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

and

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

and

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

Since

(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.

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s