Question

In a system of linear equations, three equations are given as $5x+4y+2z=13;4x{-}_y-kz=6;2x+3y+3z=16$ . What should be the value of $k$ so that the equations have no solution?

• A. $3$
• B. $5$
• C. $-3$
• D. $-5$

Answer 1

Answer: (B)

Given system of equations is
$5x+4y+2z=13$
$4x-y-kz=6$
$2x+3y+3z=16$
which can be written in matrix form as
$\left[\begin{array}{ccc}5& 4& 2\\ 4& -1& -k\\ 2& 3& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}13\\ 6\\ 16\end{array}\right]$
i.e., $AX=B$
Now, $\left|A\right|=\left|\begin{array}{ccc}5& 4& 2\\ 4& -1& -k\\ 2& 3& 3\end{array}\right|$
$=5\left(-3+3k\right)-4\left(12+2k\right)+2\left(12+2\right)$
$=-15+15k-48-8k+28$
$=-35+7k$
Since, given system has no solution
$\therefore \left|A\right|=0$
$\Rightarrow -35+7k=0$
$\Rightarrow k=\frac{35}{7}$
$=5$