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

34. Let A and P be square matrices, with P invertible. Show that \(det\left( {PA{P^{ - {\bf{1}}}}} \right) = det{\rm{ }}A\).

According totheorem 6 of themultiplicative property, if A andB are square matrices, then thedeterminant of the product matrix AB is equal to the product of the determinant of A and the determinant of B.

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

As P is invertible, its determinant cannot be 0.

Apply the multiplicative propertyon the left side of the equation \(\det \left( {PA{P^{ - 1}}} \right) = \det {\rm{ }}A\)by substituting \(A = P\)and \(B = A{P^{ - 1}}\), as shown below:

\(\det \left( {P\left( {A{P^{ - 1}}} \right)} \right) = \det \left( P \right)\det \left( {A{P^{ - 1}}} \right)\)

Again, apply the property by using\(B = {P^{ - 1}}\).

\(\begin{aligned}{}\det \left( {P\left( {A{P^{ - 1}}} \right)} \right) &= \det \left( P \right)\det \left( {A\left( {{P^{ - 1}}} \right)} \right)\\ &= \det \left( P \right)\det \left( A \right)\det \left( {{P^{ - 1}}} \right)\\ &= \det \left( P \right)\det \left( {{P^{ - 1}}} \right)\det \left( A \right)\end{aligned}\)

Since P is invertible, use the property \(\det {P^{ - 1}} = \frac{1}{{\det P}}\), as shown below:

\(\begin{aligned}{}\det \left( {PA{P^{ - 1}}} \right) &= \det \left( P \right)\det \left( {{P^{ - 1}}} \right)\det \left( A \right)\\ &= \det \left( P \right)\left( {\frac{1}{{\det P}}} \right)\det \left( A \right)\\ &= \det \left( A \right)\end{aligned}\)

Hence, it is proved that \(\det \left( {PA{P^{ - 1}}} \right) = \det {\rm{ }}A\).