142k views
3 votes
Let A be a 5x5 matrix, det(A)= -3. Find

a) det(-A)

b) det(A^5)

c) det(-2A^T)

d) det(A^-3)

1 Answer

2 votes

Some properties of the determinant:

  • For any two square matrices
    A,B, we have
    \det(AB)=\det A\det B.
  • For an
    n* n matrix
    A,
    \det(kA)=k^n\det A for
    k\in\Bbb R.

  • \det(A^\top)=\det A
  • For an invertible matrix
    A,
    \det(A^(-1))=\frac1{\det A}.

So

a.
\det(-A)=(-1)^5\det A=3

b.
\det(A^5)=(\det A)^5=243

c.
\det(-2A^\top)=(-2)^5\det(A^\top)=(-2)^5\det A=96

d. Since
\det A\\eq0, the inverse
A^(-1) exists, so
A^(-n)=(A^(-1))^n.


\det(A^(-3))=\det((A^(-1))^3)=(\det(A^(-1)))^3=\frac1{(\det A)^3}=-\frac1{27}

User IvanJazz
by
5.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.