Question

The system of equations $3x+2y+z=6$, $3x+4y+3z=14$ and $6x+10y+8z=a$, has infinite number of solutions, if $a$ is equal to

• A. 8
• B. 12
• C. 24
• D. 36

Answer 1

Answer: (D)

Given system of equation is
$3x+2y+z=6$
$3x+4y+3z=14$
$6x+10y+8z=a$
Here, $A=\left[\begin{array}{ccc}3& 2& 1\\ 3& 4& 3\\ 6& 10& 8\end{array}\right],B=\left[\begin{array}{c}6\\ 14\\ a\end{array}\right]$
$C_{11}=(32-30)=2,C_{12}=-(24-18)=-6$,
$C_{13}=(30-24)=6$
$C_{21}=-(16-10)=-6,C_{22}=(24-6)=18$,
$C_{23}=-(30-12)=-18$
$C_{31}=(6-4)=2,C_{32}=-(9-3)=-6$,
$C_{33}=(12-6)=6$
$adjA={\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]}^{\prime }={\left[\begin{array}{ccc}2& -6& 6\\ -6& 18& -18\\ 2& -6& 6\end{array}\right]}^{\prime }$
$=\left[\begin{array}{ccc}2& -6& 2\\ -6& 18& -6\\ 6& -18& 6\end{array}\right]$
So, (adj A) $B=\left[\begin{array}{ccc}2& -6& 2\\ -6& 18& -6\\ 6& -18& 6\end{array}\right]\left[\begin{array}{c}6\\ 14\\ a\end{array}\right]$
$=\left[\begin{array}{c}12-84+2a\\ -36+252-6a\\ 36-252+6a\end{array}\right]=\left[\begin{array}{c}-72+2a\\ 216-6a\\ -216+6a\end{array}\right]$
and $|A|=\left|\begin{array}{ccc}3& 2& 1\\ 3& 4& 3\\ 6& 10& 8\end{array}\right|$
$=3(32-30)-2(24-18)+1(30-24)$
$=3(2)-2(6)+6=6-12+6=0$
We know that, if $|A|=0$ and $(adjA)\cdot B=0$, then the system of equations is consistent and has an infinite number of solutions.
$(adjA)\cdot B=0$
$\Rightarrow \left[\begin{array}{c}-72+2a\\ 216+6a\\ -216+6a\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
On comparing, we get
$2a-72=0$
$\therefore a=36$