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Found in: Page 93

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

Suppose A, B, and C are invertible $$n \times n$$ matrices. Show that ABC is also invertible by producing a matrix D such that $$\left( {ABC} \right)D = I$$ and $$D\left( {ABC} \right) = I$$.

It is proved that $$\left( {ABC} \right)D = I$$ and $$D\left( {ABC} \right) = I$$.

See the step by step solution

Step 1: Condition for an invertible matrix

Theorem 6 states that if A and B are $$n \times n$$ invertible matrices, the inverse of AB is the product of the inverses of A and B in the reverse order. That is, $${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$$.

Step 2: Show that $$\left( {ABC} \right)D = I$$ and $$D\left( {ABC} \right) = I$$

Consider $$D = {C^{ - 1}}{B^{ - 1}}{A^{ - 1}}$$. Then,

\begin{aligned}{c}\left( {ABC} \right){C^{ - 1}}{B^{ - 1}}{A^{ - 1}} = AB\left( {C{C^{ - 1}}} \right){B^{ - 1}}A\\ = ABI{B^{ - 1}}{A^{ - 1}}\\ = AI{A^{ - 1}}\\ = I\end{aligned}

And

\begin{aligned}{c}{C^{ - 1}}{B^{ - 1}}{A^{ - 1}}\left( {ABC} \right) = {C^{ - 1}}{B^{ - 1}}{A^{ - 1}}ABC\\ = {C^{ - 1}}{B^{ - 1}}IBC\\ = {C^{ - 1}}IC\\ = I\end{aligned}

Hence, it is proved that $$\left( {ABC} \right)D = I$$ and $$D\left( {ABC} \right) = I$$.