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Q7. Linear Dependence and Independence

Definitions:

Linear Dependence:
A set of vectors $\{\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_k\}$ in a vector space $V$ is said to be linearly dependent if there exist scalars $c_1, c_2, \ldots, c_k$, not all zero, such that their linear combination equals the zero vector:

$c_1\vec{v}_1 + c_2\vec{v}_2 + \ldots + c_k\vec{v}_k = \vec{0}$

This implies that at least one vector in the set can be expressed as a linear combination of the others.
Linear Independence:
A set of vectors $\{\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_k\}$ in a vector space $V$ is said to be linearly independent if the only way to form the zero vector as a linear combination of these vectors is by setting all scalars to zero:

$c_1\vec{v}_1 + c_2\vec{v}_2 + \ldots + c_k\vec{v}_k = \vec{0} \quad \text{implies that} \quad c_1 = c_2 = \ldots = c_k = 0$

This means that no vector in the set can be expressed as a linear combination of the others.

Test for Linear Dependence/Independence:

Given the set of vectors: $\vec{v}_1 = (1,2,3)$, $\vec{v}_2 = (2,4,6)$, $\vec{v}_3 = (3,6,9)$.

We need to determine if there exist scalars $c_1, c_2, c_3$, not all zero, such that:

$c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 = \vec{0}$

Substitute the given vectors:

$c_1(1,2,3) + c_2(2,4,6) + c_3(3,6,9) = (0,0,0)$

This expands into a system of linear equations:

$1c_1 + 2c_2 + 3c_3 = 0$
$2c_1 + 4c_2 + 6c_3 = 0$
$3c_1 + 6c_2 + 9c_3 = 0$

We can observe a relationship between the vectors directly. Notice that:

$\vec{v}_2 = (2,4,6) = 2(1,2,3) = 2\vec{v}_1$
$\vec{v}_3 = (3,6,9) = 3(1,2,3) = 3\vec{v}_1$

From $\vec{v}_2 = 2\vec{v}_1$, we can rearrange this to $2\vec{v}_1 - \vec{v}_2 + 0\vec{v}_3 = \vec{0}$.

Here, we have scalars $c_1=2$, $c_2=-1$, $c_3=0$. Since not all these scalars are zero (e.g., $c_1=2 \neq 0$ and $c_2=-1 \neq 0$), the set of vectors is linearly dependent.

Alternatively, using the system of equations:

We can write the augmented matrix for this system:

$\begin{pmatrix}
  1 & 2 & 3 & | & 0 \\
  2 & 4 & 6 & | & 0 \\
  3 & 6 & 9 & | & 0
\end{pmatrix}$

Perform Row Operations:

$R_2 \to R_2 - 2R_1$:
$R_3 \to R_3 - 3R_1$:

$\begin{pmatrix}
  1 & 2 & 3 & | & 0 \\
  0 & 0 & 0 & | & 0 \\
  0 & 0 & 0 & | & 0
\end{pmatrix}$

The system reduces to a single equation: $c_1 + 2c_2 + 3c_3 = 0$.

This system has infinitely many solutions (non-trivial solutions). We can choose $c_2$ and $c_3$ as free variables.

Let $c_2 = s$ and $c_3 = t$, where $s, t$ are any real numbers (scalars).

Then $c_1 = -2s - 3t$.

For example, if we choose $s=1$ and $t=0$, then $c_1 = -2$, $c_2 = 1$, $c_3 = 0$.

This gives the linear combination: $-2(1,2,3) + 1(2,4,6) + 0(3,6,9) = (-2,-4,-6) + (2,4,6) + (0,0,0) = (0,0,0)$.

Since we found scalars $c_1=-2, c_2=1, c_3=0$ (not all zero) that satisfy the equation, the vectors are linearly dependent.

Conclusion:

The given set of vectors $\{(1,2,3), (2,4,6), (3,6,9)\}$ is linearly dependent because there exist non-zero scalars (e.g., $c_1=2, c_2=-1, c_3=0$) such that their linear combination equals the zero vector.

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