Determinants Matrices Question 146
Question: The number of solutions of the matrix equation $ X^{2}= \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ \end{bmatrix} $ is
Options:
A) more than 2
B) 2
C) 0
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Let $ X=( \begin{matrix} a & b \\ c & d \\ \end{matrix} ) $
$ \Rightarrow X^{2}=( \begin{matrix} a^{2}+bc & ab+bd \\ ac+cd & bc+d^{2} \\ \end{matrix} ) $
$ \Rightarrow a^{2}+bc=1 $ and $ ab+bd=1 $
$ \Rightarrow b(a+d)=1 $
$ ac+cd=2\Rightarrow c(a+d)=2\Rightarrow 2b=c $ Also. $ bc+d^{2}=3\Rightarrow d^{2}-a^{2}=2 $
$ \Rightarrow (d-a)(a+d)=2\Rightarrow d-a=2b $ (Using $ bc=1-a^{2} $ )
$ \Rightarrow 2d=2b+1/b, $
$ 2a=1/b-2b $
$ d=b+1/2b, $
$ a=1/(2b)-b $
$ c=2b $
$ \Rightarrow ( b^{2}+\frac{1}{4b^{2}}+1 )+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{matrix} 0 & 1/\sqrt{2} \\ \sqrt{2} & \sqrt{2} \\ \end{matrix} ),( \begin{matrix} 0 & -1\sqrt{2} \\ -\sqrt{2} & -\sqrt{2} \\ \end{matrix} ),( \begin{matrix} 2\sqrt{6} & -1/\sqrt{6} \\ 2/\sqrt{6} & 4\sqrt{6} \\ \end{matrix} ) $
BETA
