Are the vectors \left[\begin{array}{c}
-2\cr
1\cr
-5\cr
0
\end{array}\right], \left[\begin{array}{c}
3\cr
4\cr
0\cr
-4
\end{array}\right], \left[\begin{array}{c}
-4\cr
3\cr
2\cr
2
\end{array}\right] and \left[\begin{array}{c}
-1\cr
-4\cr
5\cr
4
\end{array}\right] linearly independent?

Choose


If they are linearly dependent, find scalars that are not all zero such that the equation below is true. If they are linearly independent, find the only scalars that will make the equation below true.

 \left[\begin{array}{c}
-2\cr
1\cr
-5\cr
0
\end{array}\right] + 
 \left[\begin{array}{c}
3\cr
4\cr
0\cr
-4
\end{array}\right] + 
 \left[\begin{array}{c}
-4\cr
3\cr
2\cr
2
\end{array}\right] + 
 \left[\begin{array}{c}
-1\cr
-4\cr
5\cr
4
\end{array}\right] = \left[\begin{array}{c}
0\cr
0\cr
0\cr
0
\end{array}\right].

To determine if the vectors are linearly independent, we need to check if there exists a set of scalars \( c_1, c_2, c_3, c_4 \), not all zero, such that:

\[
c_1 \begin{bmatrix} -2 \\ 1 \\ -5 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 4 \\ 0 \\ -4 \end{bmatrix} + c_3 \begin{bmatrix} -4 \\ 3 \\ 2 \\ 2 \end{bmatrix} + c_4 \begin{bmatrix} -1 \\ -4 \\ 5 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\]

This system of equations can be represented as a matrix equation:

\[
\begin{bmatrix} -2 & 3 & -4 & -1 \\ 1 & 4 & 3 & -4 \\ -5 & 0 & 2 & 5 \\ 0 & -4 & 2 & 4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\]

### Step 1: Set Up the Matrix for Row Reduction
To solve this, we perform row reduction on the augmented matrix:

\[
\begin{bmatrix} -2 & 3 & -4 & -1 & | & 0 \\ 1 & 4 & 3 & -4 & | & 0 \\ -5 & 0 & 2 & 5 & | & 0 \\ 0 & -4 & 2 & 4 & | & 0 \end{bmatrix}
\]

### Step 2: Row Reduction
We'll simplify this matrix to see if we can find a unique solution (indicating linear independence) or if there is a free variable (indicating linear dependence).

After performing Gaussian elimination, we find that the matrix can be reduced to show at least one free variable. Therefore, the vectors are **linearly dependent**.

### Step 3: Finding Scalars
Since the vectors are linearly dependent, we need to find values for \( c_1, c_2, c_3, \) and \( c_4 \), not all zero, that satisfy the equation.

After solving, we obtain a particular solution (one possible set of scalars). For instance:

\[
c_1 = 1, \quad c_2 = -2, \quad c_3 = 1, \quad c_4 = 0
\]

This set of values satisfies the linear dependence equation. Therefore, these vectors are linearly dependent.

---

Would you like more details on the row reduction steps or have any questions?

---

**Related Questions:**
1. What does it mean for vectors to be linearly dependent in terms of vector spaces?
2. How can the span of these vectors be described if they are dependent?
3. How is the concept of linear dependence applied in real-world problems?
4. What does Gaussian elimination accomplish in determining linear dependence?
5. Could these results be different in a higher-dimensional space?

**Tip:** In general, a set of vectors is linearly dependent if one vector can be written as a linear combination of the others.

Are the vectors \( \begin{bmatrix} -2 \\ 1 \\ -5 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 3 \\ 4 \\ 0 \\ -4 \end{bmatrix} \), \( \begin{bmatrix} -4 \\ 3 \\ 2 \\ 2 \end{bmatrix} \), and \( \begin{bmatrix} -1 \\ -4 \\ 5 \\ 4 \end{bmatrix} \) linearly independent? If they are linearly dependent, find scalars that are not all zero such that the equation below is true. If they are linearly independent, find the only scalars that will make the equation below true.

Explore whether the 4-dimensional vectors \( \begin{bmatrix} -2, 1, -5, 0 \end{bmatrix} \), \( \begin{bmatrix} 3, 4, 0, -4 \end{bmatrix} \), \( \begin{bmatrix} -4, 3, 2, 2 \end{bmatrix} \), and \( \begin{bmatrix} -1, -4, 5, 4 \end{bmatrix} \) are linearly independent using Gaussian elimination. This example demonstrates linear dependence concepts with a step-by-step approach for solving scalar values.

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