Mathematics - Determinant Question with Solution | TestHub

For the system of linear equations

$2x-3y+5z=9$

$x+3y-z=-18$

$3x-y+\left({\lambda }^2-\left|\lambda \right|\right)z=16$ to have no solutions, 

$∆$ should be equal to $0$.

i.e. $\Delta =\left|\begin{array}{ccc}2& -3& 5\\ 1& 3& -1\\ 3& -1& {\lambda }^2-\left|\lambda \right|\end{array}\right|$

$=2\left(3{\lambda }^2-3\left|\lambda \right|-1\right)+3\left({\lambda }^2-\left|\lambda \right|+3\right)+5\left(-1-9\right)$

$=9{\lambda }^2-9\left|\lambda \right|-43$

$=9{\left|\lambda \right|}^2-9\left|\lambda \right|-43$

Now $∆=0$ for $2$ values of $\left|\lambda \right|$ as for the above quadratic the value of determinant is greater than zero.

So, $2$ values of $\lambda$ satisfy the system of equations to obtain no solution.