Question

The number of solutions of the matrix equation $X^2=\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]$ is

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

Answer 1

Answer: (A)

Let $X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$
$\Rightarrow X^2=\left(\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right)$
$\Rightarrow a^2+bc=1$ 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
$\left(\begin{array}{cc}0& 1/\sqrt{2}\\ \sqrt{2}& \sqrt{2}\end{array}\right),\left(\begin{array}{cc}0& -1/\sqrt{2}\\ -\sqrt{2}& -\sqrt{2}\end{array}\right),\left(\begin{array}{cc}2/\sqrt{6}& -1/\sqrt{6}\\ 2/\sqrt{6}& 4/\sqrt{6}\end{array}\right)$