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Expert-verifiedFound in: Page 93

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 \(D\left( {ABC} \right) = I\).

**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, \({\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\).

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 \(D\left( {ABC} \right) = I\).

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