Thread: List Matrices

Find all 2 x 2 matrices such that $\displaystyle A^{-1}=A$.

Is there a systematic way to do this?

Let $\displaystyle \displaystyle A=\left(\begin{array}{ll}a & b\\c & d\end{array}\right),a,b,c,d\in\mathbf{R}$
$\displaystyle A=A^{-1}\Leftrightarrow A^2=I_2$.
Then $\displaystyle \displaystyle A=\left(\begin{array}{ll}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{array}\right)=\left(\begin{array}{ll}1 & 0\\0 & 1\end{array}\right)$.
So $\displaystyle \displaystyle \left\{\begin{array}{l}a^2+bc=1\\b(a+d)=0\\c(a+d)= 0\\d^2+bc=1\end{array}\right.$.
I. If $\displaystyle a+d\neq 0\Rightarrow b=c=0\Rightarrow a^2=d^2=1\Rightarrow a=1,d=1$ or $\displaystyle a=-1,d=-1\Rightarrow$
$\displaystyle \displaystyle A=\left(\begin{array}{ll}1 & 0\\0 & 1\end{array}\right)=I_2$ or $\displaystyle A=\left(\begin{array}{ll}-1 & 0\\0 & -1\end{array}\right)=-I_2$.
II. If $\displaystyle a+d=0\Rightarrow d=-a\Rightarrow A=\left(\begin{array}{ll}a & b\\c & -a\end{array}\right)$, so that $\displaystyle a^2+bc=1$

This is a rather more straightforward way.
If $\displaystyle A = \left[ {\begin{array}{lr}
a & b \\
c & d \\
\end{array}} \right]\quad $ then $\displaystyle A^{ - 1} = \left[ {\begin{array}{lr}
{\frac{d}{\Delta }} & {\frac{{ - b}}{\Delta }} \\
{\frac{{ - c}}{\Delta }} & {\frac{a}{\Delta }} \\
\end{array}} \right],\quad \Delta = ad - bc \not= 0$

That means that if $\displaystyle A = A^{ - 1} \Rightarrow \quad \begin{array}{lr}
{\frac{d}{\Delta } = a} & {\frac{{ - b}}{\Delta } = b} \\
{\frac{{ - c}}{\Delta } = c} & {\frac{a}{\Delta } = d} \\
\end{array}$.

Thus $\displaystyle \Delta = - 1\quad \Rightarrow \quad a = - d$; that any two-by-two matrix with determinant equal –1 and a=-d has the property.

But doesn't $\displaystyle a=-d$ exclude the possibility of the

[1 0]
[0 1]

matrix? or the

[-1 0]
[0 -1]

or are they just exceptions?

Originally Posted by DivideBy0

But doesn't $\displaystyle a=-d$ exclude the possibility of the

[1 0]
[0 1]

matrix? or the

[-1 0]
[0 -1]

or are they just exceptions?

[quote=red_dog;58719]
I. If $\displaystyle a+d\neq 0\Rightarrow b=c=0\Rightarrow a^2=d^2=1\Rightarrow a=1,d=1$ or $\displaystyle a=-1,d=-1\Rightarrow$
This is one of the two cases that red_dog gave and covers your answer.

-Dan