# Find all real matrices A , such that A²= I(2) (A is a matrix of second order)?

Jul 3, 2016

Solutions:

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right)$, $\left(\begin{matrix}1 & 0 \\ c & - 1\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 \\ c & 1\end{matrix}\right) , \left(\begin{matrix}a & b \\ \frac{1 - {a}^{2}}{b} & - a\end{matrix}\right)$

#### Explanation:

Suppose $A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

Then:

${A}^{2} = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) = \left(\begin{matrix}{a}^{2} + b c & b \left(a + d\right) \\ c \left(a + d\right) & {d}^{2} + b c\end{matrix}\right)$

So if we want ${A}^{2} = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$ then we get the following equations:

$\left\{\begin{matrix}{a}^{2} + b c = 1 \\ b \left(a + d\right) = 0 \\ c \left(a + d\right) = 0 \\ {d}^{2} + b c = 1\end{matrix}\right.$

From the second equation, we have $b = 0$ and/or $a + d = 0$

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Case $\boldsymbol{b = 0}$

$a = \pm 1$, $d = \pm 1$

If additionally $a = - d$ then $c$ can have any value. Otherwise $c = 0$.

So the case $b = 0$ results in solutions:

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right)$, $\left(\begin{matrix}1 & 0 \\ c & - 1\end{matrix}\right)$, $\left(\begin{matrix}- 1 & 0 \\ c & 1\end{matrix}\right)$

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Case $\boldsymbol{a + d = 0}$

$b c = 1 - {a}^{2}$

This results in solutions:

$\left(\begin{matrix}a & b \\ \frac{1 - {a}^{2}}{b} & - a\end{matrix}\right)$

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All of these solutions work.

Jul 3, 2016

See below

#### Explanation:

Given

$A = \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right)$

we need all $A$ such that

$A \cdot A = I \left(2\right)$

where

$I \left(2\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

Solving the system of resulting equations

{(a_(11)^2 + a_(12) a_(21) = 1), (a_(11) a_(12) + a_(12) a_(22) = 0), (a_(11) a_(21) + a_(21) a_(22) = 0), (a_(12) a_(21) + a_(22)^2 = 1) :}

we have

${A}_{1} = \left(\begin{matrix}{\lambda}_{1} & {\lambda}_{2} \\ \frac{1 - {\lambda}_{1}^{2}}{\lambda} _ 2 & - {\lambda}_{1}\end{matrix}\right)$
${A}_{2} = \left(\begin{matrix}- 1 & 0 \\ {\lambda}_{1} & 1\end{matrix}\right)$
${A}_{3} = - {A}_{2}$
${A}_{4} = I \left(2\right)$
${A}_{5} = - I \left(2\right)$

For ${\lambda}_{1} \in \mathbb{R}$ and $\left({\lambda}_{2} \ne 0\right) \in \mathbb{R}$