x+4y-6z=-1 2x-y+2z=-7 -x+2y-4z=5

1. Write the system of equations in matrix form<br /> The system can be written as $AX = B$, where $A = \begin{bmatrix} 1 & 4 & -6 \\ 2 & -1 & 2 \\ -1 & 2 & -4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} -1 \\ -7 \\ 5 \end{bmatrix}$.<br /><br />2. Use Gaussian elimination to find the row-echelon form<br /> Perform row operations to simplify $A$ to upper triangular form. <br /><br />1. Swap $R_1$ and $R_3$: <br /> $\begin{bmatrix} -1 & 2 & -4 \\ 2 & -1 & 2 \\ 1 & 4 & -6 \end{bmatrix}$<br /><br />2. Multiply $R_1$ by $-1$: <br /> $\begin{bmatrix} 1 & -2 & 4 \\ 2 & -1 & 2 \\ 1 & 4 & -6 \end{bmatrix}$<br /><br />3. Subtract $2 \times R_1$ from $R_2$: <br /> $\begin{bmatrix} 1 & -2 & 4 \\ 0 & 3 & -6 \\ 1 & 4 & -6 \end{bmatrix}$<br /><br />4. Subtract $R_1$ from $R_3$: <br /> $\begin{bmatrix} 1 & -2 & 4 \\ 0 & 3 & -6 \\ 0 & 6 & -10 \end{bmatrix}$<br /><br />5. Divide $R_2$ by $3$: <br /> $\begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 6 & -10 \end{bmatrix}$<br /><br />6. Subtract $6 \times R_2$ from $R_3$: <br /> $\begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 2 \end{bmatrix}$<br /><br />3. Back substitution to solve for variables<br /> Solve from bottom to top using the upper triangular matrix.<br /><br />1. From $R_3$: $2z = -22 \Rightarrow z = -11$<br /><br />2. Substitute $z = -11$ into $R_2$: $y - 2(-11) = -\frac{25}{3} \Rightarrow y = -\frac{25}{3} + \frac{22}{3} = -1$<br /><br />3. Substitute $y = -1$ and $z = -11$ into $R_1$: $x - 2(-1) + 4(-11) = 1 \Rightarrow x = 1 + 2 + 44 = 47$