Question

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) …(3)

1. Explain why B is invertible.
2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

Short answer

1. Bis invertible since it is the product of two invertible matrices.
2. \(X = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}A\).

Step-by-step solution

Explanation of B is invertible

(a)

Multiply each side of the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) by \(X\):

\(\begin{aligned}{c}{\left( {A - AX} \right)^{ - 1}}X = X{X^{ - 1}}B\\{A^{ - 1}}X - {X^{ - 1}}{A^{ - 1}}X = IB\\{A^{ - 1}}X - {X^{ - 1}}{A^{ - 1}}X = B\end{aligned}\)

It is the product of two invertible matrices. Thus, Bis invertible

Solve the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) for X

(b)

Use theorem 6 about the inverse of a product to invert each side of the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\):

\(\begin{aligned}{c}A - AX = {\left( {{X^{ - 1}}B} \right)^{ - 1}}\\ = {B^{ - 1}}{\left( {{X^{ - 1}}} \right)^{ - 1}}\\ = {B^{ - 1}}X\end{aligned}\)

Then,

\(\begin{aligned}{c}A - AX = {B^{ - 1}}X\\A = AX + {B^{ - 1}}X\\A = \left( {A + {B^{ - 1}}} \right)X\end{aligned}\)

The product \(\left( {A + {B^{ - 1}}} \right)X\) is invertible since Ais invertible. Xis invertible since the other factor \(\left( {A + {B^{ - 1}}} \right)\) is invertible. Therefore,

\(\begin{aligned}{c}{\left( {A + {B^{ - 1}}} \right)^{ - 1}}A = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}\left( {A + {B^{ - 1}}} \right)X\\ = IX\\ = X\end{aligned}\)

Thus, \(X = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}A\).