Let M be the vector space of 2 times 2 real-valued matrices.M=begin{bmatrix}a & b c & d end{bmatrix}and define M^{#}=begin{bmatrix}d & b c & a end{bmatrix}

Let M be the vector space of $$2 \times 2$$ real-valued matrices.
$$M=\begin{bmatrix}a & b \\c & d \end{bmatrix}$$
and define $$M^{\#}=\begin{bmatrix}d & b \\c & a \end{bmatrix}$$ Characterize the matrices M such that $$M^{\#}=M^{-1}$$

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Step 1
Let M be the vector space of $$2 \times 2$$ real-valued matrices.
$$M=\begin{bmatrix}a & b \\c & d \end{bmatrix}$$
and define $$M^{\#}=\begin{bmatrix}d & b \\c & a \end{bmatrix}$$
To characterize matrix M such that $$M^{\#}=M^{-1}$$
Therefore for findiny $$M^{-1}$$ we want $$det(M) \neq 0$$
Therefore $$det(M)=ad-bc \neq 0$$
If $$det(M) \neq 0$$
then $$M^{-1}=\frac{1}{ad-bc} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}$$
So, $$M^{\#}=M^{-1}$$
$$\Rightarrow \begin{bmatrix}d & b \\c & a \end{bmatrix}= \frac{1}{(ad-bc)} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix}(ad-bc)d & (ad-bc)b \\ (ad-bc)c & (ad-bc)a \end{bmatrix} = \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}$$
Case (1)
By comparing if either a=0 pr d=0 of both then $$ab-bc=-bc \neq 0 (\text{as } ad-bc \neq 0 )$$
So, b and c both non zero
$$\Rightarrow -b=b(ad-bc) \neq ab-bc=-1$$
Step 2
$$M^{\#}=M^{-1}$$ and if either a=0 or d=0 of both then ad-bc=-1
Case ( 2 )
If $$a,d \neq 0$$ then
$$a=(ad-bc)a \Rightarrow ad-bc=1$$
So, $$b=-b \text{ and } c=-c$$
$$\Rightarrow 2b=0\ and\ 2c=0$$
$$\Rightarrow b=0 \Rightarrow c=0$$
So, $$M=\begin{bmatrix}a & 0 \\ 0 & d \end{bmatrix}$$ with ad=1