Question

Given that $A^{-1}=\frac{1}{\textup{det}(A)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$  with $\textup{det}(A)\neq 0$, consider the following statements:

(i) $\frac{1}{\textup{det}(A)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\frac{1}{\textup{det}(A)}\textup{adj}(A)$ ;

(ii) $A=\begin{bmatrix}a&c\\b&d\end{bmatrix}^T$ ;

(iii) $\textup{adj}(A)=\begin{bmatrix}d&-c\\-b&a\end{bmatrix}^T$

Which of the above are true?

Answer 1

We know that for an invertible matrix A

$A^{-1}=\frac{1}{det(A)}.adJ(A)....(I)$

For a 2×2 matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$A^{-1}=\frac{1}{det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}......(2)$

We gave

$A^{-1}=\frac{1}{det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\space with A\#0$

Therefore from equition 1 we can conclude that

$A^{-1}=\frac{1}{det(A)}.adJ(A)=\frac{1}{det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.....(2)$

Since we have

$A^{-1}=\frac{1}{det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Therefore

$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{T}....(4)$

Also

adj(A)$=\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}=\begin{bmatrix} d & -c \\ -b & a \end{bmatrix}^{T}....(5)$

Hence, proved that (I)(ii) and (iii) are all true.