Compute the products
$\begin{bmatrix} a & b\\ c &d \end{bmatrix}\begin{bmatrix} d &-b \\ -c&a \end{bmatrix} and \begin{bmatrix} d &-b \\ -c& a \end{bmatrix}\begin{bmatrix} a & b\\ c &d \end{bmatrix}$
Show that when ad-bc= 0, A given by $(*)$ cannot have an inverse. Show that when $ad-bc\neq 0$, that A given by $(*)$ does have an inverse, the one given by the formula in Theorem 5.
Theorem 5 Let A be a 2x2 matrix given as in $(*)$. If ad-bc= 0, then $A^-$$^1$ does not exist. If $ad-bc\neq 0 $ then $A^-1=\frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c& a \end{bmatrix}$
$(*)$ is A=$\begin{bmatrix} a & b\\ c &d \end{bmatrix}$
This is what I have done...
I have multiplied $\begin{bmatrix} a & b\\ c &d \end{bmatrix}\begin{bmatrix} d &-b \\ -c&a \end{bmatrix} $ to get $=[ad-bc]I$
I have also multipled $\begin{bmatrix} d &-b \\ -c& a \end{bmatrix}\begin{bmatrix} a & b\\ c &d \end{bmatrix}$ to get $=[ad-bc]I$
But I don't really known where to go from here. What else am I suppose to do?
Let $A, \tilde{A}$ be the two matrices.
If $ad-bc \neq 0$ then we see that ${1 \over ad-bc } A \tilde{A} = {1 \over ad-bc } \tilde{A} A = I$, hence $A^{-1} = {1 \over ad-bc } \tilde{A}$.
Now suppose $ab-bc = 0$. Then $A \tilde{A} = 0$. If $\tilde{A} = 0$ then $A=0$, which is not invertible since $A(0,1)^T = A (1,0)^T$. If $\tilde{A} \neq 0$, then there is some $V$ such that $v' = \tilde{A}v \neq 0$, but $A \tilde{A} v = A v' = A 0 = 0$ and so $A$ is not invertible.