Suppose the vectors x_1,x_2 and x_3 are the solutions of the system of linear equations, Ax=b when the vector b on the right side is equal to b_1,b_2…

Let $A=[\begin{matrix}{\alpha }_{1} & {\alpha }_{2} & {\alpha }_{3} \\ {\beta }_{1} & {\beta }_{2} & {\beta }_{3} \\ {\gamma }_{1} & {\gamma }_{2} & {\gamma }_{3}\end{matrix}]$A=[α1​β1​γ1​​α2​β2​γ2​​α3​β3​γ3​​]. Now satisfying ${x}_{1},{x}_{2},{x}_{3}$x1​,x2​,x3​ in $Ax=b$Ax=b

$$A{x}_{1}={b}_{1}\Rightarrow [\begin{matrix}{\alpha }_{1} & {\alpha }_{2} & {\alpha }_{3} \\ {\beta }_{1} & {\beta }_{2} & {\beta }_{3} \\ {\gamma }_{1} & {\gamma }_{2} & {\gamma }_{3}\end{matrix}][\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=[\begin{matrix}1 \\ 0 \\ 0\end{matrix}]$$Ax1​=b1​⇒[α1​β1​γ1​​α2​β2​γ2​​α3​β3​γ3​​][111​]=[100​]

\Rightarrow {\begin{matrix}{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=1 \\ {\beta }_{1}+{\beta }_{2}+{\beta }_{3}=0 \\ {\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}=0\end{matrix}\ldots (i)

Now

$$A{x}_{2}={b}_{2}\Rightarrow [\begin{matrix}{\alpha }_{1} & {\alpha }_{2} & {\alpha }_{3} \\ {\beta }_{1} & {\beta }_{2} & {\beta }_{3} \\ {\gamma }_{1} & {\gamma }_{2} & {\gamma }_{3}\end{matrix}][\begin{matrix}0 \\ 2 \\ 1\end{matrix}]=[\begin{matrix}0 \\ 2 \\ 0\end{matrix}]$$Ax2​=b2​⇒[α1​β1​γ1​​α2​β2​γ2​​α3​β3​γ3​​][021​]=[020​]

\Rightarrow {\begin{matrix}2{\alpha }_{2}+{\alpha }_{3}=0 \\ 2{\beta }_{2}+{\beta }_{3}=2 \\ 2{\gamma }_{2}+{\gamma }_{3}=0\end{matrix}\ldots (\mathrm{ii})

And satisfying ${x}_{3}$x3​ gives,

${\alpha }_{3}=0$α3​=0, ${\beta }_{3}=0$β3​=0 and ${\gamma }_{3}=2$γ3​=2

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

${\alpha }_{2}=0,{\beta }_{2}=1,{\gamma }_{2}=-1$α2​=0,β2​=1,γ2​=−1 and ${\alpha }_{1}=1,{\beta }_{1}=-1,{\gamma }_{1}=-1$α1​=1,β1​=−1,γ1​=−1

Thus, $A=[\begin{matrix}1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & -1 & 2\end{matrix}]\Rightarrow |A|=2$A=[1−1−1​01−1​002​]⇒∣A∣=2