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Chapter 2: Q16Q (page 93)

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

Expert verified

It is proved that Ais invertible.

Step by step solution

01

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 \(x = {A^{ - 1}}b\).

02

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.

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