Solve the system x1 - 3x2 + 4x3 = -4, 3x1 - 7x2 + 7x3 = -8, and -4x1 + 6x2 + 2x3 = 4.

Step-by-step solution

1. 1

   Pick the pivot that avoids fractions. The first equation already has a coefficient of $1$ on $x_1$, so use it to clear $x_1$ from the other two. Choosing any other pivot here would introduce thirds or quarters immediately.

2. 2

   Eliminate x1 from the second equation. Subtract $3\times$(equation 1) from equation 2:

   $(3x_1-7x_2+7x_3)-3(x_1-3x_2+4x_3)=-8-3(-4)$
   $2x_2-5x_3=4.$

3. 3

   Eliminate x1 from the third equation. Add $4\times$(equation 1) to equation 3:

   $(-4x_1+6x_2+2x_3)+4(x_1-3x_2+4x_3)=4+4(-4)$
   $-6x_2+18x_3=-12\quad\Longrightarrow\quad x_2-3x_3=2.$

4. 4

   Solve the reduced 2x2 system. From $x_2=2+3x_3$, substitute into $2x_2-5x_3=4$:

   $2(2+3x_3)-5x_3=4\ \Longrightarrow\ 4+x_3=4\ \Longrightarrow\ x_3=0,$

   and therefore $x_2=2$.

5. 5

   Back-substitute for x1. Using equation 1:

   $x_1=-4+3x_2-4x_3=-4+6-0=2.$

6. 6

   Verify in all three original equations. $2-6+0=-4$ ✓; $6-14+0=-8$ ✓; $-8+12+0=4$ ✓. The coefficient determinant is $6\neq 0$, so this solution $(x_1,x_2,x_3)=(2,2,0)$ is the only one.