=3x+y+az=b$
Now,
$\Delta =\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ 1& 1& a\end{array}\right|=1-3a$
For no solution,
$\Delta =0\Rightarrow a=\frac{1}{3}$
Now, checking
${\Delta }_1=\left|\begin{array}{ccc}5& 1& 1\\ 3& -1& 1\\ b& 1& \frac{1}{3}\end{array}\right|=2b-\frac{14}{3}{\Delta }_2=\left|\begin{array}{ccc}2& 5& 1\\ 1& 3& 1\\ 1& b& \frac{1}{3}\end{array}\right|=-b+\frac{7}{3}{\Delta }_3=\left|\begin{array}{ccc}2& 1& 5\\ 1& -1& 3\\ 1& 1& b\end{array}\right|=7-3b$
Also,
${\Delta }_1\ne 0\text{or}{\Delta }_2\ne 0\text{or}{\Delta }_3\ne 0\therefore b\ne \frac{7}{3}$
Hence, for no solution
$a=\frac{1}{3},b\ne \frac{7}{3}$