Question

The number of solutions of the matrix equation $X^2$ = I other than I, is (where I is the 2 x 2 unit matrix)

• A. 0
• B. 1
• C. 2
• D. More than 2

Answer 1

Answer: (D)

More than 2

Answer 2

Let the $2 \times 2$ matrix be $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The given equation is $X^2 = I$, where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

First, calculate $X^2$:

$X^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2+bc & ab+bd \\ ca+dc & cb+d^2 \end{pmatrix}$

Equating this to $I$, we get the following system of equations:

1. $a^2+bc = 1$
2. $ab+bd = 0 \implies b(a+d) = 0$
3. $ca+dc = 0 \implies c(a+d) = 0$
4. $cb+d^2 = 1$

From equations (2) and (3), we have two cases:

Case 1: $a+d \neq 0$

If $a+d \neq 0$, then from $b(a+d)=0$ and $c(a+d)=0$, it must be that $b=0$ and $c=0$. In this case, $X$ is a diagonal matrix: $X = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$.

Substitute $b=0$ and $c=0$ into equations (1) and (4):

1. $a^2+0 = 1 \implies a^2=1 \implies a = \pm 1$
2. $0+d^2 = 1 \implies d^2=1 \implies d = \pm 1$

This gives us four possible diagonal matrices:

• If $a=1, d=1$: $X = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. This is the identity matrix.
• If $a=1, d=-1$: $X = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. This is a solution.
• If $a=-1, d=1$: $X = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$. This is a solution.
• If $a=-1, d=-1$: $X = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$. This is a solution.

Among these four, $I$ is the one to be excluded. So we have 3 solutions from this case: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.

Case 2: $a+d = 0$

If $a+d = 0$, then $d = -a$. Substitute $d=-a$ into equations (1) and (4):

1. $a^2+bc = 1$
2. $cb+(-a)^2 = 1 \implies cb+a^2 = 1$.

Both equations are identical ($a^2+bc=1$).

So, any matrix of the form $X = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}$ that satisfies $a^2+bc=1$ is a solution. We need to find solutions other than $I$. Note that $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ has $a=1, d=1$, so $a+d=2 \neq 0$. Thus $I$ is not included in this case.

Let's look for examples in this case:

• If $a=0$, then $bc=1$. We can choose any non-zero real number for $b$ and set $c=1/b$. For example, if $b=1$, $c=1$: $X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Check: $X^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. This is a solution. If $b=2$, $c=1/2$: $X = \begin{pmatrix} 0 & 2 \\ 1/2 & 0 \end{pmatrix}$. This is also a solution. Since $b$ can be any non-zero real number, there are infinitely many solutions of this form.

• If $a=1$, then $1+bc=1 \implies bc=0$. Since $a+d=0$, we have $d=-1$. If $b=0$, then $c$ can be any real number. $X = \begin{pmatrix} 1 & 0 \\ c & -1 \end{pmatrix}$. For example, if $c=1$: $X = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}$. Check: $X^2 = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1-1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. This is a solution. Note that if $c=0$, this gives $X = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, which was already found in Case 1. Since $c$ can be any real number (e.g., $c=1, 2, 3, \dots$), there are infinitely many solutions of this form (excluding $c=0$).

• If $a=-1$, then $1+bc=1 \implies bc=0$. Since $a+d=0$, we have $d=1$. If $b=0$, then $c$ can be any real number. $X = \begin{pmatrix} -1 & 0 \\ c & 1 \end{pmatrix}$. For example, if $c=1$: $X = \begin{pmatrix} -1 & 0 \\ 1 & 1 \end{pmatrix}$. This is a solution. Note that if $c=0$, this gives $X = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, which was already found in Case 1. Since $c$ can be any real number, there are infinitely many solutions of this form (excluding $c=0$).

From Case 2, we have already found infinitely many solutions other than $I$. For example, the matrices $\begin{pmatrix} 0 & b \\ 1/b & 0 \end{pmatrix}$ for any $b \in \mathbb{R}, b \neq 0$ are solutions. Also, $\begin{pmatrix} 1 & 0 \\ c & -1 \end{pmatrix}$ for any $c \in \mathbb{R}, c \neq 0$ are solutions. And $\begin{pmatrix} -1 & 0 \\ c & 1 \end{pmatrix}$ for any $c \in \mathbb{R}, c \neq 0$ are solutions.

Since we have found infinitely many solutions (e.g., $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 0 & 2 \\ 1/2 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}$, etc.), the number of solutions other than $I$ is more than 2.

The solutions are:

1. The three diagonal matrices found in Case 1: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.
2. Infinitely many non-diagonal matrices of the form $\begin{pmatrix} a & b \\ c & -a \end{pmatrix}$ where $a^2+bc=1$ and at least one of $b,c$ is non-zero. For example, $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 0 & 2 \\ 1/2 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 5 & -1 \end{pmatrix}$, $\begin{pmatrix} -1 & 7 & \\ 0 & 1 \end{pmatrix}$, etc.

Thus, the total number of solutions other than $I$ is infinite.