Mathematics - Determinant Question with Solution | TestHub

Let $A=\left[\begin{array}{lll}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right]$. Now satisfying $x_1, x_2, x_3$ in $Ax=b$
$A{x}_1=b_1\Rightarrow \left[\begin{array}{lll}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$

$\Rightarrow \left\{\begin{array}{c}{\alpha }_1+{\alpha }_2+{\alpha }_3=1\\ {\beta }_1+{\beta }_2+{\beta }_3=0\\ {\gamma }_1+{\gamma }_2+{\gamma }_3=0\end{array}\right.\dots \left(\mathrm{i}\right)$

Now 

$A{x}_2=b_2\Rightarrow \left[\begin{array}{lll}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right]\left[\begin{array}{l}0\\ 2\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 2\\ 0\end{array}\right]$

$\Rightarrow \left\{\begin{array}{c}2{\alpha }_2+{\alpha }_3=0\\ 2{\beta }_2+{\beta }_3=2\\ 2{\gamma }_2+{\gamma }_3=0\end{array}\right.\dots \left(\mathrm{ii}\right)$

And satisfying $x_3$ gives,

${\alpha }_3=0$, ${\beta }_3=0$ and ${\gamma }_3=2$

Now using $\left(\mathrm{i}\right) \text{and} \left(\mathrm{ii}\right)$

${\alpha }_2=0, {\beta }_2=1, {\gamma }_2=-1$ and ${\alpha }_1=1, {\beta }_1=-1, {\gamma }_1=-1$
Thus, $A=\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -1& -1& 2\end{array}\right]\Rightarrow \left|A\right|=2$