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
$\left[\begin{matrix}5&4&2\\ 4&-1&-k\\ 2&3&3\end{matrix}\right] \left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}13\\ 6\\ 16\end{matrix}\right]$
i.e., $AX=B$
Now, $\left|A\right|=\left|\begin{matrix}5&4&2\\ 4&-1&-k\\ 2&3&3\end{matrix}\right|$
$=5\left(-3+3k\right)-4\left(12+2k\right)+2\left(12+2\right)$
$=-15+15k-48-8k+28$
$=-35+7k$
Since, given system has no solution
$\therefore \left|A\right|=0$
$\Rightarrow -35+7k=0$
$\Rightarrow k=\frac{35}{7}$
$=5$