To determine whether the given vectors are linearly independent, we need to check if one vector can be expressed as a scalar multiple of the other. This is because two vectors are linearly independent if and only if neither is a scalar multiple of the other.
Let's examine the given vectors:
[tex]\[
\mathbf{v}_1 = \left[\begin{array}{r}
12 \\
-6 \\
3
\end{array}\right]
\][/tex]
[tex]\[
\mathbf{v}_2 = \left[\begin{array}{r}
8 \\
-4 \\
2
\end{array}\right]
\][/tex]
We will check if there exists a scalar [tex]\( k \)[/tex] such that [tex]\( \mathbf{v}_1 = k \mathbf{v}_2 \)[/tex].
1. Compare the first components:
[tex]\[
12 = k \cdot 8 \implies k = \frac{12}{8} = \frac{3}{2}
\][/tex]
2. Compare the second components:
[tex]\[
-6 = k \cdot (-4) \implies -6 = \frac{3}{2} \cdot (-4) \implies -6 = -6
\][/tex]
3. Compare the third components:
[tex]\[
3 = k \cdot 2 \implies 3 = \frac{3}{2} \cdot 2 \implies 3 = 3
\][/tex]
The scalar [tex]\( k = \frac{3}{2} \)[/tex] works consistently for all components of the vectors. This means that [tex]\( \mathbf{v}_1 \)[/tex] is indeed a scalar multiple of [tex]\( \mathbf{v}_2 \)[/tex].
Since [tex]\( \mathbf{v}_1 \)[/tex] can be expressed as [tex]\( k \mathbf{v}_2 \)[/tex] with [tex]\( k = \frac{3}{2} \)[/tex], the vectors are linearly dependent. Therefore, they are not linearly independent.
To summarize:
- [tex]\( \mathbf{v}_1 = \frac{3}{2} \mathbf{v}_2 \)[/tex]
- The vectors are linearly dependent.
Hence:
- Linearly independent: False
- Linearly dependent: True
