If $P=\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]$ and $Q=\left[\begin{array}{lll}2 & -4 & 1\end{array}\right]$ are two matrices, then $(P Q)^{\prime}$ will be :

Let's find $(PQ)'$ where $P=\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$ and $Q=\begin{bmatrix} 2 & -4 & 1 \end{bmatrix}$.

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When multiplying a $3 \times 1$ matrix with a $1 \times 3$ matrix, we get a $3 \times 3$ matrix.

$PQ = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 2 & -4 & 1 \end{bmatrix}$

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

$= \begin{bmatrix} -2 & 4 & -1 \\ 4 & -8 & 2 \\ 2 & -4 & 1 \end{bmatrix}$

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To find $(PQ)'$, we transpose the matrix by converting rows to columns:

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

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Alternatively, we can use the property: $(PQ)' = Q'P'$

$P' = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

$Q' = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}$

$Q'P' = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}$