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 linear combination

Let's assume \( \mathbf{w} = a\mathbf{u} + b\mathbf{v} \), where \( a \) and \( b \) are scalars. So we have the system of equations:

$$
LATEXBLOCK0
$$

Step2: Solve first two equations

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

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

$$ 4 = 6 - 4b + 3b $$

$$ 4 = 6 - b $$

$$ b = 6 - 4 = 2 $$

Then \( a = 3 - 2(2) = 3 - 4 = -1 \).

Step3: Check third equation

Substitute \( a = -1 \) and \( b = 2 \) into the third equation:

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

Which matches the third component of \( \mathbf{w} \). So \( \mathbf{w} = -\mathbf{u} + 2\mathbf{v} \). Also, we can check linear dependence by seeing if the determinant of the matrix formed by the vectors (as columns or rows) is zero. The matrix \( A =

$$\begin{bmatrix}1&2&3\\2&3&4\\5&6&7\end{bmatrix}$$

\). The determinant of \( A \) is calculated as:

$$ \det(A) = 1(3\times7 - 4\times6) - 2(2\times7 - 4\times5) + 3(2\times6 - 3\times5) $$

$$ = 1(21 - 24) - 2(14 - 20) + 3(12 - 15) $$

$$ = 1(-3) - 2(-6) + 3(-3) $$

$$ = -3 + 12 - 9 = 0 $$

Since the determinant is zero, the vectors are linearly dependent.

Answer:

The vectors are linearly dependent. One linear combination is \( \mathbf{w} = -\mathbf{u} + 2\mathbf{v} \) (or other valid combinations like \( \mathbf{u} = 2\mathbf{v} - \mathbf{w} \), \( \mathbf{v} = \frac{1}{2}(\mathbf{u} + \mathbf{w}) \) etc.).