Question

1. \\(\

$$\begin{bmatrix} 2 \\\\ 0 \\\\ 6 \\end{bmatrix}$$

\

$$\begin{bmatrix} 1 & -3 & 4 \\end{bmatrix}$$

\\)

Explanation:

Step1: Recall matrix multiplication rule

When multiplying a column matrix (size \( m \times 1 \)) by a row matrix (size \( 1 \times n \)), the resulting matrix is of size \( m \times n \), and each element \( (i,j) \) is the product of the \( i \)-th element of the column matrix and the \( j \)-th element of the row matrix.

The column matrix is \(

$$\begin{bmatrix} 2 \\ 0 \\ 6 \end{bmatrix}$$

\) (size \( 3 \times 1 \)) and the row matrix is \(

$$\begin{bmatrix} 1 & -3 & 4 \end{bmatrix}$$

\) (size \( 1 \times 3 \)).

Step2: Calculate each element of the product matrix

• For the first row, first column: \( 2\times1 = 2 \)
• For the first row, second column: \( 2\times(-3)= -6 \)
• For the first row, third column: \( 2\times4 = 8 \)
• For the second row, first column: \( 0\times1 = 0 \)
• For the second row, second column: \( 0\times(-3)= 0 \)
• For the second row, third column: \( 0\times4 = 0 \)
• For the third row, first column: \( 6\times1 = 6 \)
• For the third row, second column: \( 6\times(-3)= -18 \)
• For the third row, third column: \( 6\times4 = 24 \)

So the product matrix is \(

$$\begin{bmatrix} 2 & -6 & 8 \\ 0 & 0 & 0 \\ 6 & -18 & 24 \end{bmatrix}$$

\)

Answer:

\(

$$\begin{bmatrix} 2 & -6 & 8 \\ 0 & 0 & 0 \\ 6 & -18 & 24 \end{bmatrix}$$

\)