- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.16 Exercises
12. $\quad$ The equation $A^2 = I$ is satisfied by each of the 2 x 2 matrices $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 \\ c & -1 \end{bmatrix}, \quad \begin{bmatrix} 1 & b \\ 0 & -1 \end{bmatrix},$$ where $b$ and $c$ are arbitrary real numbers. Find all 2 x 2 matrices $A$ such that $A^2 = I.$
Solution. $\quad$ Let $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ be a $2 \times 2$ matrix such that
\begin{align*}
A^2 &= \begin{bmatrix}
a^2 + bc & ab + bd
\\
ac + cd & cb + d^2
\end{bmatrix}
\\
&= \begin{bmatrix}
1 & 0
\\
0 & 1
\end{bmatrix}.
\end{align*}
This gives us the following system of equations
\begin{align*}
a^2 + bc &= 1
\\
(a + d)b &= 0
\\
(a + d)c &= 0
\\
cb + d^2 &= 1
\end{align*}
The middle two equations give us two potential cases:
(1) $\quad$ $(a + d) \neq 0.$ Then, $b = c = 0,$ and $a = d = 1$ or $a = d = -1.$ In other words, if $a + d \neq 0,$ $A = I$ or $A = -I.$
(2) $\quad$ $(a + d) = 0.$ $\quad$ Then, $d = -a$ and $a$ is a solution to $a^2 = 1 - bc$ for arbitrary $b$ and $c. \quad \blacksquare$