If $A = \ \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$ is $n^{th}$ root of $I_{2}$, then choose the correct statement: If $n$ is odd, $a = 1\ ,\ b = 0$ If $n$ is odd, $a = - 1\ ,\ b = 0$ If $n$ is even, $a = 1\ ,\ b = 0$ If $n$ is even, $a = - 1\ ,\ b = 0$

Question

Dr. Abubacker · Accepted Answer

(d)
If A is $n^{th}$ root of $I_{2},$ then$A^{n} = I_{2}$. Now,
$A^{2} = \begin{bmatrix}
a & b \
0 & a
\end{bmatrix}\begin{bmatrix}
a & b \
0 & a
\end{bmatrix} = \begin{bmatrix}
a^{2} & 2ab \
0 & a^{2}
\end{bmatrix}$
$A^{3} = A^{2}A = \begin{bmatrix}
a^{2} & 2ab \
0 & a^{2}
\end{bmatrix}\begin{bmatrix}
a & b \
0 & a
\end{bmatrix} = \begin{bmatrix}
a^{3} & 3a^{2}b \
0 & a^{3}
\end{bmatrix}$
Thus,
$A^{n} = \begin{bmatrix}
a^{n} & na^{n - 1}b \
0 & a^{n}
\end{bmatrix}$
Now,
$A^{n} = I \Rightarrow \begin{bmatrix}
a^{n} & na^{n - 1}b \
0 & a^{n}
\end{bmatrix} = \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix}$
$\Rightarrow a^{n} = 1,\ b = 0$

Q. If $A = \ \begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$ is $n^{th}$ root of $I_{2}$, then choose the correct statement: If $n$ is odd, $a = 1\ ,\ b = 0$ If $n$ is odd, $a = - 1\ ,\ b = 0$ If $n$ is even, $a = 1\ ,\ b = 0$ If $n$ is even, $a = - 1\ ,\ b = 0$

Explanation

Reviewed by Dr. Abubacker

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