Suppose AB = AC, where Band Care \(n \times p\) matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible.

Short Answer

Expert verified

It is proved that \(B = C\).

Step by step solution

01

Condition for an invertible matrix

A \(n \times n\) matrixAis said to be invertible if there is an equation \(n \times n\) matrix C such that \(CA = I\) and \(AC = I\).

02

Show that B = C

A matrix that is not invertible is called asingular matrix,and an invertible matrix is called anon-singular matrix.

Multiply both sides of the equation \(AB = AC\) by \({A^{ - 1}}\) as shown below:

\(\begin{aligned}{c}{A^{ - 1}}AB = {A^{ - 1}}AC\\IB = IC\\B = C\end{aligned}\)

This conclusion is not always true when Ais singular.

Hence, it is proved that \(B = C\).

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