Question

Let $2x+ay+6z=8$ , $x+2y+bz=5$ and $x+y+3z=4$ be three equations. If these $3$ equations are consistent, then

• A. $b=3,a\neq 2$
• B. $a=2, \, b\neq 3$
• C. $a\neq 2,b\neq 3$
• D. $a\neq 2,b=4$

Answer 1

Answer: (B)

From Cramer’s rule,
for consistent system $\triangle =\triangle _{1}=\triangle _{2}=\triangle _{3}=0$
$\triangle = \begin{vmatrix} 2 & a & 6 \\ 1 & 2 & b \\ 1 & 1 & 3 \end{vmatrix} = 2 \left(6 - b\right) - a \left(3 - b\right) + 6 \left(1 - 2\right)$
$=12-2b-3a+ab-6$
$=6-2b-3a+ab$
$=\left(a - 2\right)\left(b - 3\right)$
$\left(\triangle \right)_{1}=\begin{vmatrix} 8 & a & 6 \\ 5 & 2 & b \\ 4 & 1 & 3 \end{vmatrix}=8\left(6 - b\right)-a\left(15 - 4 b\right)+6\left(5 - 8\right)$
$=48-8b-15a+4ab-18$
$=30-15a-8b+4ab$
$=\left(\right.4b-15\left.\right)\left(\right.a-2\left.\right)$
$\left(\triangle \right)_{2}=\begin{vmatrix} 2 & 8 & 6 \\ 1 & 5 & b \\ 1 & 4 & 3 \end{vmatrix}=2\left(15 - 4 b\right)-8\left(3 - b\right)+6\left(4 - 5\right)$
$=30-8b-24+8b-6=0$
$\left(\triangle \right)_{3}=\begin{vmatrix} 2 & a & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{vmatrix}=2\left(8 - 5\right)-a\left(4 - 5\right)+8\left(1 - 2\right)=a-2$
For consistent system $a=2$