Question

Question 39: Let \(A\) and \(B\) be \(3 \times 3\) matrices, with \(\det A = - 3\)and \(\det B = 4\). Use properties of determinants (in the text and in the exercises above) to compute:

a. \(\det AB\)

b. \(\det 5A\)

c. \(\det {B^T}\)

d. \(\det {A^{ - 1}}\)

e. \(\det {A^3}\)

Short answer

1. \(\det AB = - 12\)
2. \(\det 5A = - 375\)
3. \(\det {B^T} = 4\)
4. \(\det {A^{ - 1}} = \frac{1}{{ - 3}}\)
5. \(\det {A^3} = - 27\)

Step-by-step solution

Use the properties of determinants 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 \(\det B = 4\).

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

\(\begin{aligned}{}\det AB &= \left( {\det A} \right)\left( {\det B} \right)\\ &= \left( { - 3} \right)\left( 4 \right)\\ &= - 12\end{aligned}\)

Thus, \(\det AB = - 12\).

Use the properties of determinants to compute \(\det 5A\)

b)

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

Let \(A\) and \(B\) be \(3 \times 3\) matrices with \(\det A = - 3\)and \(\det B = 4\).

Use Exercise 36 to compute \(\det 5A\) as shown below:

\(\begin{aligned}{}\det 5A = {5^3}\det A\\ = 125 \times - 3\\ = - 375\end{aligned}\)

Thus, \(\det 5A = - 375\).

Use the properties of determinants to compute \(\det {B^T}\)

c)

Theorem 5states that if \(A\) is an\(n \times n\) matrix, then \(\det {A^T} = \det A\).

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

\(\begin{aligned}{}\det {B^T} &= \det B\\ &= 4\end{aligned}\)

Thus, \(\det {B^T} = 4\).

Use the properties of determinants to compute \(\det {A^{ - 1}}\)

d)

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

Use Exercise 31 to compute \(\det {A^{ - 1}}\) as shown below.

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

Thus, \(\det {A^{ - 1}} = \frac{1}{{ - 3}}\).

Use the properties of determinants to compute \(\det {A^3}\)

e)

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 \(\det {A^3}\) as shown below:

\(\begin{aligned}{}\det {A^3} &= {\left( {\det A} \right)^3}\\ &= {\left( { - 3} \right)^3}\\ &= - 27\end{aligned}\)

Thus, \(\det {A^3} = - 27\).