Question

1. are the vectors below linearly dependent? if so, write one of them as a linear combination of the others:

$mathbf{u} = \begin{bmatrix} 1 \\ 2 \\ 5 end{bmatrix}$, $mathbf{v} = \begin{bmatrix} 2 \\ 3 \\ 6 end{bmatrix}$, $mathbf{w} = \begin{bmatrix} 3 \\ 4 \\ 7 end{bmatrix}$

Explanation:

Step1: Assume a linear combination

Let \( \mathbf{w} = a\mathbf{u} + b\mathbf{v} \), where \( a \) and \( b \) are scalars. So we have the system of equations:
For the first component: \( 3 = a(1) + b(2) \)
For the second component: \( 4 = a(2) + b(3) \)
For the third component: \( 7 = a(5) + b(6) \)

Step2: Solve the first two equations

From the first equation: \( a + 2b = 3 \) (Equation 1)
From the second equation: \( 2a + 3b = 4 \) (Equation 2)
Multiply Equation 1 by 2: \( 2a + 4b = 6 \) (Equation 3)
Subtract Equation 2 from Equation 3: \( (2a + 4b) - (2a + 3b) = 6 - 4 \)
Simplify: \( b = 2 \)
Substitute \( b = 2 \) into Equation 1: \( a + 2(2) = 3 \)
So \( a = 3 - 4 = -1 \)

Step3: Verify with the third component

Check if \( -1\mathbf{u} + 2\mathbf{v} \) gives \( \mathbf{w} \) for the third component.
\( -1(5) + 2(6) = -5 + 12 = 7 \), which matches the third component of \( \mathbf{w} \).

So \( \mathbf{w} = -1\mathbf{u} + 2\mathbf{v} \), meaning the vectors are linearly dependent, and \( \mathbf{w} \) is a linear combination of \( \mathbf{u} \) and \( \mathbf{v} \) (or we could also check other combinations, but this works).

Answer:

The vectors are linearly dependent, and \( \mathbf{w} = -\mathbf{u} + 2\mathbf{v} \) (or other valid combinations like \( \mathbf{v}=\frac{1}{2}\mathbf{u}+\frac{1}{2}\mathbf{w} \), \( \mathbf{u}=2\mathbf{v}-\mathbf{w} \) are also possible, but the above is one solution).