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

Answer 1

Answer: (B)

From Cramer’s rule,
for consistent system $△={△}_1={△}_2={△}_3=0$
$△=\left|\begin{array}{ccc}2& a& 6\\ 1& 2& b\\ 1& 1& 3\end{array}\right|=2\left(6-b\right)-a\left(3-b\right)+6\left(1-2\right)$
$=12-2b-3a+ab-6$
$=6-2b-3a+ab$
$=\left(a-2\right)\left(b-3\right)$
${\left(△\right)}_1=\left|\begin{array}{ccc}8& a& 6\\ 5& 2& b\\ 4& 1& 3\end{array}\right|=8\left(6-b\right)-a\left(15-4b\right)+6\left(5-8\right)$
$=48-8b-15a+4ab-18$
$=30-15a-8b+4ab$
$=(4b-15)(a-2)$
${\left(△\right)}_2=\left|\begin{array}{ccc}2& 8& 6\\ 1& 5& b\\ 1& 4& 3\end{array}\right|=2\left(15-4b\right)-8\left(3-b\right)+6\left(4-5\right)$
$=30-8b-24+8b-6=0$
${\left(△\right)}_3=\left|\begin{array}{ccc}2& a& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\right|=2\left(8-5\right)-a\left(4-5\right)+8\left(1-2\right)=a-2$
For consistent system $a=2$