Find the values of and x,y and z that satisfy the vector equation: (x multplied by a column vector (-2,3-1)) + (y multplied by a column vector (2,-2,1)) + (z multplied by a column vector (1,1,-2)) = (1 multplied by a column vector (3,-2,1)) + (3 multplied by a column vector (1,-1,1)) + 4(this is not a column vector). options are: z = 3y + 1 = -x/4 - 1/2 z = y + 6 = 2x + 2 No solutions. x = -3, y = 1, z = 6

We are given the vector equation:

\[
x \begin{pmatrix} -2 \\ 3 \\ -1 \end{pmatrix} + y \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} + z \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + 3 \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} + 4
\]

### Step 1: Simplify the right-hand side

The right-hand side consists of two vector terms plus a scalar value 4. First, calculate the sum of the vectors:

\[
\begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + 3 \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -5 \\ 4 \end{pmatrix}
\]

Next, we add the scalar 4 to each component of the resulting vector:

\[
\begin{pmatrix} 6 \\ -5 \\ 4 \end{pmatrix} + 4 = \begin{pmatrix} 10 \\ -1 \\ 8 \end{pmatrix}
\]

So the simplified right-hand side is:

\[
\begin{pmatrix} 10 \\ -1 \\ 8 \end{pmatrix}
\]

### Step 2: Set up the vector equation

Now we have the equation:

\[
x \begin{pmatrix} -2 \\ 3 \\ -1 \end{pmatrix} + y \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} + z \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 10 \\ -1 \\ 8 \end{pmatrix}
\]

We can now break this into three separate equations based on the components:

1. \( -2x + 2y + z = 10 \)
2. \( 3x - 2y + z = -1 \)
3. \( -x + y - 2z = 8 \)

### Step 3: Solve the system of equations

We have the system:

1. \( -2x + 2y + z = 10 \)
2. \( 3x - 2y + z = -1 \)
3. \( -x + y - 2z = 8 \)

We can solve this using substitution or elimination. Let’s solve it step by step.

#### Step 3.1: Solve for \( z \) from equation 1
From equation (1), solve for \( z \):

\[
z = 10 + 2x - 2y \tag{4}
\]

#### Step 3.2: Substitute \( z \) into equation (2)

Substitute \( z = 10 + 2x - 2y \) into equation (2):

\[
3x - 2y + (10 + 2x - 2y) = -1
\]
Simplify:

\[
3x - 2y + 10 + 2x - 2y = -1
\]
\[
5x - 4y + 10 = -1
\]
\[
5x - 4y = -11 \tag{5}
\]

#### Step 3.3: Substitute \( z \) into equation (3)

Substitute \( z = 10 + 2x - 2y \) into equation (3):

\[
-x + y - 2(10 + 2x - 2y) = 8
\]
Simplify:

\[
-x + y - 20 - 4x + 4y = 8
\]
\[
-5x + 5y = 28
\]
\[
x - y = -\frac{28}{5} \quad (\text{Divide by } -5)
\]



Find the values of x, y, and z that satisfy the vector equation: (x multiplied by a column vector (-2, 3, -1)) + (y multiplied by a column vector (2, -2, 1)) + (z multiplied by a column vector (1, 1, -2)) = (1 multiplied by a column vector (3, -2, 1)) + (3 multiplied by a column vector (1, -1, 1)) + 4. Options are: z = 3y + 1 = -x/4 - 1/2, z = y + 6 = 2x + 2, No solutions, x = -3, y = 1, z = 6.

Learn how to solve the system of vector equations to find values of x, y, and z. The problem involves vector addition, simplifying vector expressions, and solving a system of linear equations. Ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation