Question

If $P = \begin{pmatrix}i & 0 & -i \\ 0 & -i & i \\ -i & i & 0\end{pmatrix}$ and $Q = \begin{pmatrix}-i & i \\ 0 & 0 \\ i & -i\end{pmatrix}$ then PQ is equal to
A. $\begin{pmatrix}-2 & 2 \\ -1 & 1 \\ -1 & 1 \end{pmatrix}$
B. $\begin{pmatrix}2 & -2 \\ 1 & -1 \\ 1 & -1 \end{pmatrix}$
C. $\begin{pmatrix}2 & -2 \\ -1 & 1 \end{pmatrix}$
D. $\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$

Answer 1

Hint: We must find the product of two given matrices in this question. We can look at the dimension of matrices given in the option and by discarding the not possible options we arrive at the answer.

Complete step by step solution: We have , $P = \begin{pmatrix}i & 0 & -i \\ 0 & -i & i \\ -i & i & 0\end{pmatrix}$ and $Q = \begin{pmatrix}-i & i \\ 0 & 0 \\ i & -i\end{pmatrix}$
The dimension of matrix P is 3×3 and the dimension of matrix Q is 3×2, then the dimension of matrix PQ is 3×2.
Looking at the dimension of PQ we can easily discard Option C and Option D. So, our answer will be one of the first two options.
If we observe the remaining options we can reach by finding the first element of the matrix, let us denote the first element of PQ as $a_11$. Then,
$a_{11} = i\times-i+0\times0+-i\times i$
$a_{11} = -i^2 - i^2$
$a_{11} = 1 + 1$
$a_{11} = 2$
Therefore, the first element of the matrix PQ is 2.

Option ‘B’ is correct

Note: When we deal with this kind of question, we do not find the product directly. Instead, try to look for ways in which we can discard one or more options. One such way may be looking at the dimension of the product matrix, as we have done in this question. Another may be looking at the matrix and finding the elements in certain positions of the matrix, like we have done in this question we found that the first element was different, and we found that element by matrix multiplication. In this way, we reach the correct answer quickly.