# Solve this

Question:

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

Solution:

We know that $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$.

$A=\left[\begin{array}{ccc}1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1\end{array}\right]$

$A^{-1}=\frac{1}{|A|}$ Adj. $A$

Now,

$|A|=\left|\begin{array}{ccc}1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1\end{array}\right|$

$=1(-1-8)-2(-8+3)$

$=-9+10$

$=1$

Now, to find Adj. $A$

$A_{11}=(-1)^{1+1}(-9)=-9 \quad A_{21}=(-1)^{2+1}(-8)=8 \quad A_{31}=(-1)^{3+1}(-5)=-5$

$A_{12}=(-1)^{1+2}(8)=-8 \quad A_{22}=(-1)^{2+2}(7)=7 \quad A_{32}=(-1)^{3+2}(4)=-4$

$A_{13}=(-1)^{1+3}(-2)=-2 \quad A_{23}=(-1)^{2+3}(-2)=2 \quad A_{33}=(-1)^{3+3}(-1)=-1$

Therefore,

Adj. $A=\left[\begin{array}{lll}-9 & 8 & -5 \\ -8 & 7 & -4 \\ -2 & 2 & -1\end{array}\right]$

Thus,

$A^{-1}=\left[\begin{array}{lll}-9 & 8 & -5 \\ -8 & 7 & -4 \\ -2 & 2 & -1\end{array}\right]$

$\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$

$=\left[\begin{array}{rrr}-9 & 8 & -5 \\ -8 & 7 & -4 \\ -2 & 2 & -1\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1\end{array}\right]$

Hence, $\left(A^{T}\right)^{-1}=\left[\begin{array}{ccc}-9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1\end{array}\right]$