Systems of Linear Equations

Step-by-step solution

1. 1

   Set up the augmented matrix

   Write the 3 equations in 3 unknowns as one block matrix [A ∣ b]; the last column is the right-hand side.

   1

   1

   1

   6

   2

   -1

   1

   3

   1

   2

   -1

   2
2. 2

   Partial pivot: swap R1 ↔ R2

   Bring the largest-magnitude entry of column 1 into row 1 to use it as the pivot.

   $$R_{1} \leftrightarrow R_{2}$$

   2

   -1

   1

   3

   1

   1

   1

   6

   1

   2

   -1

   2
3. 3

   Normalize R1

   Divide row 1 by 2 so the pivot becomes 1.

   $$R_{1} \to \frac{1}{2}\,R_{1}$$

   1

   -1/2

   1/2

   3/2

   1

   1

   1

   6

   1

   2

   -1

   2
4. 4

   Clear column 1

   R2 → R2 − (1)·R1 · R3 → R3 − (1)·R1

   $$R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - R_{1}$$

   1

   -1/2

   1/2

   3/2

   0

   3/2

   1/2

   9/2

   0

   5/2

   -3/2

   1/2
5. 5

   Partial pivot: swap R2 ↔ R3

   Bring the largest-magnitude entry of column 2 into row 2 to use it as the pivot.

   $$R_{2} \leftrightarrow R_{3}$$

   1

   -1/2

   1/2

   3/2

   0

   5/2

   -3/2

   1/2

   0

   3/2

   1/2

   9/2
6. 6

   Normalize R2

   Divide row 2 by 5/2 so the pivot becomes 1.

   $$R_{2} \to \frac{1}{\frac{5}{2}}\,R_{2}$$

   1

   -1/2

   1/2

   3/2

   0

   1

   -3/5

   1/5

   0

   3/2

   1/2

   9/2
7. 7

   Clear column 2

   R1 → R1 − (-1/2)·R2 · R3 → R3 − (3/2)·R2

   $$R_{1} \to R_{1} + \frac{1}{2}\,R_{2} \\ R_{3} \to R_{3} - \frac{3}{2}\,R_{2}$$

   1

   0

   1/5

   8/5

   0

   1

   -3/5

   1/5

   0

   0

   7/5

   21/5
8. 8

   Normalize R3

   Divide row 3 by 7/5 so the pivot becomes 1.

   $$R_{3} \to \frac{1}{\frac{7}{5}}\,R_{3}$$

   1

   0

   1/5

   8/5

   0

   1

   -3/5

   1/5

   0

   0

   1

   3
9. 9

   Clear column 3

   R1 → R1 − (1/5)·R3 · R2 → R2 − (-3/5)·R3

   $$R_{1} \to R_{1} - \frac{1}{5}\,R_{3} \\ R_{2} \to R_{2} + \frac{3}{5}\,R_{3}$$

   1

   0

   0

   1

   0

   1

   0

   2

   0

   0

   1

   3
10. 10

    Read off the solution

    The left half is the identity matrix, so each row gives one variable directly.

    $$\begin{aligned} x_{1} = 1 \\ x_{2} = 2 \\ x_{3} = 3 \end{aligned}$$

    1

    0

    0

    1

    0

    1

    0

    2

    0

    0

    1

    3
11. 11

    Result

    x₁ = 1, x₂ = 2, x₃ = 3