So for my linear algebra and differential equations, our professor gave us a problem to work on over the weekend. I have been able to solve this problem multiple times by using any values for $a,b,c,d$, but I have no clue how to prove it using the variables.
The statement is, prove that if $$A= \begin{bmatrix}a&b\\c&d\end{bmatrix}$$ then $$ A^{-1} = \frac{1}{\det{A}} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
The only clue that he gave me when I messaged him about it was to use the adjoint method, but from my understanding that is just the matrix flipped and negated which is the result. How do I prove this?
Hint: multiply them together, and use the fact that $\det(A)=ad-bc$
I'd write something like this:
Let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ And $$B= \frac{1}{\det{A}} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$ Then $$AB=\dots=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$$ So by definition, we have that $$A^{-1}=B=\frac{1}{\det{A}} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$