Mathematics - Determinant Question with Solution | TestHub

Given,

$2x+y-z=3$

$x-y-z=\alpha$

$3x+3y+\beta z=3$

Let $A= \left[\begin{array}{ccc}2& 1& -1\\ 1& -1& -1\\ 3& 3& \beta \end{array}\right] \& B=\left[\begin{array}{c}3\\ \alpha \\ 3\end{array}\right]$

For Equation to have infinitely many solutions $\left|A\right|=0 \& \left(Adj. A\right).B=0$

$\left|A\right|=\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& -1\\ 3& 3& \beta \end{array}\right|=0$

$\Rightarrow -3\beta -3=0$

$\Rightarrow \beta =-1$

Now,

Minor of $A=\left[\begin{array}{ccc}2\beta +3& \beta +3& 6\\ \beta +3& 2\beta +3& 3\\ -2& -1& -3\end{array}\right]$

Cofactor of $A=\left[\begin{array}{ccc}-\left(2\beta +3\right)& -\left(\beta +3\right)& -6\\ -\left(\beta +3\right)& -\left(2\beta +3\right)& -3\\ 2& 1& 3\end{array}\right]$

Adjoint of $A=\left[\begin{array}{ccc}-\left(2\beta +3\right)& -\left(\beta +3\right)& 2\\ -\left(\beta +3\right)& -\left(2\beta +3\right)& 1\\ -6& -3& 3\end{array}\right]$

Now, $\left(adj. A\right).B=0$

$\Rightarrow \left[\begin{array}{ccc}-\left(2\beta +3\right)& -\left(\beta +3\right)& 2\\ -\left(\beta +3\right)& -\left(2\beta +3\right)& 1\\ -6& -3& 3\end{array}\right]\left[\begin{array}{c}3\\ \alpha \\ 3\end{array}\right]=0$

$\left[\begin{array}{c}-6\beta -9-\alpha \beta -3\alpha +6\\ -3\beta -9-2\alpha \beta -3\alpha +3\\ -18-3\alpha +9\end{array}\right]=0$

$\Rightarrow -18-3\alpha +9=0$

$\Rightarrow \alpha =-3$

Therefore,$\left|\alpha +\beta -\alpha \beta \right|=\left|-3-1-3\right|=\left|-7\right|=7$