List Matrices

**DivideBy0** · July 2nd 2007, 07:47 AM

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

Is there a systematic way to do this?

**red_dog** · July 2nd 2007, 08:23 AM

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

**Plato** · July 2nd 2007, 10:38 AM

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

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

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

**DivideBy0** · July 2nd 2007, 11:17 PM

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

[1 0]
[0 1]

matrix? or the

[-1 0]
[0 -1]

or are they just exceptions?

**topsquark** · July 3rd 2007, 03:37 AM

Originally Posted by DivideBy0

But doesn't $a=-d$ 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 $a+d\neq 0\Rightarrow b=c=0\Rightarrow a^2=d^2=1\Rightarrow a=1,d=1$ or $a=-1,d=-1\Rightarrow$
This is one of the two cases that red_dog gave and covers your answer.

-Dan

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