Question

If the system of equations $2x+3y=8$, $7x-5y+3=0$, $4x-6y+\lambda =0$ has no solution, then $\lambda$

• A. $=4$
• B. $=8$
• C. $\ne 8$
• D. $\ne 4$

Answer 1

Answer: (C)

The given system of equations can be written as,
$2x+3y+0.Z=8\dots \left(i\right)$
$7x-5y+0.z=-3\dots \left(ii\right)$
$4x-6y+0.z=-\lambda \dots \left(iii\right)$
The above equation can be written in matrix form as,
$\left[\begin{array}{ccc}2& 3& 0\\ 7& -5& 0\\ 4& -6& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}8\\ -3\\ -\lambda \end{array}\right]$
or$AX=B$, where
$A=\left[\begin{array}{ccc}2& 3& 0\\ 7& -5& 0\\ 4& -6& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, and $B=\left[\begin{array}{c}8\\ -3\\ -\lambda \end{array}\right]$
$\left|\begin{array}{ccc}2& 3& 0\end{array}\right|$
The system has no solution if $\left(adjA\right)$ $B\ne O$
So, cofactor matrix of $A=\left[\begin{array}{ccc}0& 0& -22\\ 0& 0& 24\\ 0& 0& -31\end{array}\right]$
$\therefore \left(adjA\right)=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ -22& 24& -31\end{array}\right]$
Now, (adj A )$B\ne O$
$\Rightarrow \left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ -22& 24& -31\end{array}\right]\left[\begin{array}{c}8\\ -3\\ -\lambda \end{array}\right]\ne \left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}0+0+0\\ 0+0+0\\ -176-72+31\lambda \end{array}\right]\ne \left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow -248+31\lambda \ne 0\Rightarrow 31\lambda \ne 0\Rightarrow 31\lambda \ne 248\Rightarrow \lambda \ne 8$