Question

If $A=\left[\begin{array}{ccc}1& -2& 3\\ 0& -1& 4\\ -2& 2& 1\end{array}\right],\mathrm{find}{\left(A^T\right)}^{-1}.$

Answer 1

We know that (A^T)⁻¹ = (A⁻¹)^T.

$$\begin{gathered} A=\left[\begin{array}{ccc}1& -2& 3\\ 0& -1& 4\\ -2& 2& 1\end{array}\right] \\ {A}^{-1}=\frac{1}{\left|A\right|}\mathrm{Adj}.A \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{ccc}1& -2& 3\\ 0& -1& 4\\ -2& 2& 1\end{array}\right| \\ =1\left(-1-8\right)-2\left(-8+3\right) \\ =-9+10 \\ =1 \\ \mathrm{Now},\mathrm{to}\mathrm{find}\mathrm{Adj}.A \\ \begin{array}{ccc}{A}_{11}={\left(-1\right)}^{1+1}\left(-9\right)=-9 \\ {A}_{12}={\left(-1\right)}^{1+2}\left(8\right)=-8 \\ {A}_{13}={\left(-1\right)}^{1+3}\left(-2\right)=-2& A_{21}={\left(-1\right)}^{2+1}\left(-8\right)=8 \\ {A}_{22}={\left(-1\right)}^{2+2}\left(7\right)=7 \\ {A}_{23}={\left(-1\right)}^{2+3}\left(-2\right)=2& A_{31}={\left(-1\right)}^{3+1}\left(-5\right)=-5 \\ {A}_{32}={\left(-1\right)}^{3+2}\left(4\right)=-4 \\ {A}_{33}={\left(-1\right)}^{3+3}\left(-1\right)=-1\end{array} \\ \mathrm{Therefore}, \\ \mathrm{Adj}.A=\left[\begin{array}{ccc}-9& 8& -5\\ -8& 7& -4\\ -2& 2& -1\end{array}\right] \\ \mathrm{Thus}, \\ {A}^{-1}=\left[\begin{array}{ccc}-9& 8& -5\\ -8& 7& -4\\ -2& 2& -1\end{array}\right]. \\ {\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T \\ ={\left[\begin{array}{ccc}-9& 8& -5\\ -8& 7& -4\\ -2& 2& -1\end{array}\right]}^T \\ =\left[\begin{array}{ccc}-9& -8& -2\\ 8& 7& 2\\ -5& -4& -1\end{array}\right] \\ \mathrm{Hence},{\left(A^T\right)}^{-1}=\left[\begin{array}{ccc}-9& -8& -2\\ 8& 7& 2\\ -5& -4& -1\end{array}\right]. \end{gathered}$$