QuestionConsider the system of linear equations $A \vec{x}=$ $A=\left[\begin{array}{rrr} 1 & 2 & -3 \\ -1 & 0 & -1 \\ -3 & -9 & 15 \end{array}\right]$ (a) Use matrix multiplication to determine each (i) Suppose $\overrightarrow{x_{1}}=$ $\left[\begin{array}{c} 4 \\ -1 \\ -1 \end{array}\right]$
Then $\overrightarrow{x_{1}}$ is a solution to $A \overrightarrow{x_{1}}=\vec{b}$ when $\vec{b}=$

Studdy Solution

STEP 1

1. We are given a matrix $A$ and a vector $\overrightarrow{x_1}$ .
2. We need to perform matrix multiplication to find the vector $\vec{b}$ such that $A \overrightarrow{x_1} = \vec{b}$ .
3. Matrix multiplication involves taking the dot product of each row of the matrix with the vector.

STEP 2

1. Write down the matrix $A$ and vector $\overrightarrow{x_1}$ .
2. Perform matrix multiplication to find $\vec{b}$ .
3. Verify the calculation.

STEP 3

Write down the matrix $A$ and the vector $\overrightarrow{x_1}$ :
\[ A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & 0 & -1 \\ -3 & -9 & 15 \end{bmatrix}, \quad \overrightarrow{x_1} = \begin{bmatrix} 4 \\ -1 \\ -1 \end{bmatrix} $

STEP 4

Perform matrix multiplication to find $\vec{b}$ :
Calculate each component of $\vec{b}$ by taking the dot product of each row of $A$ with $\overrightarrow{x_1}$ .
1. First row: $ (1)(4) + (2)(-1) + (-3)(-1) = 4 - 2 + 3 = 5 \]
2. Second row: $ (-1)(4) + (0)(-1) + (-1)(-1) = -4 + 0 + 1 = -3 \]
3. Third row: $ (-3)(4) + (-9)(-1) + (15)(-1) = -12 + 9 - 15 = -18 \]
Thus, $\vec{b} = \begin{bmatrix} 5 \\ -3 \\ -18 \end{bmatrix}$ .

STEP 5

Verify the calculation by checking each component:
1. First component: $5$ is correct.
2. Second component: $-3$ is correct.
3. Third component: $-18$ is correct.
The vector $\vec{b}$ is confirmed as $\begin{bmatrix} 5 \\ -3 \\ -18 \end{bmatrix}$ .
The vector $\vec{b}$ is $\boxed{\begin{bmatrix} 5 \\ -3 \\ -18 \end{bmatrix}}$ .

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