Q107 / 180 · KCET 2015 · Mathematics
If \( A=\left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] \), then \( A^{2} \) equal to
- A\( \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
- B\( \left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right] \)
- Correct answerC\( \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
- D\( \left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right] \)
Solution
Given that \( A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
Now, \( A \cdot A=A^{2}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
\( =\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Therefore \( A^{2}=I \)
Now, \( A \cdot A=A^{2}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
\( =\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Therefore \( A^{2}=I \)
Hence, the correct answer is option: 'C'.