bject: Mathematics II 1. Let $$ v_{1}^{+}=\left[\begin{array}{l}1 \ 2 \ 3\end{array} ight], v_{2}=\left[\begin{array}{l}4 \ 5 \ 6\end{array} ight], v_{3}=\left[\begin{array}{l}2 \ 1 \ 0\end{array} ight] $$ a Determine if the set $$ \left[v_{1}, v_{2}, v_{3} ight) $$ is linearly independent b If possible, find a linear dependence relation among $$ v_{1}, v_{2}, v $$

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bject: Mathematics II 1. Let $$ v_{1}^{+}=\left[\begin{array}{l}1 \ 2 \ 3\end{array}ight], v_{2}=\left[\begin{array}{l}4 \ 5 \ 6\end{array}ight], v_{3}=\left[\begin{array}{l}2 \ 1 \ 0\end{array}ight] $$ a Determine if the set $$ \left[v_{1}, v_{2}, v_{3}ight) $$ is linearly independent b If possible, find a linear dependence relation among $$ v_{1}, v_{2}, v $$

UpStudy AI · Accepted Answer

To determine if the set of vectors $$ \{ v_1, v_2, v_3 \} $$ is linearly independent, we need to check if the only solution to the equation

$$
c_1 v_1 + c_2 v_2 + c_3 v_3 = 0
$$

is $$ c_1 = c_2 = c_3 = 0 $$.

### Step 1: Set up the equation

We have:

$$
v_1 = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 4 \ 5 \ 6 \end{bmatrix}, \quad v_3 = \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix}
$$

The equation becomes:

$$
c_1 \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} + c_2 \begin{bmatrix} 4 \ 5 \ 6 \end{bmatrix} + c_3 \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}
$$

This leads to the following system of equations:

1. $$ c_1 + 4c_2 + 2c_3 = 0 $$ (Equation 1)
2. $$ 2c_1 + 5c_2 + c_3 = 0 $$ (Equation 2)
3. $$ 3c_1 + 6c_2 + 0c_3 = 0 $$ (Equation 3)

### Step 2: Form the augmented matrix

We can represent this system as an augmented matrix:

$$
\begin{bmatrix}
1 & 4 & 2 & | & 0 \
2 & 5 & 1 & | & 0 \
3 & 6 & 0 & | & 0
\end{bmatrix}
$$

### Step 3: Row reduce the matrix

We will perform row operations to reduce this matrix to row echelon form.

1. Subtract 2 times the first row from the second row:

$$
\begin{bmatrix}
1 & 4 & 2 & | & 0 \
0 & -3 & -3 & | & 0 \
3 & 6 & 0 & | & 0
\end{bmatrix}
$$

2. Subtract 3 times the first row from the third row:

$$
\begin{bmatrix}
1 & 4 & 2 & | & 0 \
0 & -3 & -3 & | & 0 \
0 & -6 & -6 & | & 0
\end{bmatrix}
$$

3. Divide the second row by -3:

$$
\begin{bmatrix}
1 & 4 & 2 & | & 0 \
0 & 1 & 1 & | & 0 \
0 & -6 & -6 & | & 0
\end{bmatrix}
$$

4. Add 6 times the second row to the third row:

$$
\begin{bmatrix}
1 & 4 & 2 & | & 0 \
0 & 1 & 1 & | & 0 \
0 & 0 & 0 & | & 0
\end{bmatrix}
$$

### Step 4: Solve the system

From the row-reduced form, we have:

1. $$ c_1 + 4c_2 + 2c_3 = 0 $$
2. $$ c_2 + c_3 = 0 $$

From the second equation, we can express $$ c_3 $$ in terms of $$ c_2 $$:

$$
c_3 = -c_2
$$

Substituting $$ c_3 $$ into the first equation:

$$
c_1 + 4c_2 + 2(-c_2) = 0 \implies c_1 + 4c_2 - 2c_2 = 0 \implies c_1 + 2c_2 = 0 \implies c_1 = -2c_2
$$

### Step 5: Conclusion

The general solution is:

$$
c_1 = -2c_2, \quad c_2 = c_2, \quad c_3 = -c_2
$$

This means that the vectors are linearly dependent because we can find non-trivial solutions (i.e., $$ c_2 $$ can be any non-zero scalar).

### Linear Dependence Relation

If we let $$ c_2 = 1 $$, we have:

$$
c_1 = -2, \quad c_2 = 1, \quad c_3 = -1
$$

Thus, a linear dependence relation among $$ v_1, v_2, v_3 $$ is:

$$
-2v_1 + 1v_2 - 1v_3 = 0
$$

or equivalently:

$$
2v_1 - v_2 + v_3 = 0
$$

### Summary

a) The set $$ \{ v_1, v_2, v_3 \} $$ is linearly dependent.

b) A linear dependence relation is $$ 2v_1 - v_2 + v_3 = 0 $$.
a) The set $$ \{ v_1, v_2, v_3 \} $$ is linearly dependent. b) A linear dependence relation is $$ 2v_1 - v_2 + v_3 = 0 $$.

bject: Mathematics II

Let
a Determine if the set is linearly independent
b If possible, find a linear dependence relation among

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bject: Mathematics II Let a Determine if the set is linearly independent b If possible, find a linear dependence relation among