Question 7.AE.12: In order to demonstrate the entire procedure for the computa......
In order to demonstrate the entire procedure for the computation of the inverse of a matrix the following 3×3 matrix will be used
\bar{A}=\left[\begin{array}{ccc} 1 & 3 & -1 \\ 2 & -4 & 2 \\ 3 & 1 & 1 \end{array}\right]Transpose of \bar{A} is
\bar{A}^t=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 3 & -4 & 1 \\ -1 & 2 & 1 \end{array}\right]The cofactors of \bar{A}^t are
\begin{aligned} & A_{11}^t=(-1)^{(1+1)}\left|\begin{array}{cc} -4 & 1 \\ 2 & 1 \end{array}\right|=(-4-2)=-6 \\ & A_{12}^t=(-1)^{(1+2)}\left|\begin{array}{cc} 3 & 1 \\ -1 & 1 \end{array}\right|=-(3+1)=-4 \\ & A_{13}^t=(-1)^{(1+3)}\left|\begin{array}{cc} 3 & -4 \\ -1 & 2 \end{array}\right|=(6-4)=2 \\ & A_{21}^t=(-1)^{(2+1)}\left|\begin{array}{cc} 2 & 3 \\ 2 & 1 \end{array}\right|=(-2+6)=4 \\ & A_{22}^t=(-1)^{(2+2)}\left|\begin{array}{cc} 1 & 3 \\ -1 & 1 \end{array}\right|=(1+3)=4 \\ & A_{23}^t=(-1)^{(2+3)}\left|\begin{array}{cc} 1 & 2 \\ -1 & 2 \end{array}\right|=-(2+2)=-4 \\ & A_{31}^t=(-1)^{(3+1)}\left|\begin{array}{cc} 2 & 3 \\ -4 & 1 \end{array}\right|=(2+24)=14 \\ & A_{32}^t=(-1)^{(3+2)}\left|\begin{array}{cc} 1 & 3 \\ 3 & 1 \end{array}\right|=-(1-9)=8 \\ & A_{33}^t=(-1)^{(3+3)}\left|\begin{array}{cc} 1 & 2 \\ 3 & -4 \end{array}\right|=(-4-6)=-10 \end{aligned}.
The determinant det (\bar{A})=|\bar{A}| is
\begin{aligned} |\bar{A}| & =1 \cdot\left|\begin{array}{cc} -4 & 2 \\ 1 & 1 \end{array}\right|-3 \cdot\left|\begin{array}{ll} 2 & 2 \\ 3 & 1 \end{array}\right|-1 \cdot\left|\begin{array}{cc} 2 & -4 \\ 3 & 1 \end{array}\right| \\ & =-4-2-6+18-2-12=-26+18=-8 . \end{aligned}The inverse of \bar{A} is now
(\bar{A})^{-1}=\frac{\operatorname{Adj}(\bar{A})}{|\bar{A}|}=-\frac{1}{8}\left[\begin{array}{ccc} -6 & -4 & 2 \\ 4 & 4 & -4 \\ 14 & 8 & -10 \end{array}\right] .