let A = (1,1,1) B = (0,1,1) C = (2,1,1) and D = xA + yB +zC determine: the components of the vector D if D = (0,0 ,0) find x, y, z find x, y, z such that D = (1,2,3)

To determine the components of the vector D as a linear combination of vectors A, B, and C, we can express D as follows: D = xA + yB + zC Now, we have two different scenarios: **Scenario 1: D = (0, 0, 0)** In this scenario, we want to find x, y, and z such that D is the zero vector (0, 0, 0). 0 = x(1, 1, 1) + y(0, 1, 1) + z(2, 1, 1) This leads to the following system of equations: 1. x + 0 + 2z = 0 2. x + y + z = 0 3. x + y + z = 0 The third equation is the same as the second one, so we effectively have two equations: 1. x + 2z = 0 2. x + y + z = 0 Now, you can solve this system of equations to find the values of x, y, and z. Since it's a homogeneous system (the right-hand side is all zeros), it has non-unique solutions. One solution is: x = 0 y = 0 z = 0 So, if D = (0, 0, 0), then x, y, and z are all equal to 0. **Scenario 2: D = (1, 2, 3)** In this scenario, we want to find x, y, and z such that D is equal to (1, 2, 3). (1, 2, 3) = x(1, 1, 1) + y(0, 1, 1) + z(2, 1, 1) This leads to the following system of equations: 1. x + 0 + 2z = 1 2. x + y + z = 2 3. x + y + z = 3 Now, you can solve this system of equations to find the values of x, y, and z. Subtracting equation 2 from equation 3 gives: 3 - 2 = x + y + z - (x + y + z) 1 = 0 This means the equations are inconsistent, and there is no solution that makes D equal to (1, 2, 3) using the given linear combinations of A, B, and C. So, for D = (1, 2, 3), there is no solution for x, y, and z using the vectors A, B, and C.