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 form a basis of \( \mathbb{R}^4 \), we have to show that they are linearly independent and span \( \mathbb{R}^4 \). Using the Simplified Span Method, we can check span by constructing a matrix with these vectors and performing row reduction. Create a matrix \( A \) by placing the vectors as rows: \[ A = \begin{pmatrix} -8 & 2 & -3 & 2 \\ 4 & 1 & -5 & -4 \\ -3 & -1 & 4 & 3 \\ 2 & -1 & 0 & -1 \end{pmatrix} \] Row-reducing this matrix, if we find any zero rows after reaching Row Echelon Form and the leading coefficients are not in all four rows, then they do not span \( \mathbb{R}^4 \). Next, let's apply the Independence Test Method by checking if the determinant of the square matrix formed by these vectors is non-zero. However, since we have 4 vectors in \( \mathbb{R}^4 \) (a 4x4 matrix), we can compute the determinant. We perform Gaussian elimination or any method like Sarrus' rule for smaller matrices to see if the determinant is zero. A zero determinant means the vectors are linearly dependent, while a non-zero determinant indicates they are independent. If, after performing these calculations, the four vectors yield a dimension of at least four or the determinant is non-zero, they form a basis for \( \mathbb{R}^4 \).