Let's restate the problem: Let \[ v_1 = \begin{bmatrix} \\ \\ -2 \end{bmatrix}, \quad v_2 = \begin{bmatrix} \\ -3 \\ 8 \end{bmatrix}, \quad v_3 = \begin{bmatrix} 4 \\ -1 \\ -5 \end{bmatrix} \] Does \(\{v_1, v_2, v_3\}\) span \(\mathbb{R}^3\)? Why or why not? --- ## Solution Vectors span \(\mathbb{R}^3\) **if and only if they are linearly independent** (i.e., the only solution to \(c_1 v_1 + c_2 v_2 + c_3 v_3 = \) is \(c_1 = c_2 = c_3 = \)), or, equivalently, if the matrix formed by these vectors as columns has rank 3. ### Step 1: Form the Matrix \[ A = \begin{bmatrix} & & 4 \\ & -3 & -1 \\ -2 & 8 & -5 \end{bmatrix} \] ### Step 2: Check for Linear Independence We can check if the vectors are linearly independent (i.e., does \(A\) have a non-zero determinant?): \[ \det(A) = \begin{vmatrix} & & 4 \\ & -3 & -1 \\ -2 & 8 & -5 \end{vmatrix} \] Expanding along the first row: \[ \det(A) = \cdot \begin{vmatrix} -3 & -1 \\ 8 & -5 \end{vmatrix} - \cdot \begin{vmatrix} & -1 \\ -2 & -5 \end{vmatrix} + 4 \cdot \begin{vmatrix} & -3 \\ -2 & 8 \end{vmatrix} \] Only the third term is nonzero: \[ = 4 \cdot [()(8) - (-2)(-3)] = 4 \cdot [ - 6] = 4 \cdot (-6) = -24 \] This is **nonzero**, so the vectors are linearly independent. --- ## **Conclusion** **Yes, \(\{v_1, v_2, v_3\}\) span \(\mathbb{R}^3\), because they are linearly independent.**