Question

The number of solutions of the matrix equation $X^{2} =\begin{bmatrix}1&1\\ 2&3\end{bmatrix} $ is

• A. more than $2$
• B. $2$
• C. $0$
• D. $1$

Answer 1

Answer: (A)

Let $X =\begin{pmatrix}a&b\\ c&d\end{pmatrix} $
$ \Rightarrow X^{2} = \begin{pmatrix}a^{2}+bc&ab+bd\\ ac+cd&bc+d^{2}\end{pmatrix} $
$ \Rightarrow a^{2}+bc =1$ and $ ab + bd = 1$
$\Rightarrow b\left(a+b\right)=1 $
$ac + cd = 2 $
$\Rightarrow c\left(a +d\right) = 2$
$ \Rightarrow 2b =c $
Also,
$bc + d^{2} = 3 $
$ \Rightarrow d^{2} - a^{2} = 2 $
$ \Rightarrow \left(d-a\right)\left(a+d\right) = 2$
$ \Rightarrow d-a =2b $ (using $bc = 1 - a^2)$
$a + d = 1/b $
$\Rightarrow 2d = 2b + 1/b, 2a = 1/b - 2b$
$ d = b + 1/2b, a = 1/(2b) -b$
$c = 2b$
$\Rightarrow \left(b^{2} + \frac{1}{4b^{2}} + 1\right)+2b^{2} = 3 $
$ \Rightarrow 3b^{2} + \frac{1}{4b^{2}} = 2 $
$\Rightarrow 3x + \frac{1}{4x} = 2 $
or $b =\pm \frac{1}{\sqrt{6}}$ or $b = \pm \frac{1}{\sqrt{2}}$
Therefore, matrices are
$\begin{pmatrix}0&1/\sqrt{2}\\ \sqrt{2}&\sqrt{2}\end{pmatrix},\begin{pmatrix}0&-1/\sqrt{2}\\ -\sqrt{2}&-\sqrt{2}\end{pmatrix},\begin{pmatrix}2/\sqrt{6}&-1/\sqrt{6}\\ 2/\sqrt{6}&4/\sqrt{6}\end{pmatrix}$