Question

If the following system of linear equations
$2x+y+z=5$
$x-y+z=3$
$x+y+az=b$
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

Answer: (D)

Here $D=\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ 1& 1& a\end{array}\right|$
$=2(-a-1)-1(a-1)+1+1$
$=1-3a$
$D_3=\left|\begin{array}{ccc}2& 1& 5\\ 1& -1& 3\\ 1& 1& b\end{array}\right|$
$=2(-b-3)-1(b-3)+5(1+1)$
$=7-3b$
for $a=\frac{1}{3},b\ne \frac{7}{3}$, system has no solutions