Write each vector as a linear combination of the vectors in 5 . (Use \( 5_{1} \) and \( s_{2} \), respectively, for the vectors in the set. If not possible, enter IMPOSSIBLE.) \( S=\{(1,2,-2),(2,-1,1)\} \) (a) \( z=(-5,-5,5) \) \( z=\square \) (b) \( \mathbf{v}=(-2,-5,5) \) \( \mathbf{v}=\square \) (c) \( \mathbf{w}=(-2,-19,19) \) \( \mathbf{w}=\square \)

To express each vector as a linear combination of the vectors in the set \( S = \{(1, 2, -2), (2, -1, 1)\} \), we need to find scalars \( a \) and \( b \) such that: \[ z = a(1, 2, -2) + b(2, -1, 1) \] This leads to the following system of equations for each vector: 1. \( a + 2b = z_1 \) 2. \( 2a - b = z_2 \) 3. \( -2a + b = z_3 \) We will solve this system for each vector \( z \), \( \mathbf{v} \), and \( \mathbf{w} \). ### (a) For \( z = (-5, -5, 5) \) We set up the equations: 1. \( a + 2b = -5 \) (1) 2. \( 2a - b = -5 \) (2) 3. \( -2a + b = 5 \) (3) Now, we will solve this system of equations. Let's solve equations (1) and (2) first. From (1), we can express \( a \) in terms of \( b \): \[ a = -5 - 2b \] Substituting this into (2): \[ 2(-5 - 2b) - b = -5 \] \[ -10 - 4b - b = -5 \] \[ -5b = 5 \implies b = -1 \] Now substituting \( b = -1 \) back into the expression for \( a \): \[ a = -5 - 2(-1) = -5 + 2 = -3 \] Thus, for \( z = (-5, -5, 5) \): \[ z = -3(1, 2, -2) + (-1)(2, -1, 1) \] ### (b) For \( \mathbf{v} = (-2, -5, 5) \) We set up the equations: 1. \( a + 2b = -2 \) (1) 2. \( 2a - b = -5 \) (2) 3. \( -2a + b = 5 \) (3) Let's solve this system. From (1): \[ a = -2 - 2b \] Substituting into (2): \[ 2(-2 - 2b) - b = -5 \] \[ -4 - 4b - b = -5 \] \[ -5b = -1 \implies b = \frac{1}{5} \] Now substituting \( b = \frac{1}{5} \) back into the expression for \( a \): \[ a = -2 - 2\left(\frac{1}{5}\right) = -2 - \frac{2}{5} = -\frac{10}{5} - \frac{2}{5} = -\frac{12}{5} \] Thus, for \( \mathbf{v} = (-2, -5, 5) \): \[ \mathbf{v} = -\frac{12}{5}(1, 2, -2) + \frac{1}{5}(2, -1, 1) \] ### (c) For \( \mathbf{w} = (-2, -19, 19) \) We set up the equations: 1. \( a + 2b = -2 \) (1) 2. \( 2a - b = -19 \) (2) 3. \( -2a + b = 19 \) (3) Let's solve this system. From (1): \[ a = -2 - 2b \] Substituting into (2): \[ 2(-2 - 2b) - b = -19 \] \[ -4 - 4b - b = -19 \] \[ -5b = -15 \implies b = 3 \] Now substituting \( b = 3 \) back into the expression for \( a \): \[ a = -2 - 2(3) = -2 - 6 = -8 \] Thus, for \( \mathbf{w} = (-2, -19, 19) \): \[ \mathbf{w} = -8(1, 2, -2) + 3(2, -1, 1) \] ### Final Answers (a) \( z = -3(1, 2, -2) + (-1)(2, -1, 1) \) (b) \( \mathbf{v} = -\frac{12}{5}(1, 2, -2) + \frac{1}{5}(2, -1, 1) \) (c) \( \mathbf{w} = -8(1, 2, -2) + 3(2, -1, 1) \)