Question

Question 40: Let \(A\) and \(B\) be \(4 \times 4\) matrices, with \(\det A = - 3\)and \(\det B = - 1\). Use properties of determinants (in the text and in the exercises above) to compute:

a. \(\det AB\)

b. \(\det {B^5}\)

c. \(\det 2A\)

d. \(\det {A^T}BA\)

e. \(\det {B^{ - 1}}AB\)

Short answer

1. \(\det AB = 3\)
2. \(\det {B^5} = - 1\)
3. \(\det 2A = - 48\)
4. \(\det {A^T}A = 9\)
5. \(\det {B^{ - 1}}AB = - 3\)

Step-by-step solution

Use the properties of the determinant to compute \(\det AB\)

a)

Theorem 6states that if \(A\) and \(B\) are \(n \times n\) matrices, then \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).

Let \(A\) and \(B\) be \(3 \times 3\) matrices with \(\det A = - 3\) and \(\det B = - 1\).

Use the multiplicative property to compute \(\det AB\) as shown below:

\(\begin{aligned}{}\det AB & = \left( {\det A} \right)\left( {\det B} \right)\\ & = \left( { - 3} \right)\left( { - 1} \right)\\ & = 3\end{aligned}\)

Thus, \(\det AB = 3\).

Use the properties of the determinant to compute \(\det {B^5}\)

b)

Theorem 6states that if \(A\) and \(B\) are \(n \times n\) matrices, then \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\).

Use theorem 6 to compute \(\det {B^5}\) as shown below:

\(\begin{aligned}{}\det {B^5} & = {\left( { - 1} \right)^5}\\ & = - 1\end{aligned}\)

Thus, \(\det {B^5} = - 1\).

Use the properties of the determinant to compute \(\det 2A\)

c)

Exercises 36states that if \(A\) is an\(n \times n\) matrix, then \(\det rA = {r^n}\det A\).

Use exercise 36 to compute \(\det {B^T}\) as shown below:

\(\begin{aligned}{}\det 2A & = {2^4}\det A\\ & = 16 \times - 3\\ & = - 48\end{aligned}\)

Thus, \(\det 2A = - 48\).

Use the properties of the determinant to compute \(\det {A^T}A\)

d)

Theorem 6states that if \(A\) and \(B\) are \(n \times n\) matrices, then \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\). Theorem 5states that if \(A\) is an\(n \times n\) matrix, then \(\det {A^T} = \det A\).

Use theorems 5 and 6 to compute \(\det {A^T}A\) as shown below:

\(\begin{aligned}{}\det {A^T}A & = \left( {\det {A^T}} \right)\left( {\det A} \right)\\ & = \left( {\det A} \right)\left( {\det A} \right)\\ & = - 3 \times - 3\\ & = 9\end{aligned}\)

Thus, \(\det {A^T}A = 9\).

Use the properties of the determinant to compute \(\det {B^{ - 1}}AB\)

e)

Exercise 31states that if \(A\) invertible, then \(\det {A^{ - 1}} = \frac{1}{{\det A}}\).

Use theorem 6 and exercise 31 to compute \(\det {B^{ - 1}}AB\) as shown below:

\(\begin{aligned}{}\det {B^{ - 1}}AB & = \left( {\det {B^{ - 1}}} \right)\left( {\det A} \right)\left( {\det B} \right)\\ & = \left( {\frac{1}{{\det B}}} \right)\left( {\det A} \right)\left( {\det B} \right)\\ & = \det A\\ & = - 3\end{aligned}\)

Thus, \(\det {B^{ - 1}}AB = - 3\).