Question

If $P = \begin{bmatrix} 0 & i & i \\ -i & 0 & i \\ -i & -i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} 0 & 0 & -i \\ 0 & 0 & -i \\ i & i & 0 \end{bmatrix}$, then $PQ$ is equal to

• A. $\begin{bmatrix} -2 & -1 & -1 \\ 1 & -2 & -1 \\ 1 & 1 & -2 \end{bmatrix}$
• B. $\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix}$
• C. $\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Hint:

Remember $i^2 = -1$ when multiplying complex matrices.

Answer 1

The Correct Option is B

To find the product \( PQ \), where \( P = \begin{bmatrix} 0 & i & i \\ -i & 0 & i \\ -i & -i & 0 \end{bmatrix} \) and \( Q = \begin{bmatrix} 0 & 0 & -i \\ 0 & 0 & -i \\ i & i & 0 \end{bmatrix} \), we need to multiply these matrices step-by-step.

Matrix multiplication is done by taking the dot product of the rows of the first matrix with the columns of the second matrix.

1. Compute the element in the first row, first column of \( PQ \): \(0 \cdot 0 + i \cdot 0 + i \cdot i = 0 + 0 - 1 = -1\)
2. Compute the element in the first row, second column of \( PQ \): \(0 \cdot 0 + i \cdot 0 + i \cdot i = 0 + 0 - 1 = -1\)
3. Compute the element in the first row, third column of \( PQ \): \(0 \cdot -i + i \cdot -i + i \cdot 0 = 0 + 1 + 0 = 1\)
4. Compute the element in the second row, first column of \( PQ \): \(-i \cdot 0 + 0 \cdot 0 + i \cdot i = 0 + 0 - 1 = -1\)
5. Compute the element in the second row, second column of \( PQ \): \(-i \cdot 0 + 0 \cdot 0 + i \cdot i = 0 + 0 - 1 = -1\)
6. Compute the element in the second row, third column of \( PQ \): \(-i \cdot -i + 0 \cdot i + i \cdot 0 = 1 + 0 + 0 = 1\)
7. Compute the element in the third row, first column of \( PQ \): \(-i \cdot 0 + -i \cdot 0 + 0 \cdot i = 0 + 0 + 0 = 0\)
8. Compute the element in the third row, second column of \( PQ \): \(-i \cdot 0 + -i \cdot 0 + 0 \cdot i = 0 + 0 + 0 = 0\)
9. Compute the element in the third row, third column of \( PQ \): \(-i \cdot -i + -i \cdot i + 0 \cdot 0 = 1 - 1 + 0 = 0\)

Thus, the resulting product \( PQ \) is:

2	-1	-1
-1	2	-1
-1	-1	2

Therefore, the correct answer is:

\(\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix}\)