Question

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Short answer

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

Step-by-step solution

Condition for an invertible matrix

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

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