## 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.