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)

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.