Calculate the result of the matrix multiplication Ax where A = [[6, 2, 4], [-1, 4, 3], [-2, 9, 3]] and x = [[4], [-2], [1]].

Step 1: Write the augmented matrix for the system Ax = c. [ ccc|c 6 & 2 & 4 & 4 \\ 1 & 4 & 3 & -2 \\ -2 & 9 & 3 & 1 ] Step 2: Swap row 1 (R1) and row 2 (R2) to get a leading 1 in position (1,1). [ ccc|c 1 & 4 & 3 & -2 \\ 6 & 2 & 4 & 4 \\ -2 & 9 & 3 & 1 ] Step 3: Eliminate column 1 below the pivot: R2 R2 - 6R1; R3 R3 + 2R1. R2: [6-6(1),\ 2-6(4),\ 4-6(3)\ |\ 4-6(-2)] = [0,\ -22,\ -14\ |\ 16] R3: [-2+2(1),\ 9+2(4),\ 3+2(3)\ |\ 1+2(-2)] = [0,\ 17,\ 9\ |\ -3] [ ccc|c 1 & 4 & 3 & -2 \\ 0 & -22 & -14 & 16 \\ 0 & 17 & 9 & -3 ] Step 4: Simplify row 2: Divide R2 by -2 to get R2 -(1)/(2)R2. [0,\ 11,\ 7\ |\ -8] [ ccc|c 1 & 4 & 3 & -2 \\ 0 & 11 & 7 & -8 \\ 0 & 17 & 9 & -3 ] Step 5: Eliminate column 2 below the pivot: Multiply R2 by 17 and R3 by 11, then subtract. 17 × R2: [0,\ 187,\ 119\ |\ -136] 11 × R3: [0,\ 187,\ 99\ |\ -33] New R3 17R2 - 11R3: [0,\ 0,\ 20\ |\ -103] [ ccc|c 1 & 4 & 3 & -2 \\ 0 & 11 & 7 & -8 \\ 0 & 0 & 20 & -103 ] Step 6: Back-substitute starting from the bottom. From R3: 20x_3 = -103 x_3 = -(103)/(20) Substitute into R2: 11x_2 + 7(-(103)/(20)) = -8 11x_2 - (721)/(20) = -8 11x_2 = -8 + (721)/(20) = -(160)/(20) + (721)/(20) = (561)/(20) x_2 = (561)/(20 × 11) = (561)/(220) = (51)/(20) Substitute into R1: x_1 + 4((51)/(20)) + 3(-(103)/(20)) = -2 x_1 + (204)/(20) - (309)/(20) = -2 x_1 - (105)/(20) = -2 x_1 = -2 + (105)/(20) = -(40)/(20) + (105)/(20) = (65)/(20) = (13)/(4) The solution is x = x_1 \\ x_2 \\ x_3 = (13)/(4) \\ (51)/(20) \\ -(103)/(20) . (13)/(4) \\[1em] (51)/(20) \\[1em] -(103)/(20)