Back Substitution

Linear Algebra (Matrices) · Axiom Academy

Once a system is in row echelon form, solve it from the bottom up — each solved value flows into the row above. Here is our system already in row echelon form . Notice the staircase: row 1 has three unknowns, row 2 has two, and the bottom row has just one — so it solves instantly, with no algebra at all. The augmented matrix is upper-triangular The bottom row gave us z = 2 . Now send that value up into row 2, y + 2z = 8 . Substituting collapses its two unknowns to one, and y falls out directly. From the bottom row, z = 2 is already pinned down. Replace z in y + 2z = 8 , giving y + 4 = 8 . One unknown remains, so y = 4 immediately. Each upward step turns a multi-unknown row into a single-unknown one. We never solve a system all at once — only one easy equation at a time, climbing the staircase. 3. The Cascade Reaches the Top Now both knowns, y = 4 and z = 2 , flow up into the top row x + 2y + 3z = 14 . They fill its two trailing unknowns, leaving x alone — and the full solution snaps into place. z = 2 — read straight off the bottom row y = 8 - 2(2) = 4 — after substituting z x = 14 - 2(4) - 3(2) = 0 — after substituting y and z You've run back substitution end to end: solve the bottom row, then let each known value cascade upward until the whole solution is found. Scroll up to revisit any step.

This is the written version of the interactive lesson above. See the full Linear Algebra (Matrices) course.