The number of, solutions of the matrix equation X2=1123 is

Let X=a bc da bc d⇒ X2=a2+bcab+bdac+cdbc+d2⇒ a2+bc=1 and ab+bd=1⇒b(a+d)=1 ac+cd=2⇒c(a+d)=2⇒2b=cAlso,bc+d2=3⇒d2−a2=2⇒ (d−a)(a+d)=2⇒d−a=2b⇒ a+d=1/b⇒ 2d=2b+1/b, 2a=1/b−2bd=b+1/2b, a=1/(2b)−bc=2b⇒ b2+14b2+1+2b2=3⇒ 3b2+14b2=2⇒ 12b2−8b2+1=0⇒ 6b2−12b2−1=0 or b=±16 or b=±12Therefore, matrices are01/222,0−1/2−2−2,2/61/62/64/6

The number of, solutions of the matrix equation X2=1123 is

Let X=a bc da bc d⇒ X2=a2+bcab+bdac+cdbc+d2⇒ a2+bc=1 and ab+bd=1⇒b(a+d)=1 ac+cd=2⇒c(a+d)=2⇒2b=cAlso,bc+d2=3⇒d2−a2=2⇒ (d−a)(a+d)=2⇒d−a=2b⇒ a+d=1/b⇒ 2d=2b+1/b, 2a=1/b−2bd=b+1/2b, a=1/(2b)−bc=2b⇒ b2+14b2+1+2b2=3⇒ 3b2+14b2=2⇒ 12b2−8b2+1=0⇒ 6b2−12b2−1=0 or b=±16 or b=±12Therefore, matrices are01/222,0−1/2−2−2,2/61/62/64/6

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

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Detailed Solution

Let $X = (\begin{array}{ll} a b \\ c d \end{array}) (\begin{array}{ll} a b \\ c d \end{array})$

$\begin{array}{l} \Rightarrow X^{2} = (\begin{array}{ll} a^{2} + bc & ab + bd \\ ac + cd & bc + d^{2} \end{array}) \\ \Rightarrow a^{2} + bc = 1 and ab + bd = 1 \Rightarrow b (a + d) = 1 \\ ac + cd = 2 \Rightarrow c (a + d) = 2 \Rightarrow 2 b = c \end{array}$

Also,

$\begin{array}{l} bc + d^{2} = 3 \Rightarrow d^{2} - a^{2} = 2 \\ \Rightarrow (d - a) (a + d) = 2 \Rightarrow d - a = 2 b \\ \Rightarrow a + d = 1 / b \\ \Rightarrow 2 d = 2 b + 1 / b, 2 a = 1 / b - 2 b \\ d = b + 1 / 2 b, a = 1 / (2 b) - b \\ c = 2 b \end{array}$

$\begin{array}{l} \Rightarrow (b^{2} + \frac{1}{4 b^{2}} + 1) + 2 b^{2} = 3 \\ \Rightarrow 3 b^{2} + \frac{1}{4 b^{2}} = 2 \\ \Rightarrow 12 b^{2} - 8 b^{2} + 1 = 0 \\ \Rightarrow (6 b^{2} - 1) (2 b^{2} - 1) = 0 \\ or b = \pm \frac{1}{\sqrt{6}} or b = \pm \frac{1}{\sqrt{2}} \end{array}$

Therefore, matrices are

$(\begin{array}{cc} 0 & 1 / \sqrt{2} \\ \sqrt{2} & \sqrt{2} \end{array}), (\begin{array}{cc} 0 & - 1 / \sqrt{2} \\ - \sqrt{2} & - \sqrt{2} \end{array}), (\begin{array}{cc} 2 / \sqrt{6} & 1 / \sqrt{6} \\ 2 / \sqrt{6} & 4 / \sqrt{6} \end{array})$