Math · Precalculus · Worked example

Recognize infinitely many solutions in RREF

Solve x + y + z = 4, 2x + y − z = 1, 3x + 2y = 5.

[\begin{array}{cccc} 1 & 1 & 1 & 4 \\ 2 & 1 & - 1 & 1 \\ 3 & 2 & 0 & 5 \end{array}]

Clear the first column

R₂ − 2R₁ and R₃ − 3R₁ give two identical rows.

[\begin{array}{cccc} 1 & 1 & 1 & 4 \\ 0 & - 1 & - 3 & - 7 \\ 0 & - 1 & - 3 & - 7 \end{array}]

A row of zeros appears

R₃ − R₂ gives 0 = 0: the third equation carried no new information.

[\begin{array}{cccc} 1 & 1 & 1 & 4 \\ 0 & - 1 & - 3 & - 7 \\ 0 & 0 & 0 & 0 \end{array}]

Reduce to RREF

Scale R₂ by −1, then replace R₁ with R₁ − R₂.

[\begin{array}{cccc} 1 & 0 & - 2 & - 3 \\ 0 & 1 & 3 & 7 \\ 0 & 0 & 0 & 0 \end{array}]

Name the free variable

Column 3 has no pivot, so z is free: set z = t. The two rows say x − 2z = −3 and y + 3z = 7.

x = - 3 + 2 t, y = 7 - 3 t, z = t

Check one solution

t = 1 gives (−1, 4, 1), which satisfies all three equations.

\begin{array}{l} - 1 + 4 + 1 = 4 \\ 2 (- 1) + 4 - 1 = 1 \\ 3 (- 1) + 2 \cdot 4 = 5 \end{array}

Result

Infinitely many solutions: x = −3 + 2t, y = 7 − 3t, z = t for any real number t.

Your turn

Row reduction of a system gives the rows (1, 0, 2 | 5), (0, 1, −1 | 3) and (0, 0, 0 | 4). How many solutions does the system have?

Show the answer and explanation

None.

The last row says 0x + 0y + 0z = 4, which is false for every x, y and z, so the system is inconsistent.

0 x + 0 y + 0 z = 4

Keep exploring

In Matrices & linear systems, the classification reads Infinitely many solutions. Change the last constant from 5 to 6 and the zero row becomes 0 = 1: no solution.

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