Linear Algebra and Its Applications, Exercise 1.6.3

Exercise 1.6.3. Given AB = C, express A^{-1} in terms of B and C. Similar, given PA = LU, express A^{-1} in terms of P, L, and U.

Answer: Assume that both B and C are invertible (see below). We then have

AB = C \rightarrow (AB)C^{-1} = CC^{-1} \rightarrow A(BC^{-1}) = I


AB = C \rightarrow (BC^{-1})(AB)B^{-1} = (BC^{-1})CB^{-1}

\rightarrow (BC^{-1})A(BB^{-1}) =  B(CC^{-1})B^{-1} \rightarrow (BC^{-1})A = BB^{-1} = I

so that

A^{-1} = BC^{-1}

(Note that we need to assume that B is invertible, even though we do not use B^{-1} in the formula for A^{-1}. In the absence of this assumption we can at best prove that

(BC^{-1})AB = B

This equation would always be true if B were zero, in which case we couldn’t draw any conclusions about whether or not BC^{-1} was a left inverse for A.)

Assume L and U are invertible. (If P is a permutation matrix then we already know P is invertible from exercise 1.6.2.) We then have

PA = LU \rightarrow U^{-1}L^{-1}PA = U^{-1}L^{-1}LU = U^{-1}U = I


PA = LU \rightarrow PAU^{-1}L^{-1}P = LUU^{-1}L^{-1}P = LL^{-1} = P

\rightarrow P^{-1}PAU^{-1}L{-1}P = P^{-1}P \rightarrow AU^{-1}L{-1}P = I


(U^{-1}L^{-1}P)A = A(U^{-1}L^{-1}P) = I

we have

A^{-1} = U^{-1}L^{-1}P

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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