Exercise 4.1.10

For brevity, we write $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ for some $2\times 2$ matrix and $C=\left(\begin{array}{cc}\hfill d\hfill & \hfill -c\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right)$ for the corresponding classical adjoint.

1.

Directly check that

$$\mathit{CA}=\left(\begin{array}{cc}\hfill \mathit{ad}-\mathit{bc}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mathit{ad}-\mathit{bc}\hfill \end{array}\right)$$

and

$$\mathit{AC}=\left(\begin{array}{cc}\hfill \mathit{ad}-\mathit{bc}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mathit{ad}-\mathit{bc}\hfill \end{array}\right).$$

2.

Calculate that

$$\det <!--\; FUNCTION\; APPLICATION\; -->(C)=\mathit{da}-(-c)(-b)=\mathit{ad}-\mathit{bc}=\det <!--\; FUNCTION\; APPLICATION\; -->(A).$$

3.

Since the transpose matrix $A^t$ is $\left(\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right)$, the corresponding classical adjoint would be

$$\left(\begin{array}{cc}\hfill d\hfill & \hfill -b\hfill \\ \hfill -c\hfill & \hfill a\hfill \end{array}\right)=C^t.$$

4.

If $A$ is invertible, we have that $\det <!--\; FUNCTION\; APPLICATION\; -->(A)\ne 0$ by Theorem 4.2. So we can write

$${[\det <!--\; FUNCTION\; APPLICATION\; -->(A)]}^{-1}\mathit{CA}=A{[\det <!--\; FUNCTION\; APPLICATION\; -->(A)]}^{-1}C=I$$

and get the desired result.