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: Define linear dependence

Three vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) in \( \mathbb{R}^n \) are linearly dependent if there exist scalars \( a, b, c \), not all zero, such that \( a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0} \). We can also check if one vector is a linear combination of the other two. Let's assume \( \mathbf{w} = x\mathbf{u} + y\mathbf{v} \), then we have the system of equations:

$$
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$$

Step2: Solve the first two equations

From the first equation \( 3 = x + 2y \), we can express \( x = 3 - 2y \). Substitute \( x = 3 - 2y \) into the second equation \( 4 = 2x + 3y \):

$$ 4 = 2(3 - 2y) + 3y $$

$$ 4 = 6 - 4y + 3y $$

$$ 4 = 6 - y $$

$$ y = 6 - 4 = 2 $$

Then \( x = 3 - 2 \cdot 2 = 3 - 4 = -1 \).

Step3: Check the third equation

Now check if \( x = -1 \) and \( y = 2 \) satisfy the third equation \( 7 = 5x + 6y \):

$$ 5(-1) + 6(2) = -5 + 12 = 7 $$

Which matches the third component of \( \mathbf{w} \). So \( \mathbf{w} = -1\mathbf{u} + 2\mathbf{v} \), or \( \mathbf{w} = 2\mathbf{v} - \mathbf{u} \). This shows the vectors are linearly dependent.

Answer:

The vectors are linearly dependent, and \( \mathbf{w} = 2\mathbf{v} - \mathbf{u} \) (or \( \mathbf{w} = - \mathbf{u} + 2\mathbf{v} \)).