Question

Verify the boxed statement preceding Example 1 i.e., Let A and B be square matrices. If \[AB = I\], then Aand B are both invertible, with \[B = {A^{ - {\bf{1}}}}\] and \[A = {B^{ - {\bf{1}}}}\].

Short answer

Square matrices A and B are both invertible with \[B = {A^{ - 1}}\] and \[A = {B^{ - 1}}\].

Step-by-step solution

Use the inverse matrix theorem

Given that \[AB = I\]. Then A is invertible based on statements (a) and (k) of the inverse matrix theorem.

Use the fact of the invertible matrix

\[{A^{ - 1}}\]exists.

Multiply by \[{A^{ - 1}}\] on both sides of \[AB = I\].

\[\begin{array}{c}{A^{ - 1}}AB = {A^{ - 1}}I\\IB = {A^{ - 1}}\\B = {A^{ - 1}}\end{array}\]

This implies B is invertible. The inverse of B is

\[\begin{array}{c}{B^{ - 1}} = {\left( {{A^{ - 1}}} \right)^{ - 1}}\\ = A.\end{array}\]

Draw a conclusion

Hence, A and B are both invertible with\[B = {A^{ - 1}}\]and\[A = {B^{ - 1}}\].