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Solve any linear system with rref( in under a minute

Your calculator does full Gauss-Jordan elimination on command. Two menus, one function, and whatever system you throw at it comes back solved.

Systems of linear equations show up everywhere from algebra to physics to econ. Hand-elimination works for 2×2 but gets painful fast. The calculator's rref( (reduced row echelon form) function does the whole thing in one step.

Step 1: Enter the augmented matrix

Press 2nd + x⁻¹ (this is the MATRIX menu — the label above x⁻¹).

  1. Arrow right to EDIT.
  2. Pick any matrix name ([A] is fine).
  3. Enter the dimensions. For a system of n equations in m unknowns, the augmented matrix is n × (m+1).

Example — solve this system:

2x + y − z = 8
-3x − y + 2z = -11
-2x + y + 2z = -3

This is a 3×4 matrix:

[ 2   1  -1 |  8 ]
[-3  -1   2 |-11 ]
[-2   1   2 | -3 ]

Enter dimensions 3 4, then type the 12 values row by row. Press 2nd + mode to quit out when done.

Step 2: Run rref(

On the home screen:

  1. 2nd + x⁻¹ → arrow right to MATH → scroll to B:rref( (letter may vary by firmware).
  2. Press enter. The function rref( appears on the home screen.
  3. Tell it which matrix: 2nd + x⁻¹NAMES1:[A]. Close the paren: ). Press enter.

Reading the output

The calculator returns the matrix in reduced row echelon form:

[ 1  0  0 |  2 ]
[ 0  1  0 |  3 ]
[ 0  0  1 | -1 ]

Last column is your solution: x=2, y=3, z=-1.

Interpreting special outputs

  • Last row is all zeros including the augmented column (e.g. [0 0 0 | 0]): infinite solutions. The system is under-determined; express one variable in terms of the others.
  • Last row is [0 0 0 | c] with c ≠ 0: no solution. The system is inconsistent.
  • Non-integer entries that look weird: apply ►Frac to see exact rational answers. rref([A])►Frac.

Why this beats substitution

For 3×3 and larger, Gauss-Jordan elimination is algorithmic — no insight required, no chance of an arithmetic slip halfway through. The calculator does the book-keeping perfectly.

Bonus: ref( vs rref(

  • ref( gives row echelon form (upper-triangular, leading 1's). You still have to back-substitute.
  • rref( goes all the way to reduced row echelon form — leading 1's with zeros above and below. The answer falls out.

Unless a teacher specifically asks for ref, use rref.

Gotcha

If your matrix has mixed fractions and decimals, the calculator may return floating-point results that look non-exact but are. Append ►Frac to force the exact fraction representation whenever you enter a system with fractional coefficients.