The number of solutions of the matrix equatio | vedclass

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(A) Let $X = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Then $X^2 = \begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix} = \begin

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more than $2$

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(A) Let $X = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Then $X^2 = \begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 2 & 3 \end{bmatrix}$.Equating the elements,we get:$1) a^2 + bc = 1$$2) b(a + d) = 1$$3) c(a + d) = 2$$4) bc + d^2 = 3$From $(2)$ and $(3)$,we have $\frac{c}{b} = 2$,so $c = 2b$.Subtracting $(1)$ from $(4)$,we get $d^2 - a^2 = 2$,which implies $(d - a)(d + a) = 2$.From $(2)$,$a + d = \frac{1}{b}$. Substituting this into the previous equation,we get $d - a = 2b$.Adding and subtracting $d - a = 2b$ and $d + a = \frac{1}{b}$,we find $d = b + \frac{1}{2b}$ and $a = \frac{1}{2b} - b$.Substituting $a$ and $c$ into $a^2 + bc = 1$,we get $(\frac{1}{2b} - b)^2 + b(2b) = 1$.$\frac{1}{4b^2} - 1 + b^2 + 2b^2 = 1 \implies 3b^2 + \frac{1}{4b^2} = 2$.Let $u = b^2$. Then $3u + \frac{1}{4u} = 2 \implies 12u^2 - 8u + 1 = 0$.Solving this quadratic,$(6u - 1)(2u - 1) = 0$,so $u = \frac{1}{6}$ or $u = \frac{1}{2}$.Since $b^2 = u$,we have $b = \pm \frac{1}{\sqrt{6}}$ or $b = \pm \frac{1}{\sqrt{2}}$.Each value of $b$ gives a unique matrix $X$. Thus,there are $4$ solutions,which is more than $2$.

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

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