Answer:
Verified
Step-by-step explanation:
Let the 2x2 matrix A be in the form of:
[tex]\left[\begin{array}{cc}a&b\\c&d\end{array}\right][/tex]
Where det(A) = ad - bc # 0 so A is nonsingular:
Then the transposed version of A is
[tex]A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right][/tex]
Then the inverted version of transposed A is
[tex](A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right][/tex]
The inverted version of A is:
[tex]A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right][/tex]
The transposed version of inverted A is:
[tex](A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right][/tex]
We can see that
[tex] (A^T)^{-1} = (A^{-1})^T[/tex]
So this theorem is true for 2 x 2 matrices
