To check that a matrix B is the inverse of a matrix A, the definition says you have to check two conditions: AB = I and BA = I. It turns out that you need check only one of these conditions; the other follows automatically. Here's why.
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Proof: Suppose A has reduced row-echelon form R; then for some elementary matrices E1, E2, ..., Ek, we have that Ek...E2E1A = R. Set C = Ek...E2E1; then C is invertible and CA = R. Multiply on the right by BC–1
and simplify: Since I doesn't have a row of zeroes, neither can R, so R must be the identity matrix. Since R was the reduced row-echelon form of A, A is invertible. Since AB = I, A–1(AB)A = A–1IA, i.e. BA = I. |
In some applications of matrices, it's useful to be able to decompose a matrix into the product of simpler matrices.
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Proof: If A is the product of elementary matrices, then it is invertible, since each elementary matrix is invertible. |
If A is invertible, then Ek...E2E1A = I for some elementary matrices E1, E2, ..., Ek. Solve for A: (Ek...E2E1)–1(Ek...E2E1)A = (Ek...E2E1)–1I, so A = (Ek...E2E1)–1 = E1–1E2–1...Ek–1. A is then the product of the elementary matrices E1–1, E2–1, ..., Ek–1. |
This gives you a method for decomposing any invertible matrix into elementary matrices.
To decompose an invertible matrix A into the product of elementary matrices: |
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