Search any question & find its solution
Question:
Answered & Verified by Expert
If $A=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right], B=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right], C=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$,
then which one of the following is not correct?
Options:
then which one of the following is not correct?
Solution:
2614 Upvotes
Verified Answer
The correct answer is:
$A B=B A$
Let $\mathrm{A}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right], \mathrm{C}=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots . .$ (1)
$\mathrm{B}^{2}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots \ldots . .(2)$
From (1) and (2), we have $\mathrm{A}^{2}=\mathrm{B}^{2}$
Now, $C^{2}=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
$\ldots \ldots . .(3)$
From (2) and (3), we have $\mathrm{B}^{2}=\mathrm{C}^{2}$
Now, $\mathrm{AB}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\mathrm{C}$
Now, we find
$\mathrm{BA}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right] \neq \mathrm{C}$
Hence $\mathrm{AB} \neq \mathrm{BA}$
Now, $A^{2}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots . .$ (1)
$\mathrm{B}^{2}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \ldots \ldots . .(2)$
From (1) and (2), we have $\mathrm{A}^{2}=\mathrm{B}^{2}$
Now, $C^{2}=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
$\ldots \ldots . .(3)$
From (2) and (3), we have $\mathrm{B}^{2}=\mathrm{C}^{2}$
Now, $\mathrm{AB}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]=\left[\begin{array}{cc}0 & -i \\ -i & 0\end{array}\right]=\mathrm{C}$
Now, we find
$\mathrm{BA}=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right] \neq \mathrm{C}$
Hence $\mathrm{AB} \neq \mathrm{BA}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.