For Part (b), we are given the following system of linear equations:
2x₁ + 3x₂ – 4x₃ = -4
x₁ – x₂ + x₃ = 2
x₁ + 2x₂ + 3x₃ = 14
We need to solve this system using matrices. First, we write the system in matrix form AX = B, where A is the coefficient matrix, X is the column matrix of variables, and B is the constant matrix. Thus:
A = [2 3 -4; 1 -1 1; 1 2 3], X = [x₁; x₂; x₃], B = [-4; 2; 14].
To solve for X, we can use the method of Gaussian elimination or find the inverse of A. Using Gaussian elimination, we form the augmented matrix [A|B]:
[2 3 -4 | -4]
[1 -1 1 | 2]
[1 2 3 | 14]
We perform row operations to convert this into row-echelon form. First, swap row 1 and row 2 to have a 1 in the top-left position:
[1 -1 1 | 2]
[2 3 -4 | -4]
[1 2 3 | 14]
Next, replace row 2 with row 2 – 2(row 1): [2-2(1) 3-2(-1) -4-2(1) | -4-2(2)] = [0 5 -6 | -8]
Replace row 3 with row 3 – row 1: [1-1 2-(-1) 3-1 | 14-2] = [0 3 2 | 12]
Now the augmented matrix is:
[1 -1 1 | 2]
[0 5 -6 | -8]
[0 3 2 | 12]
Next, we eliminate the second column below row 2. Multiply row 2 by 1/5 to simplify, but it’s easier to eliminate row 3 using row 2. Replace row 3 with 5(row 3) – 3(row 2): 5[0 3 2 | 12] – 3[0 5 -6 | -8] = [0 15 10 | 60] – [0 15 -18 | -24] = [0 0 28 | 84]. So row 3 becomes [0 0 28 | 84]. Divide row 3 by 28: [0 0 1 | 3]. Now we have:
[1 -1 1 | 2]
[0 5 -6 | -8]
[0 0 1 | 3]
From row 3, we get x₃ = 3. Substitute into row 2: 5x₂ – 6(3) = -8 → 5x₂ – 18 = -8 → 5x₂ = 10 → x₂ = 2. Substitute x₂ = 2 and x₃ = 3 into row 1: x₁ – 2 + 3 = 2 → x₁ + 1 = 2 → x₁ = 1. Therefore, the solution is x₁ = 1, x₂ = 2, x₃ = 3. We can verify by substituting back into the original equations: 2(1)+3(2)-4(3)=2+6-12=-4; 1-2+3=2; 1+4+9=14, all correct.