Question

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

Short answer

The equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) has the solution \(X = CB - A\).

Step-by-step solution

Determine whether the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) has a solution

Let \(X\) satisfy the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = I\).

\(\begin{aligned}{l}C{C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = I\\I\left( {A + X} \right){B^{ - 1}} = C\\\end{aligned}\)

Multiply both sides of the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = I\) by \(C\):

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

Multiply both sides of the obtained equation by \(B\):

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

Expand the left side, and then subtract A from both sides of the equation \(\left( {A + X} \right)I = CB\):

\(\begin{aligned}{l}AI + XI = CB\\A + X - A = CB - A\\X = CB - A\end{aligned}\)

If there exists a solution, it must be \(CB - A\).

Show that \(CB - A\) is a solution

Substitute \(X = CB - A\) in equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}}\):

\(\begin{aligned}{c}{C^{ - 1}}\left( {A + CB - A} \right){B^{ - 1}} = {C^{ - 1}}\left( {CB} \right){B^{ - 1}}\\ = {C^{ - 1}}CB{B^{ - 1}}\\ = II\\ = I\end{aligned}\)

Thus, the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) has a solution \(X = CB - A\).