question_answer

Find \[{{A}^{-1}},\] if \[A=\left[ \begin{matrix}    0 & 1 & 1  \\    1 & 0 & 1  \\    1 & 1 & 0  \\ \end{matrix} \right]\] and show that \[{{A}^{-1}}=\frac{{{A}^{2}}-3I}{2}.\]

OR

If \[A=\left[ \begin{matrix}    1 & -\,1 & 1  \\    2 & 1 & -\,3  \\    1 & 1 & 1  \\ \end{matrix} \right],\] find \[{{A}^{-1}}\] and hence solve the system of linear equation \[x+2y+z=4,\] \[-\,x+y+z=0,\] \[x-3y+z=2.\]

Answer:

We have,

Clearly,

\[\therefore \] \[{{A}^{-1}}\] exists.

Now, let us evaluate the cofactors of elements of \[|A|.\]

Clearly, \[{{C}_{11}}={{(-\,1)}^{1+1}}\left| \begin{matrix}    0 & 1  \\    1 & 0  \\ \end{matrix} \right|=-\,1\]

\[{{C}_{21}}={{(-\,1)}^{2+1}}\left| \begin{matrix}    1 & 1  \\    1 & 0  \\ \end{matrix} \right|=1,\] \[{{C}_{31}}={{(-\,1)}^{3+1}}\left| \begin{matrix}    1 & 1  \\    0 & 1  \\ \end{matrix} \right|=1,\]

\[{{C}_{12}}={{(-\,1)}^{1+2}}\left| \begin{matrix}    1 & 1  \\    1 & 0  \\ \end{matrix} \right|=1,\] \[{{C}_{22}}={{(-\,1)}^{2+2}}\left| \begin{matrix}    0 & 1  \\    1 & 0  \\ \end{matrix} \right|=1,\]

\[{{C}_{32}}={{(-\,1)}^{3+2}}\left| \begin{matrix}    0 & 1  \\    1 & 1  \\ \end{matrix} \right|=1,\] \[{{C}_{13}}={{(-\,1)}^{1+3}}\left| \begin{matrix}    1 & 0  \\    1 & 1  \\ \end{matrix} \right|=1,\]

\[{{C}_{23}}={{(-\,1)}^{2+3}}\left| \begin{matrix}    0 & 1  \\    1 & 1  \\ \end{matrix} \right|=1\]

and \[{{C}_{33}}={{(-\,1)}^{3+3}}\left| \begin{matrix}    0 & 1  \\    1 & 0  \\ \end{matrix} \right|=-1\]

\[\therefore \]adj

Thus, \[{{A}^{-1}}=\frac{1}{|A|}\] adj

Now, consider

Hence proved.

OR

We have,

Clearly,

\[=4+5+1=10\ne 0\]

\[\therefore \] \[{{A}^{-1}}\] exist.

Now, let us find the cofactors \[{{C}_{ij}}\] of elements \[{{a}_{ij}}\]in A.

Clearly, \[{{C}_{11}}={{(-\,1)}^{1+1}}\left| \begin{matrix}    1 & -\,3  \\    1 & 1  \\ \end{matrix} \right|=4,\]

\[{{C}_{12}}={{(-1)}^{2+1}}\left| \begin{matrix}    2 & -\,3  \\    1 & 1  \\ \end{matrix} \right|=-\,5,\] \[{{C}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix}    2 & 1  \\    1 & 1  \\ \end{matrix} \right|=1,\]

\[{{C}_{21}}={{(-\,1)}^{2+1}}\left| \begin{matrix}    -\,1 & 1  \\    1 & 1  \\ \end{matrix} \right|=2,\] \[{{C}_{22}}={{(-\,1)}^{2+2}}\left| \begin{matrix}    1 & 1  \\    1 & 1  \\ \end{matrix} \right|=0,\]

\[{{C}_{23}}={{(-\,1)}^{2+3}}\left| \begin{matrix}    1 & -\,1  \\    1 & 1  \\ \end{matrix} \right|=-\,2,\] \[{{C}_{31}}={{(-\,1)}^{3+1}}\left| \begin{matrix}    -\,1 & 1  \\    1 & -\,3  \\ \end{matrix} \right|=2,\] \[{{C}_{32}}={{(-\,1)}^{3+2}}\left| \begin{matrix}    1 & 1  \\    2 & -\,3  \\ \end{matrix} \right|=5,\] \[{{C}_{33}}={{(-\,1)}^{3+3}}\left| \begin{matrix}    1 & -\,1  \\    2 & 1  \\ \end{matrix} \right|=3\]

\[\therefore \] adj

\[\Rightarrow \]        ?(i)

Now, let us solve the given system of linear equations which can be written in matrix form as

OR

\[{{A}^{T}}X=B,\] where  and

Since, \[|{{A}^{T}}|\,\,=\,\,|A|\,\,=10\ne 0.\] So, the given system of equations is consistent and have a unique solution given by \[X={{({{A}^{T}})}^{-1}}B={{({{A}^{-1}})}^{T}}B\]

\[\Rightarrow \] \[x=\frac{9}{5},\] \[y=\frac{2}{5}\] and \[z=\frac{7}{5},\] which is the required solution.