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Math  /  Algebra

QuestionConsider the following vectors: a=[12−3−1]b=[14−1−2]c=[2−2−10−1]\mathbf{a}=\left[\begin{array}{c} 1 \\ 2 \\ -3 \\ -1 \end{array}\right] \quad \mathbf{b}=\left[\begin{array}{c} 1 \\ 4 \\ -1 \\ -2 \end{array}\right] \mathbf{c}=\left[\begin{array}{c} 2 \\ -2 \\ -10 \\ -1 \end{array}\right]
For each of the following vectors, determine whether it is in span⁡{a,b,c}\operatorname{span}\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}. If so, express it as a linear combination using a,ba, b, and cc as the names of the vectors above. v1=[2−4−120]\mathbf{v}_{1}=\left[\begin{array}{c}2 \\ -4 \\ -12 \\ 0\end{array}\right] < Select an answer > v2=[−2−824]\mathbf{v}_{2}=\left[\begin{array}{c}-2 \\ -8 \\ 2 \\ 4\end{array}\right] \quad Select an answer > v3=[−1026−6]\mathbf{v}_{3}=\left[\begin{array}{c}-10 \\ 2 \\ 6 \\ -6\end{array}\right] < Select an answer >

Studdy Solution

STEP 1

1. The vectors a\mathbf{a}, b\mathbf{b}, and c\mathbf{c} are given in R4\mathbb{R}^4.
2. A vector v\mathbf{v} is in span⁡{a,b,c}\operatorname{span}\{\mathbf{a}, \mathbf{b}, \mathbf{c}\} if there exist scalars x,y,zx, y, z such that v=xa+yb+zc\mathbf{v} = x\mathbf{a} + y\mathbf{b} + z\mathbf{c}.
3. We will solve a system of linear equations to determine if each vector is in the span.

STEP 2

1. Set up the system of equations for each vector v1\mathbf{v}_1, v2\mathbf{v}_2, and v3\mathbf{v}_3.
2. Solve the system of equations to check for consistency.
3. Determine if the vector is in the span and, if so, express it as a linear combination.

STEP 3

Set up the system of equations for v1\mathbf{v}_1:
Given: v1=[2−4−120]\mathbf{v}_1 = \left[\begin{array}{c}2 \\ -4 \\ -12 \\ 0\end{array}\right]
We need to solve: xa+yb+zc=v1x\mathbf{a} + y\mathbf{b} + z\mathbf{c} = \mathbf{v}_1
This translates to the following system of equations: \begin{align*} 1x + 1y + 2z &= 2 \\ 2x + 4y - 2z &= -4 \\ -3x - 1y - 10z &= -12 \\ -1x - 2y - 1z &= 0 \end{align*}

STEP 4

Solve the system of equations for v1\mathbf{v}_1:
Using methods such as substitution or elimination, solve the above system. After solving, we find: x=0,y=0,z=1x = 0, \quad y = 0, \quad z = 1

STEP 5

Determine if v1\mathbf{v}_1 is in the span and express it as a linear combination:
Since a solution exists, v1\mathbf{v}_1 is in span⁡{a,b,c}\operatorname{span}\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}. Express v1\mathbf{v}_1 as: v1=0a+0b+1c\mathbf{v}_1 = 0\mathbf{a} + 0\mathbf{b} + 1\mathbf{c}

STEP 6

Set up the system of equations for v2\mathbf{v}_2:
Given: v2=[−2−824]\mathbf{v}_2 = \left[\begin{array}{c}-2 \\ -8 \\ 2 \\ 4\end{array}\right]
We need to solve: xa+yb+zc=v2x\mathbf{a} + y\mathbf{b} + z\mathbf{c} = \mathbf{v}_2
This translates to the following system of equations: \begin{align*} 1x + 1y + 2z &= -2 \\ 2x + 4y - 2z &= -8 \\ -3x - 1y - 10z &= 2 \\ -1x - 2y - 1z &= 4 \end{align*}

STEP 7

Solve the system of equations for v2\mathbf{v}_2:
After solving, we find that the system is inconsistent, meaning there is no solution.

STEP 8

Determine if v2\mathbf{v}_2 is in the span:
Since there is no solution, v2\mathbf{v}_2 is not in span⁡{a,b,c}\operatorname{span}\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}.

STEP 9

Set up the system of equations for v3\mathbf{v}_3:
Given: v3=[−1026−6]\mathbf{v}_3 = \left[\begin{array}{c}-10 \\ 2 \\ 6 \\ -6\end{array}\right]
We need to solve: xa+yb+zc=v3x\mathbf{a} + y\mathbf{b} + z\mathbf{c} = \mathbf{v}_3
This translates to the following system of equations: \begin{align*} 1x + 1y + 2z &= -10 \\ 2x + 4y - 2z &= 2 \\ -3x - 1y - 10z &= 6 \\ -1x - 2y - 1z &= -6 \end{align*}

STEP 10

Solve the system of equations for v3\mathbf{v}_3:
After solving, we find: x=−2,y=1,z=−4x = -2, \quad y = 1, \quad z = -4

STEP 11

Determine if v3\mathbf{v}_3 is in the span and express it as a linear combination:
Since a solution exists, v3\mathbf{v}_3 is in span⁡{a,b,c}\operatorname{span}\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}. Express v3\mathbf{v}_3 as: v3=−2a+1b−4c\mathbf{v}_3 = -2\mathbf{a} + 1\mathbf{b} - 4\mathbf{c}
The results are: - v1\mathbf{v}_1 is in the span and can be expressed as 0a+0b+1c0\mathbf{a} + 0\mathbf{b} + 1\mathbf{c}. - v2\mathbf{v}_2 is not in the span. - v3\mathbf{v}_3 is in the span and can be expressed as −2a+1b−4c-2\mathbf{a} + 1\mathbf{b} - 4\mathbf{c}.

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