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 $A\mathbf{x} = \mathbf{c}$.

$$\left[ \begin{array}{ccc|c} 6 & 2 & 4 & 4 \\ 1 & 4 & 3 & -2 \\ -2 & 9 & 3 & 1 \end{array} \right]$$

Step 2: Swap row 1 (R1) and row 2 (R2) to get a leading 1 in position (1,1).

$$\left[ \begin{array}{ccc|c} 1 & 4 & 3 & -2 \\ 6 & 2 & 4 & 4 \\ -2 & 9 & 3 & 1 \end{array} \right]$$

Step 3: Eliminate column 1 below the pivot: R2 $\leftarrow$ R2 $- 6$R1; R3 $\leftarrow$ R3 $+ 2$R1.

R2: $[6-6(1),\ 2-6(4),\ 4-6(3)\ |\ 4-6(-2)] = [0,\ -22,\ -14\ |\ 16]$

R3: $[-2+2(1),\ 9+2(4),\ 3+2(3)\ |\ 1+2(-2)] = [0,\ 17,\ 9\ |\ -3]$

$$\left[ \begin{array}{ccc|c} 1 & 4 & 3 & -2 \\ 0 & -22 & -14 & 16 \\ 0 & 17 & 9 & -3 \end{array} \right]$$

Step 4: Simplify row 2: Divide R2 by $-2$ to get R2 $\leftarrow -\frac{1}{2}$R2.

$[0,\ 11,\ 7\ |\ -8]$

$$\left[ \begin{array}{ccc|c} 1 & 4 & 3 & -2 \\ 0 & 11 & 7 & -8 \\ 0 & 17 & 9 & -3 \end{array} \right]$$

Step 5: Eliminate column 2 below the pivot: Multiply R2 by 17 and R3 by 11, then subtract.

17 $\times$ R2: $[0,\ 187,\ 119\ |\ -136]$

11 $\times$ R3: $[0,\ 187,\ 99\ |\ -33]$

New R3 $\leftarrow$ 17R2 $-$ 11R3: $[0,\ 0,\ 20\ |\ -103]$

$$\left[ \begin{array}{ccc|c} 1 & 4 & 3 & -2 \\ 0 & 11 & 7 & -8 \\ 0 & 0 & 20 & -103 \end{array} \right]$$

Step 6: Back-substitute starting from the bottom.

From R3: $20x_3 = -103 \implies x_3 = -\frac{103}{20}$

Substitute into R2: $11x_2 + 7\left(-\frac{103}{20}\right) = -8$

$11x_2 - \frac{721}{20} = -8$

$11x_2 = -8 + \frac{721}{20} = -\frac{160}{20} + \frac{721}{20} = \frac{561}{20}$

$x_2 = \frac{561}{20 \times 11} = \frac{561}{220} = \frac{51}{20}$

Substitute into R1: $x_1 + 4\left(\frac{51}{20}\right) + 3\left(-\frac{103}{20}\right) = -2$

$x_1 + \frac{204}{20} - \frac{309}{20} = -2$

$x_1 - \frac{105}{20} = -2$

$x_1 = -2 + \frac{105}{20} = -\frac{40}{20} + \frac{105}{20} = \frac{65}{20} = \frac{13}{4}$

The solution is $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \frac{13}{4} \\ \frac{51}{20} \\ -\frac{103}{20} \end{bmatrix}$.

$$\boxed{\begin{bmatrix} \dfrac{13}{4} \\[1em] \dfrac{51}{20} \\[1em] -\dfrac{103}{20} \end{bmatrix}}$$