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

Thread: List Matrices

LinkBack

Thread Tools

Display

List Matrices

Similar Math Help Forum Discussions

how to list all of 7C4

List

help with list the

list of tables and list of figures

List Two Different Matrices A s.t. A^2=I

Search Tags