Question

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

Short answer

It is proved that Ais invertible.

Step-by-step solution

Condition for an invertible matrix

Theorem 5states that Ais an invertible \(n \times n\) matrix, then for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has a unique solution \(x = {A^{ - 1}}b\).

Show that A is invertible

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

Consider \(C = AB\).

Multiply each side of the equation \(C = AB\) by \({B^{ - 1}}\):

\(\begin{aligned}{c}C{B^{ - 1}} = AB{B^{ - 1}}\\ = AI\\ = A\end{aligned}\)

According to theorem 6, Ais the product of invertible matrices; thus, Ais invertible.

Hence, it is proved that Ais invertible.