Question

If the system of linear equations
$x + ay + z = 3 $
$x + 2y + 2z = 6 $
$x + 5y + 3z = b$
has no solution, then :

• A. $a = - 1, b = 9$
• B. $a=-1, b \neq 9$
• C. $a \neq -1, b = 9$
• D. $a = 1 , b \neq 9 $

Answer 1

Answer: (B)

Given: $x+a y+z=3, x+2 y+2 z=6, x+5 y+3 z=b$
$\Delta=\begin{vmatrix}1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3\end{vmatrix}=1(6-10)-a(3-2)+1(5-2)$
$=-4-a+z=-(a+1) $
$\Delta_{1}=\begin{vmatrix}3 & a & 1 \\ 6 & 2 & 2 \\ b & 5 & 3\end{vmatrix}=3(6-10)-a(18-2 b)+(30-2 b)$
$=-12-18 a+2 a b+30-2 b$
$=18-2 b-18 a+2 a b$
$=18(1-a)-2 b(1-a)$
$=2(g-b)(1-a)$
$\Delta_{2}=\begin{vmatrix}1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3\end{vmatrix}=1(18-2 b)-3(3-2)+1(b-6)$
$=18-2 b-3+b-6$
$=(g-b)$
$\Delta_{3}=\begin{vmatrix}1 & a & 3 \\ 1 & 2 & b \\ 1 & 5 & b\end{vmatrix}=1(2 b-30)-a(b-6)+3(5-2)$
$=2 b-30-a b+6 a+9$
$=b(2-a)-3(7-2 a)$
For no solution, $\Delta=0$ and at least one of $\Delta_{1}, \Delta_{2}$ and $\Delta_{3}$ is non zero.
$\Delta=0 \Rightarrow a=-1$
$\Delta_{1}=0 \Rightarrow a=1, b=9$
$\Delta_{2}=0 \Rightarrow b=9$
Therefore, for no solution, $a=-1$ and $b \neq 9$.