Question

1. \\(\

$$\begin{bmatrix}-5\\\\6\\\\0\\end{bmatrix}$$

\cdot\

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

\\)

Explanation:

Step1: Recall matrix multiplication rule

To multiply a \( 3 \times 1 \) matrix (column vector) \(

$$\begin{bmatrix} a \\ b \\ c \end{bmatrix}$$

\) with a \( 1 \times 2 \) matrix \(

$$\begin{bmatrix} d & e \end{bmatrix}$$

\), the result is a \( 3 \times 2 \) matrix where each element \( (i,j) \) is \( (\text{row } i \text{ of first matrix}) \times (\text{column } j \text{ of second matrix}) \).

Given the first matrix \(

$$\begin{bmatrix} -5 \\ 6 \\ 0 \end{bmatrix}$$

\) (let's call it \( A \)) and the second matrix \(

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

\) (let's call it \( B \)).

Step2: Perform the multiplication

For the first row of the resulting matrix:

• First element (row 1, column 1): \( (-5) \times 3 = -15 \)
• Second element (row 1, column 2): \( (-5) \times (-1) = 5 \)

For the second row of the resulting matrix:

• First element (row 2, column 1): \( 6 \times 3 = 18 \)
• Second element (row 2, column 2): \( 6 \times (-1) = -6 \)

For the third row of the resulting matrix:

• First element (row 3, column 1): \( 0 \times 3 = 0 \)
• Second element (row 3, column 2): \( 0 \times (-1) = 0 \)

So the resulting matrix is \(

$$\begin{bmatrix} -15 & 5 \\ 18 & -6 \\ 0 & 0 \end{bmatrix}$$

\)

Answer:

\(

$$\begin{bmatrix} -15 & 5 \\ 18 & -6 \\ 0 & 0 \end{bmatrix}$$

\)