let A be invertiable $2\times 2$ matrix then $A^{-1}=\frac1{\det A} [\operatorname{tr}(A)I-A]$ Let A be invertiable $2\times 2$ matrix. Show that 
$$A^{-1}=\frac{1}{\det A} [\operatorname{tr}(A)I-A]$$
Is we prove this formula  $A^{-1}=\frac{1}{\det A} \operatorname{adj}(A)$
 A: $$A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$$
$$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
$$detA = ad-bc$$
$$tr(A) = a+d$$
$$tr(A)I-A = \begin{bmatrix}a+d&0\\0&a+d\end{bmatrix}-\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
A: This follows directly from Cayley-Hamilton theorem.
A: Another answer.
$$\det(A-\lambda I) = 0 \implies (a-\lambda)(d-\lambda)-bc = 0 \implies \lambda^2 -(a+d)\lambda +ad-bc = 0$$
Note that $\lambda = A$ satisfies the equation because $\det(A-AI) = 0$. Using this and the fact that $A$ is invertible we get,
$$A^2 - \operatorname{tr}(A)A+\det A = 0 \implies A^{-1}A^2-A^{-1}\operatorname{tr}(A)A + A^{-1}\det(A) = A^{-1}0 = 0 \\ \implies A-\operatorname{tr}(A)I + A^{-1}\det(A) = 0 \implies A^{-1} = \frac{1}{\det A}(\operatorname{tr}(A)I-A)$$
A: Cayley–Hamilton implies that $A$ satisfies its characteristic polynomial, so
$$ A^2 - A(\operatorname{tr}A) + I\det{A} = 0, $$
Multiplying by $A^{-1}/\det{A}$ gives the result. This, of course, gives the generalisation to higher dimensions: if $p(t)=\det{(tI-A)}$ is the characteristic polynomial of the $n \times n$ matrix $A$,
$$ 0 = p(A) = q(A)A+(-1)^n \det{A}, $$
where $q(A)$ is a monic polynomial of degree $n-1$ (if one wants to get fancy, this is using the quotient theorem in the domain $F[A]$ to divide $p(A)$ by $A$), and then multiplying by $A^{-1}/\det{A}$ gives
$$ A^{-1} = (-1)^{n+1}\frac{q(A)}{\det{A}}. $$
One can express the coefficients of $q$ in terms of $\operatorname{tr}(A^k)$ using the Faddeev–LeVerrier algorithm if so desired.
