Solve the system x2 + 4x3 = -4, x1 + 3x2 + 3x3 = -2, 3x1 + 7x2 + 5x3 = 6.

Step-by-step solution

1. 1

   Write the augmented matrix and fix the variable order. With columns for $x_1,x_2,x_3$ and the constants, the first equation has no $x_1$, so it contributes a leading $0$: $\left[\begin{array}{ccc|c} 0&1&4&-4\\ 1&3&3&-2\\ 3&7&5&6 \end{array}\right]$

2. 2

   Swap rows so the first pivot is a 1. Exchanging $R_1$ and $R_2$ puts a leading $1$ in the top-left corner and avoids fractions later: $\left[\begin{array}{ccc|c} 1&3&3&-2\\ 0&1&4&-4\\ 3&7&5&6 \end{array}\right]$

3. 3

   Clear below the first pivot. $R_3 \leftarrow R_3-3R_1$ gives $[3,7,5\mid 6]-3[1,3,3\mid -2] = [0,-2,-4\mid 12]$, so the third row becomes $0,-2,-4\mid 12$. The second pivot is already $1$ in $R_2$.

4. 4

   Clear below the second pivot. $R_3 \leftarrow R_3+2R_2$ gives $[0,-2,-4\mid12]+2[0,1,4\mid-4] = [0,0,4\mid4]$. Scaling by $\tfrac14$ gives $[0,0,1\mid1]$, so $x_3 = 1$ - the matrix is now in row echelon form.

5. 5

   Back-substitute by clearing above the pivots. $R_2 \leftarrow R_2-4R_3$ turns $[0,1,4\mid-4]$ into $[0,1,0\mid-8]$; $R_1 \leftarrow R_1-3R_3$ turns $[1,3,3\mid-2]$ into $[1,3,0\mid-5]$; finally $R_1 \leftarrow R_1-3R_2$ gives $[1,0,0\mid19]$.

6. 6

   Read off and verify the unique solution. The reduced matrix is the identity augmented by $(19,-8,1)$, so $x_1=19$, $x_2=-8$, $x_3=1$. Checking all three originals: $-8+4(1)=-4$; $19+3(-8)+3(1)=19-24+3=-2$; $3(19)+7(-8)+5(1)=57-56+5=6$. Three pivots for three unknowns means the solution is unique.