Question

Question: In Exercises 31–36, mention an appropriate theorem in your explanation.

33. Let A and B be square matrices. Show that even thoughABand BAmay not be equal, it is always true that\(det{\rm{ }}AB = det{\rm{ }}BA\).

Short answer

It is proved that \(\det {\rm{ }}AB = \det {\rm{ }}BA\).

Step-by-step solution

Write the multiplicative property

Based on theorem 6 of multiplicative property, if A andB are square matrices, then the determinant of the product matrix AB is equal to the product of the determinant of A and the determinant of B.

\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)

Prove the statement

From the theorem,

\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\). … (1)

Replace A by B and B by A.

\(\begin{aligned}{}\det BA &= \left( {\det B} \right)\left( {\det A} \right)\\\det BA &= \left( {\det A} \right)\left( {\det B} \right){\rm{ }}...{\rm{ }}\left( 2 \right)\end{aligned}\)

From equations (1) and (2),

\(\det {\rm{ }}AB = \det {\rm{ }}BA\)

Hence, it is proved that \(\det {\rm{ }}AB = \det {\rm{ }}BA\).