Computing a 3×3 Inverse

Linear Algebra (Matrices) · Axiom Academy

EXAMPLE Computing a Matrix Inverse Finding A^ -1 by row-reducing the augmented matrix into Find the inverse of the matrix We adjoin the identity to form and apply row operations until the left block becomes I . Whatever the right block becomes is A^ -1 , because the same operations that turn A into I turn I into A^ -1 . An inverse is only correct if multiplying it back recovers the identity. Multiply A by the A^ -1 we found: Every product equals the identity, so A^ -1 is correct. ✓︎ Nice work — you inverted a matrix by row reduction. The pieces that make the method work: Augment with the identity: start from and aim for . The right block is a ledger that records every operation. Forward elimination: create zeros below each pivot, top-left to bottom-right, to reach an upper-triangular left block. Back-substitution: work upward, clearing entries above each pivot, until the left block is the identity. Operate on both blocks: every row operation hits the left and right blocks together — that is what builds A^ -1 on the right. Always verify: confirm AA^ -1 = I . A row of zeros on the left instead would mean A is singular (no inverse). Here , which is why A^ -1 came out with whole-number entries — a clean check that the arithmetic held together.

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