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

Answer 1

Answer: (D)

Here $D =\begin{vmatrix}2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix}$
$=2(- a -1)-1( a -1)+1+1$
$=1-3a$
$D _{3}=\begin{vmatrix}2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix}$
$=2(- b -3)-1( b -3)+5(1+1)$
$=7-3b$
for $a=\frac{1}{3}, b \neq \frac{7}{3}$, system has no solutions