Question

The solution set of the following system of equations $2x-3y+6=0$ and $6x+y+8=0$ also satisfies the equation :

• A. $2x+y+2=0$
• B. $2x+3=0,y-2=0$
• C. $4x-3y+9=0$
• D. $2x+2y+1=0$

Answer 1

Answer: (A), (C), (D)

$2x+2y+1=0$

​​​​​​Given:
$2x-3y+6=0\Rightarrow 2x-3y=-6$
$6x+y+8=0\Rightarrow 6x+y=-8$
The corresponding matrix equation is
$\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-6\\ -8\end{array}\right]$
i.e., $AX=B$
Where, $A=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}-6\\ -8\end{array}\right]$

$AX=B$

$\Rightarrow X=A^{-1}B$

Now, $A=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]\Rightarrow \left|A\right|=2+18=20$

Adjoint of $A=adjA=\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]$

$A^{-1}=\frac{1}{\left|A\right|}\left(adjA\right)=\frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]X=\frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]\left[\begin{array}{c}-6\\ -8\end{array}\right]$
$=\frac{1}{20}\left[\begin{array}{c}-30\\ 20\end{array}\right]=\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]$

So we have, $(x,y)=(-\frac{3}{2},1)$ which also satisfies $2x+y+2=0,4x-3y+9=0$ and $2x+2y+1=0$