Question

\(\text{ If } P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \text{ and } Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix} \text{ are two matrices, then } (PQ)^T \text{ will be:}\)

• A. \(\begin{bmatrix} 4 & 5 & 7 \\ -3 & -3 & 0 \\ 0 & -3 & -2 \end{bmatrix}\)
• B. \(\begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 5 & 5 & 2 \\ 7 & 6 & 7 \\ -9 & -7 & 0 \end{bmatrix}\)
• D. \(\begin{bmatrix} -2 & 4 & 8 \\ 7 & 5 & 7 \\ -8 & -2 & 6 \end{bmatrix}\)

Hint:

When performing matrix multiplication, especially with a column matrix and a row matrix, remember that the resulting matrix's dimensions are determined by the row count of the column matrix and the column count of the row matrix. The outer product of a \( m \times 1 \) and \( 1 \times n \) matrix results in a matrix of dimensions \( m \times n \). For transposing a matrix, simply swap the rows with columns, and this operation is essential for many matrix-related problems. Don't forget that matrix multiplication follows the distributive property, so you can multiply each element of the rows and columns accordingly.

Answer 1

The Correct Option is B

To find \( (PQ)' \), we first compute the product \( PQ \) where:

• \( P \) is a column matrix of order \( 3 \times 1 \)
• \( Q \) is a row matrix of order \( 1 \times 3 \)

The product \( PQ \) will be a \( 3 \times 3 \) matrix given by:

\[PQ = \begin{bmatrix}-1 \\2 \\1\end{bmatrix}\times\begin{bmatrix}2 & -4 & 1\end{bmatrix}=\begin{bmatrix}-2 & 4 & -1 \\4 & -8 & 2 \\2 & -4 & 1\end{bmatrix}\]

Next, we find the transpose \( (PQ)' \):

\[(PQ)' = \begin{bmatrix}-2 & 4 & -1 \\4 & -8 & 2 \\2 & -4 & 1\end{bmatrix}'=\begin{bmatrix}-2 & 4 & 2 \\4 & -8 & -4 \\-1 & 2 & 1\end{bmatrix}\]

Thus, the correct answer is:

\[\begin{bmatrix}-2 & 4 & 2 \\4 & -8 & -4 \\-1 & 2 & 1\end{bmatrix}\]

Answer 2

To solve for \((PQ)^T\), we need to perform several steps. First, compute the product of matrices \(P\) and \(Q\), then transpose the resulting matrix. 

Given matrices:

\(P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\)

\(Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix}\)

Step 1: Calculate \(PQ\).

Since \(P\) is a column matrix (3x1) and \(Q\) is a row matrix (1x3), \(PQ\) will be a 3x3 matrix:

\(PQ = \begin{bmatrix} -1 \times 2 & -1 \times -4 & -1 \times 1 \\ 2 \times 2 & 2 \times -4 & 2 \times 1 \\ 1 \times 2 & 1 \times -4 & 1 \times 1 \end{bmatrix}\)

Simplifying each element gives:

\(PQ = \begin{bmatrix} -2 & 4 & -1 \\ 4 & -8 & 2 \\ 2 & -4 & 1 \end{bmatrix}\)

Step 2: Transpose the matrix \(PQ\).

The transpose of a matrix is obtained by swapping its rows and columns, so \((PQ)^T\) is:

\((PQ)^T = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)

Therefore, the final result for \((PQ)^T\) is:

\(\begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)