If $\mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right]$, find A^-1. Hence, solve the system equations x+y+z=6, x+2z=7,
3x+y+z=12

$\mathbf{OR}$

Find the inverse of the following matrix using elementary operations. $\mathrm{A}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]$

Given : $\mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right]$

To find $A^{\mathit{-}\mathit{1}}\mathit{,}$ we need cofactors of each element of matrix A.

cofactor of ${\mathrm{a}}_{11}={(-1)}^{1+1}\left|\begin{array}{cc}0& 2\\ 1& 1\end{array}\right|=-2$

cofactor of ${\mathrm{a}}_{12}={(-1)}^{1+2}\left|\begin{array}{cc}1& 2\\ 3& 1\end{array}\right|=-(1-6)=5$

cofactor of ${\mathrm{a}}_{13}={(-1)}^{1+3}\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|=1$

cofactor of ${\mathrm{a}}_{21}={(-1)}^{2+1}\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=0$

cofactor of ${\mathrm{a}}_{22}={(-1)}^{2+2}\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|=(1-3)=-2$

cofactor of ${\mathrm{a}}_{23}={(-1)}^{2+3}\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|=-(1-3)=2$

cofactor of ${\mathrm{a}}_{31}={(-1)}^{3+1}\left|\begin{array}{cc}1& 1\\ 0& 2\end{array}\right|=2$

cofactor of ${\mathrm{a}}_{32}={(-1)}^{3+2}\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=-(2-1)=-1$

cofactor of ${\mathrm{a}}_{33}={(-1)}^{3+3}\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|=-1$

So,

cofactor of matrix of $\mathrm{A}=\left[\begin{array}{ccc}-2& 5& 1\\ 0& -2& 2\\ 2& -1& 1\end{array}\right]$

Now, we know that transpose of cofactor matrix A is adj (A).

$\therefore \mathrm{adj} \left(\mathrm{A}\right)=\left[\begin{array}{ccc}-2& 0& 2\\ 5& -2& -1\\ 1& 2& -1\end{array}\right]$

For, |A|

$$\begin{gathered} \left|\mathrm{A}\right|=1(0-2)-1(1-6)+(1-0) \\ =-2+5+1 \\ =4 \\ \& {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|} \mathrm{adj} \left(\mathrm{A}\right) \end{gathered}$$

$\therefore {\mathrm{A}}^{-1}=\frac{1}{4}\left[\begin{array}{ccc}-2& 0& 2\\ 5& -2& -1\\ 1& 2& -1\end{array}\right]$

Now, given system equations are, 

$$\begin{gathered} \mathrm{x}+\mathrm{y}+\mathrm{z}=6, \\ \mathrm{x}+2\mathrm{z}=7, \\ 3\mathrm{x}+\mathrm{y}+\mathrm{z}=12 \end{gathered}$$

Writing the above equation in matrix form

$$\begin{gathered} \left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right] \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}6\\ 7\\ 12\end{array}\right] \\ AX=B \\ \Rightarrow X=A^{-1}B \\ \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}-2& 0& 2\\ 5& -2& -1\\ 1& 2& -1\end{array}\right] \left[\begin{array}{c}6\\ 7\\ 12\end{array}\right] \end{gathered}$$

$$\begin{gathered} \therefore \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}-12& +0& +24\\ 30& -14& -12\\ 6& +14& -12\end{array}\right] \\ \therefore \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}12\\ 4\\ 8\end{array}\right] \\ \therefore \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 2\end{array}\right] \end{gathered}$$

Therefore, solutions are x=3, y=1 and z=2.

                                          OR

We know that A=IA, hence

$$\begin{gathered} \left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\mathrm{A} \\ \Rightarrow \left[\begin{array}{ccc}1& 2& -2\\ 0& 5& -2\\ 0& -2& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right] \mathrm{A} \left[\mathrm{Applying} {\mathrm{R}}_2\to {\mathrm{R}}_2+{\mathrm{R}}_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 2& -2\\ 0& 1& 0\\ 0& -2& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 2\\ 0& 0& 1\end{array}\right] \mathrm{A} \left[\mathrm{Applying} {\mathrm{R}}_2\Rightarrow {\mathrm{R}}_2+2{\mathrm{R}}_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-1& -2& -4\\ 1& 1& 2\\ 2& 2& 5\end{array}\right] \mathrm{A} \left[\mathrm{Applying} {\mathrm{R}}_1\to {\mathrm{R}}_1+(-2){\mathrm{R}}_2, {\mathrm{R}}_3\to {\mathrm{R}}_3+2{\mathrm{R}}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right] \mathrm{A} \left[\mathrm{Applying} {\mathrm{R}}_1\to {\mathrm{R}}_1+2{\mathrm{R}}_3\right] \end{gathered}$$

Therefore, ${\mathrm{A}}^{-1}=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right].$ Which is the required answer.