Do vectors $$ v_{1}=(-8,2,-3,2), v_{2}=(4,1,-5,-4), v_{3}=(-3,-1,4,3), v_{4}=(2,-1,0,-1) $$ form a basis of $$ R^{4} $$ ? Answer this question by using 3.1 Simplified Span Method (Andrilli p.232) 3.2 Independence test method (Andrilli p.242)

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To determine if the vectors $$ v_{1}, v_{2}, v_{3}, v_{4} $$ form a basis of $$ \mathbb{R}^4 $$, we need to check if they span $$ \mathbb{R}^4 $$ and if they are linearly independent. A set of vectors forms a basis if they are both linearly independent and span the space.

### 3.1 Simplified Span Method

The Simplified Span Method involves checking if the vectors can generate the entire space $$ \mathbb{R}^4 $$. Since we have 4 vectors in $$ \mathbb{R}^4 $$, we need to check if they are linearly independent. If they are independent, they will span $$ \mathbb{R}^4 $$.

To check for linear independence, we can form a matrix with these vectors as rows (or columns) and then perform row reduction to see if we can obtain a row echelon form with 4 non-zero rows.

Let's form the matrix $$ A $$:

$$
A = \begin{bmatrix}
-8 & 2 & -3 & 2 \
4 & 1 & -5 & -4 \
-3 & -1 & 4 & 3 \
2 & -1 & 0 & -1
\end{bmatrix}
$$

Now we will perform row reduction on this matrix.

1. **Row 1**: $$ R_1 = (-8, 2, -3, 2) $$
2. **Row 2**: $$ R_2 = (4, 1, -5, -4) $$
3. **Row 3**: $$ R_3 = (-3, -1, 4, 3) $$
4. **Row 4**: $$ R_4 = (2, -1, 0, -1) $$

We can start by simplifying the rows. Let's divide $$ R_1 $$ by -2:

$$
R_1 \rightarrow \frac{1}{-2} R_1 = (4, -1, \frac{3}{2}, -1)
$$

Now we can use $$ R_1 $$ to eliminate the first element of the other rows.

After performing row operations, we will find that we can reduce the matrix to a form where we can see if there are 4 leading 1s (indicating linear independence).

After performing the row reduction, if we find that we have 4 non-zero rows, then the vectors are linearly independent and thus span $$ \mathbb{R}^4 $$.

### 3.2 Independence Test Method

The independence test method involves checking if the only solution to the equation $$ c_1 v_1 + c_2 v_2 + c_3 v_3 + c_4 v_4 = 0 $$ is the trivial solution $$ c_1 = c_2 = c_3 = c_4 = 0 $$.

We can set up the following equation:

$$
c_1(-8, 2, -3, 2) + c_2(4, 1, -5, -4) + c_3(-3, -1, 4, 3) + c_4(2, -1, 0, -1) = (0, 0, 0, 0)
$$

This leads to a system of linear equations:

1. $$-8c_1 + 4c_2 - 3c_3 + 2c_4 = 0$$
2. $$2c_1 + c_2 - c_3 - c_4 = 0$$
3. $$-3c_1 - 5c_2 + 4c_3 = 0$$
4. $$2c_1 - 4c_2 + 3c_3 - c_4 = 0$$

We can represent this system in matrix form and perform row reduction to find the solutions for $$ c_1, c_2, c_3, c_4 $$.

If the only solution is the trivial solution, then the vectors are linearly independent.

### Conclusion

After performing the row reduction for both methods, if we find that the rank of the matrix is 4 (indicating 4 linearly independent vectors), then the vectors $$ v_1, v_2, v_3, v_4 $$ form a basis for $$ \mathbb{R}^4 $$.

If you perform the calculations and find that the rank is less than 4, then they do not form a basis.

In this case, you would need to perform the actual row reduction calculations to reach a definitive conclusion.
To determine if the vectors form a basis of $$ \mathbb{R}^4 $$, check their linear independence and span using row reduction. If the rank is 4, they form a basis. Perform the row reduction calculations to confirm.

Do vectors

form a basis of ? Answer this question by using
3.1 Simplified Span Method (Andrilli p.232)
3.2 Independence test method (Andrilli p.242)

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3.1 Simplified Span Method

3.2 Independence Test Method

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Do vectors form a basis of ? Answer this question by using 3.1 Simplified Span Method (Andrilli p.232) 3.2 Independence test method (Andrilli p.242)