Question

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