Compute each of the following quantities

Question

Assume that A and B are 3 x 3 matrices with det(A) = 2 and det(B) = 4

1. det(AB)
2. det(2A²)
3. det(A^-1B^T)
4. rank(A)
5. dim(null(B))

Answer 1

1) Here you need to use det(AB)=det(A)det(B). Since det(A) = 2 and det(B) = 4 we get det(AB)=2·4=8.

2) Note that 2A² =2I · A², where I is the 3×3 identity matrix. Using the formula det(AB)=det(A)det(B) we get det(2A²) = det(2I · A²)= det(2I) · det(A²). Since det(2I)=2³=8 and det(A²)= det(A) · det(A) = 2 · 2=4, you get det(2A²) =8 · 4 =32.

3) Since A·A^-1=I we have det(A·A^-1)=det(I). Since det(A·A^-1)=det(A) ·det(A^-1) and det(I)=1, we get

det(A) ·det(A^-1) = 1. This implies det(A^-1) =1/ det(A). Also, det(B^T)=det(B) since the determinant of a  matrix is the same as the determinant of its transpose.

Thus det(A^-1B^T)= det(A^-1) · det(B^T)= (1/ det(A) ) · det(B) = (1/2) · 4 = 2.

4) The rank of a matrix is defined as the maximum number of linearly independent column vectors in the matrix. Here A is a 3×3 matrix, so it has 3 column vectors. Since det(A)=2 is nonzero its 3 column vectors are linearly independent. Thus rank(A) = 3.

5) null(B) denotes the nullspace of B. It is the set of all vectors X such that BX=0. Since det(B) =4 is nonzero, B is invertible. Let B^-1 denote its inverse. We multiply both sides of the equation BX=0 by B^-1.

⇒ B^-1BX=B^-10 ⇒ IX =0 ⇒ X=0. Thus null(B) = {0} and dim(null(B)) = 0.

Note. If a matrix M has nonzero determinant then it is invertible and dim(null(M))=0.