Answer:
The given system of equation can be written in the matrix form as \[AX=B\] Where, \[A=\left[ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{matrix} \right]B=\left[ \begin{matrix} 5 \\ 3 \\ 4 \\ \end{matrix} \right]\text{and}\,\,X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]\] its solution is given by\[X={{A}^{-\,1}}B\] ?(i) Now, \[|A|\,\,=1\,(1-0)-1\,(0-1)+0\] \[=1+1=2\] \[\Rightarrow \]\[|A|\,\,\ne 0,\]hence unique solution exists. Now, cofactors of A are \[{{A}_{11}}={{(-\,1)}^{1\,\,+\,\,1}}\left| \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right|=1\,(1-0)=1\] \[{{A}_{12}}={{(-\,1)}^{1\,\,+\,\,2}}\left| \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right|=(-\,1)\,\,(0-1)=1\] \[{{A}_{13}}={{(-\,1)}^{1\,\,+\,\,3}}\left| \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right|=1\,(0-1)=-\,1\] \[{{A}_{21}}={{(-\,1)}^{2\,\,+\,\,1}}\left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right|=(-\,1)\,\,(1-0)=-\,1\] \[{{A}_{22}}={{(-\,1)}^{2\,\,+\,\,2}}\left| \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right|=1\,\,(1-0)=1\] \[{{A}_{23}}={{(-\,1)}^{2\,\,+\,\,3}}\left| \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right|=(-\,1)\,\,(0-1)=1\] \[{{A}_{31}}={{(-\,1)}^{3\,\,+\,\,1}}\left| \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right|=1\,\,(1-0)=1\] \[{{A}_{32}}={{(-\,1)}^{3\,\,+\,\,2}}\left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right|=(-\,1)\,\,(1-0)=-\,1\] \[{{A}_{33}}={{(-\,1)}^{3\,\,+\,\,3}}\left| \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right|=1\,\,(1-0)=1\] \[\therefore \] \[adj\,(A)={{\left[ \begin{matrix} {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\ \end{matrix} \right]}^{T}}\] \[={{\left[ \begin{matrix} 1 & 1 & -\,1 \\ -\,1 & 1 & 1 \\ 1 & -\,1 & 1 \\ \end{matrix} \right]}^{T}}=\left[ \begin{matrix} 1 & -\,1 & 1 \\ 1 & 1 & -\,1 \\ -\,1 & 1 & 1 \\ \end{matrix} \right]\] \[\therefore \]\[{{A}^{-\,1}}=\frac{1}{|A|}adj\,(A)=\frac{1}{2}=\left[ \begin{matrix} 1 & -\,1 & 1 \\ 1 & 1 & -\,1 \\ -\,1 & 1 & 1 \\ \end{matrix} \right]\] From Eq. (i), we get \[\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\frac{1}{2}\left[ \begin{matrix} 1 & -\,1 & 1 \\ 1 & 1 & -\,1 \\ -\,1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 5 \\ 3 \\ 4 \\ \end{matrix} \right]\] \[=\frac{1}{2}\left[ \begin{matrix} 5-3+4 \\ 5+3-4 \\ -\,5+3+4 \\ \end{matrix} \right]=\frac{1}{2}\left[ \begin{matrix} 6 \\ 4 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ 1 \\ \end{matrix} \right]\] \[\therefore \] \[x=3,\]\[y=2\]and \[z=1\]
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