Question-9

What are the consequences of a square matrix being equal to its inverse?

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• We have \(A=A^{-1}\). Multiplying both sides by \(A\), we get \(A^{2}=I\). This is the first observation.

• Since \(\text{det}(A^{-1})=\frac{1}{\text{det}(A)}\), we have:

\[ \text{det}(A)^{2}=1\implies\text{det}(A)=\pm1 \]

• \(A=\pm I\) are two obvious solutions to \(A^{2}=I\). But these two are not the only matrices. Let us quickly construct a non-trivial solution:

\[ \begin{aligned} A & =\begin{bmatrix}a & b\\ c & d \end{bmatrix}\\ A^{2} & =\begin{bmatrix}a & b\\ c & d \end{bmatrix}\begin{bmatrix}a & b\\ c & d \end{bmatrix}\\ & =\begin{bmatrix}a^{2}+bc & b(a+d)\\ c(a+d) & d^{2}+bc \end{bmatrix}\\ & =\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix} \end{aligned} \]

We can set \(a=1,b=c=0,d=-1\). Thus the matrix \(A=\begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}\) satisfies \(A^{2}=I\). Besides, note that it is its own inverse and its determinant is \(-1\).