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Question Number 119124 by abdullahquwatan last updated on 22/Oct/20

$$\mathrm{if}\mathrm{matrix}{\mathrm{A}}^2=\left(\begin{array}{c}13\\ 01\end{array}\right)\mathrm{then}\mathrm{matrix}\mathrm{A}=...$$

Answered by benjo_mathlover last updated on 22/Oct/20

$$\iff A=\left(\begin{array}{c}1\frac{3}{2}\\ 01\end{array}\right)$$$$A^2=\left(\begin{array}{c}1\frac{3}{2}\\ 01\end{array}\right)\left(\begin{array}{c}1\frac{3}{2}\\ 01\end{array}\right)=\left(\begin{array}{c}1\frac{6}{2}\\ 01\end{array}\right)$$

Commented by abdullahquwatan last updated on 22/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mindispower last updated on 23/Oct/20

$$det\left(A^2\right)=1,ifdetA=1$$$$det(A^2-{aI}_2)=0\Rightarrow a=1$$$$ifdet(A-I_2)=0$$$$A=\left(\begin{array}{c}ab\\ cd\end{array}\right)$$$$X^2-(a+d)X+det\left(A\right)=P\left(X\right)$$$$P\left(1\right)=1-(a+d)+1=0\Rightarrow a+d=2$$$$A^2=\left(\begin{array}{c}{a}^2+bcba+bd\\ ac+cdbc+d^2\end{array}\right)$$$$ac+cd=c(a+d)=0\Rightarrow c=0$$$$ba+bd=b(a+d)=2b=3\Rightarrow b=\frac{3}{2}$$$$bc+d^2=1\Rightarrow d=\underset{-}{+}1$$$$a=\underset{-}{+}1,a+d=2\Rightarrow a=d=1$$$$A=\left(\begin{array}{c}1\frac{3}{2}\\ 01\end{array}\right)$$

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