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Math  /  Algebra

QuestionOfficial Time: 10:14:12
Question 3 [10 points] If u\mathbf{u} and v\mathbf{v} are the vectors below, solve the following equation for the vector x\mathbf{x}. u=[−5−3−2]v=[−555]x−2u+v=3(−u−x)−2vx=[000]\begin{array}{l} \mathbf{u}=\left[\begin{array}{l} -5 \\ -3 \\ -2 \end{array}\right] \mathbf{v}=\left[\begin{array}{c} -5 \\ 5 \\ 5 \end{array}\right] \\ \mathbf{x}-2 \mathbf{u}+\mathbf{v}=3(-\mathbf{u}-\mathbf{x})-2 \mathbf{v} \\ \mathbf{x}=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{array} SUBMIT AND MARK

Studdy Solution

STEP 1

What is this asking? Find the vector x\mathbf{x} that satisfies the given equation involving vectors u\mathbf{u} and v\mathbf{v}. Watch out! Vector arithmetic follows specific rules – be careful with those negative signs and scalar multiplications!

STEP 2

1. Expand the equation
2. Group like terms
3. Isolate x\mathbf{x}
4. Substitute u\mathbf{u} and v\mathbf{v}
5. Calculate x\mathbf{x}

STEP 3

Let's **distribute** those scalars on both sides of the equation to get rid of the parentheses.
Remember, this means multiplying each component of the vector by the scalar.
So, we have: x−2u+v=3(−u−x)−2v \mathbf{x} - 2\mathbf{u} + \mathbf{v} = 3(-\mathbf{u} - \mathbf{x}) - 2\mathbf{v} becomes x−2u+v=−3u−3x−2v \mathbf{x} - 2\mathbf{u} + \mathbf{v} = -3\mathbf{u} - 3\mathbf{x} - 2\mathbf{v} Great!

STEP 4

Now, let's get all the x\mathbf{x} terms on one side and all the u\mathbf{u} and v\mathbf{v} terms on the other.
We want to **isolate** x\mathbf{x}, after all!
Adding 3x3\mathbf{x} to both sides and adding 2u2\mathbf{u} and subtracting v\mathbf{v} from both sides gives us: x+3x=−3u+2u−2v−v \mathbf{x} + 3\mathbf{x} = -3\mathbf{u} + 2\mathbf{u} - 2\mathbf{v} - \mathbf{v}

STEP 5

This simplifies to: 4x=−u−3v 4\mathbf{x} = -\mathbf{u} - 3\mathbf{v} Awesome!

STEP 6

To completely **isolate** x\mathbf{x}, we need to divide both sides by 44: x=14(−u−3v) \mathbf{x} = \frac{1}{4}(-\mathbf{u} - 3\mathbf{v}) Looking good!

STEP 7

Now, let's **substitute** the given values of u\mathbf{u} and v\mathbf{v} into our equation: x=14(−[−5−3−2]−3[−555]) \mathbf{x} = \frac{1}{4}\left(-\begin{bmatrix} -5 \\ -3 \\ -2 \end{bmatrix} - 3\begin{bmatrix} -5 \\ 5 \\ 5 \end{bmatrix}\right)

STEP 8

First, we'll handle the **scalar multiplications**: x=14([532]+[15−15−15]) \mathbf{x} = \frac{1}{4}\left(\begin{bmatrix} 5 \\ 3 \\ 2 \end{bmatrix} + \begin{bmatrix} 15 \\ -15 \\ -15 \end{bmatrix}\right)

STEP 9

Next, let's **add** those vectors component-wise: x=14[20−12−13] \mathbf{x} = \frac{1}{4}\begin{bmatrix} 20 \\ -12 \\ -13 \end{bmatrix}

STEP 10

Finally, we **multiply** by 14\frac{1}{4}: x=[14⋅2014⋅(−12)14⋅(−13)]=[5−3−134] \mathbf{x} = \begin{bmatrix} \frac{1}{4} \cdot 20 \\ \frac{1}{4} \cdot (-12) \\ \frac{1}{4} \cdot (-13) \end{bmatrix} = \begin{bmatrix} 5 \\ -3 \\ -\frac{13}{4} \end{bmatrix}

STEP 11

So, the solution is x=[5−3−134]\mathbf{x} = \begin{bmatrix} 5 \\ -3 \\ -\frac{13}{4} \end{bmatrix}.

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