Solve 2x + y + z = 3, −2x − y − z = −3, 3x + y + z = 4
Worked out line by line the way a teacher would write it.
Answer
| Solution | infinitely many — the equations are dependent |
Step-by-step solution
9 steps-
1 The system, with the equations numbered\begin{aligned}2 x + y + z &= 3 & (1)\\ - 2 x - y - z &= -3 & (2)\\ 3 x + y + z &= 4 & (3)\end{aligned}
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2 Write the augmented matrix (coefficients | constants)
The augmented matrix keeps just the numbers: one row per equation, one column per unknown and the constants after the bar. Row operations on it do exactly what the same moves would do to the equations.
\left[\begin{array}{ccc|c} 2 & 1 & 1 & 3\\ -2 & -1 & -1 & -3\\ 3 & 1 & 1 & 4 \end{array}\right] -
3 Divide row 1 by 2 so the pivot is 1
Multiplying a whole row by a non-zero number is like multiplying both sides of that equation by it, so the solutions stay the same. Multiplying by 1 / pivot turns the pivot into 1.
\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\ -2 & -1 & -1 & -3\\ 3 & 1 & 1 & 4 \end{array}\right] -
4 Row 2 ← Row 2 + 2 · Row 1
Adding a multiple of the pivot row to another row is like adding equal amounts to both sides of an equation. The multiple is chosen so the entry in the pivot column becomes 0.
\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\ 0 & 0 & 0 & 0\\ 3 & 1 & 1 & 4 \end{array}\right] -
5 Row 3 ← Row 3 − 3 · Row 1\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\ 0 & 0 & 0 & 0\\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \end{array}\right]
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6 Swap rows 3 and 2 to get a non-zero pivot
Swapping two rows only changes the order of the equations, so the solutions stay the same. It brings a usable entry into the pivot position.
\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2}\\ 0 & 0 & 0 & 0 \end{array}\right] -
7 Divide row 2 by -1/2 so the pivot is 1\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\ 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right]
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8 Row 1 ← Row 1 − 1/2 · Row 2\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right]
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9 Fewer pivots than unknowns: z can be anything
Each free variable gets a parameter that can be any number, and the pivot variables are written in terms of those parameters. Every choice gives a solution.
x = 1,\quad y = 1 - zInfinitely many solutions, one for every value of the free variable(s).
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.