Solve the system of equations by using row operations to write a matrix in REDUCED row-echelon form.
$4x+2y+3z=6$
$x+y+z=3$$5x+y+2z=5$

• A. $x=0,y=-3,z=4$
• B. $x=-\frac14,y=-\frac{19}{4},z=\frac{11}{2}$
• C. $x=1,y=4,z=-2$
• D. $x=\frac12,y=\frac12,z=1$

Answer: (C)

Verified step by step guidance

1. Write the system of equations as an augmented matrix. The system is: \( \begin{align*} 4x + 2y + 3z &= 6 \\ x + y + z &= 3 \\ 5x + y + 2z &= 5 \end{align*} \). The augmented matrix is: \( \begin{bmatrix} 4 & 2 & 3 & | & 6 \\ 1 & 1 & 1 & | & 3 \\ 5 & 1 & 2 & | & 5 \end{bmatrix} \).
2. Use row operations to get a leading 1 in the first row, first column. This can be done by swapping Row 1 and Row 2: \( \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 4 & 2 & 3 & | & 6 \\ 5 & 1 & 2 & | & 5 \end{bmatrix} \).
3. Eliminate the first column below the leading 1 by replacing Row 2 with Row 2 minus 4 times Row 1, and Row 3 with Row 3 minus 5 times Row 1: \( \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & -2 & -1 & | & -6 \\ 0 & -4 & -3 & | & -10 \end{bmatrix} \).
4. Get a leading 1 in the second row, second column by dividing Row 2 by -2: \( \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & \frac{1}{2} & | & 3 \\ 0 & -4 & -3 & | & -10 \end{bmatrix} \).
5. Eliminate the second column below the leading 1 by replacing Row 3 with Row 3 plus 4 times Row 2: \( \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & \frac{1}{2} & | & 3 \\ 0 & 0 & -1 & | & 2 \end{bmatrix} \).