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Show that if ABis invertible, so is A. You cannot use Theorem 6(b), because you cannot assumethat Aand Bare invertible. (Hint:There is a matrix Wsuch that \(ABW = I\). Why?)

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BothABand A are invertible.

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01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed by using the augmented matrix\(\left( {\begin{array}{*{20}{c}}A&I\end{array}} \right)\), where\(I\)is the identity matrix. Matrix Ahas an inverse only if \(\left( {\begin{array}{*{20}{c}}A&I\end{array}} \right)\) is row equivalent to \(\left( {\begin{array}{*{20}{c}}I&{{A^{ - 1}}}\end{array}} \right)\).

02

Show that A is invertible

The product AB is invertible, so there should be an inverse of matrix AB. Let W be the inverse matrix of AB.

Then, it can be represented as shown below:

\(\begin{array}{c}\left( {AB} \right)W = I\\A\left( {BW} \right) = I\end{array}\)

The equation\(A\left( {BW} \right) = I\) shows that matrix\(BW\)is the inverse of matrix A.

Therefore, A is invertible.

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