If $A$ is a $3 \times 3$ matrix and $\det(A) = 4$, then compute $\det(((-9A)^4)^T)$.

Question

Given a $3\times3$ matrix $A$ $$A= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} $$ and $\det(A)=4$

Calculate $\det(((-9)\cdot A)^4)^T$.

Answer 1

Hints: If $M$ is an $n \times n$ matrix and $k \in \mathbb N$, then:

• $\det(M^T) = \det M$
• $\det(M^k) = (\det M)^k$
• $\det(kM) = k^n\det M$

Answer 2

Use the properties of determinant:

1. $\det A=\det A^T$
2. If $A$ is $n\times n$ matrix then $\det cA=c^n\det A$
3. $\det (AB)=(\det A)(\det B)$

Answer 3

HINT: The answer is $4^4\cdot{(-9)}^{12}$.

Answer 4

First recall

1. $det(A^p) = det(A)^p$

2. $det(A^T) = det(A)^T$

3. $det(kA) = k^ndet(A)$ where n is the size of the matrix (3 in your case: 3 rows and 3 columns).

$$\begin{split} det((((−9)⋅A)^4)^T &= det(((−9)⋅A)^4$ \text{ by property 2}\\ &= [det((−9)⋅A)]^4$ \text{ by property 1} \\ &= [(−9)^3⋅det(A)]^4$ \text{ by property 3} \\ &= [(−9)^3⋅4]^4$ \text{ by your note det(A) =4}. \end{split} $$