| If \[A=\left[ \begin{matrix} a & b \\ 0 & a \\ \end{matrix} \right]\] is nth root of \[{{I}_{2}}\], then choose the correct statements: |
| (i) if n is odd, \[a=1,\text{ }b=0\] |
| (ii) in n is odd, \[a=-1,\text{ }b=0\] |
| (iii) if n is even, \[a=1,\text{ }b=0\] |
| (iv) if n is even, \[a=-1,\text{ }b=0\] |
A) i, ii, iii
B) ii, iii, iv
C) i, ii, iii, iv
D) i, iii, iv
Correct Answer: D
Solution :
[d] if A is nth root of \[{{I}_{2}}\], then \[{{A}^{n}}={{I}_{2}}\]. Now, \[{{A}^{2}}=\left[ \begin{matrix} a & b \\ 0 & a \\ \end{matrix} \right]\left[ \begin{matrix} a & b \\ 0 & a \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}} & 2ab \\ 0 & {{a}^{2}} \\ \end{matrix} \right]\] \[{{A}^{3}}={{A}^{2}}A=\left[ \begin{matrix} {{a}^{2}} & 2ab \\ 0 & {{a}^{2}} \\ \end{matrix} \right]\left[ \begin{matrix} a & b \\ 0 & a \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{3}} & 3{{a}^{2}}b \\ 0 & {{a}^{3}} \\ \end{matrix} \right]\]Thus, \[{{A}^{n}}=\left[ \begin{matrix} {{a}^{n}} & n{{a}^{n-1}}b \\ 0 & {{a}^{n}} \\ \end{matrix} \right]\] Now, \[{{A}^{n}}=I\Rightarrow \left[ \begin{matrix} {{a}^{n}} & n{{a}^{n-1}}b \\ 0 & {{a}^{n}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow {{a}^{n}}=1,\,\,b=0\]
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