Question

Question: Show that if \(A\) and \(B\) are similar, then \(\det A = \det B\).

Short answer

It is proved that \({\rm{det}}A = {\rm{det}}B\).

Step-by-step solution

Write the given condition

Consider that\(P\) is invertible. Then, use Theorem 3 (Property of Determinant): \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).

\[\begin{array}I = {\rm{det}}I\\ = {\rm{det}}\left( {P{P^{ - 1}}} \right)\\ = \left( {{\rm{det}}P} \right)\left( {{\rm{det}}{P^{ - 1}}} \right)\end{array}\]

Show that \({\rm{det}}A = {\rm{det}}B\)

Assume that \(A = PB{P^{ - 1}}\). Then use theorem 3 (Property of Determinant): \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).

\[\begin{array}{\rm{det}}A = {\rm{det}}\left( {PB{P^{ - 1}}} \right)\\ = {\rm{det}}\left( P \right){\rm{det}}\left( B \right){\rm{det}}\left( {{P^{ - 1}}} \right)\\ = {\rm{det}}\left( B \right){\rm{det}}\left( P \right){\rm{det}}\left( {{P^{ - 1}}} \right)\\ = {\rm{det}}\left( B \right){\rm{det}}\left( I \right)\\ = {\rm{det}}B\end{array}\]

Thus, this implies that \({\rm{det}}A = {\rm{det}}B\).