How to find the inverse of a 3x3 matrix using row operations?

Question

I'm studying linear algebra and I can find the inverse of a $2 \times 2$ matrix easily using the formula $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\-c & a \end{pmatrix}$ .

But for $3 \times 3$ matrices, the formula involving cofactors and the adjugate is very tedious. I've heard that Gauss-Jordan elimination (row reduction) is more efficient.

Could someone show the complete row reduction process to find the inverse of:

A = \begin{pmatrix} 1 & 2 & 3 \\0 & 1 & 4 \\5 & 6 & 0 \end{pmatrix}

Amara Okafor · Accepted Answer

The Gauss-Jordan method: augment A with the identity matrix [ A ∣ I ] and row-reduce until the left side becomes I. The right side will be A − 1. [ A ∣ I ] = ( 1 2 3 ∣ 1 0 0 0 1 4 ∣ 0 1 0 5 6 0 ∣ 0 0 1 ) Step 1: R 3 ← R 3 − 5 R 1 ( 1 2 3 ∣ 1 0 0 0 1 4 ∣ 0 1 0 0 − 4 − 15 ∣ − 5 0 1 ) Step 2: R 3 ← R 3 + 4 R 2 ( 1 2 3 ∣ 1 0 0 0 1 4 ∣ 0 1 0 0 0 1 ∣ − 5 4 1 ) Step 3: R 2 ← R 2 − 4 R 3 ( 1 2 3 ∣ 1 0 0 0 1 0 ∣ 20 − 15 − 4 0 0 1 ∣ − 5 4 1 ) Step 4: R 1 ← R 1 − 2 R 2 − 3 R 3 ( 1 0 0 ∣ − 24 18 5 0 1 0 ∣ 20 − 15 − 4 0 0 1 ∣ − 5 4 1 ) Therefore: A − 1 = ( − 24 18 5 20 − 15 − 4 − 5 4 1 ) Verification: A ⋅ A − 1 should equal I.

How to find the inverse of a 3x3 matrix using row operations?

1 Answer

How to find the inverse of a 3x3 matrix using row operations?

1 Answer