Question

Does anyone know where I would start with this and what method I would use?

Q. Suppose det A = 3, B = -2/3 and C = 1/2. Calculate det (A^2B^-1C).

Answer 1

Am I supposed to use detA^-1 formula = 1/detA

so then my answer becomes det (3^2 * 1 / -2/3 * 1/2 )
= -27/4

Answer 2

yes

for matrices A and B
det(A*B)= det(A)* det(B)
also, det(A_inv) = 1/det(A)
so you get
det(A)*det(A)*det(B_inv)*det©
3*3* * -3/2 * ½ = -27/4
which is what you got.

Answer 3

thank you!
for b) det (A^-1B^2C^-1)^-1
my workings are:

det ( 1 / ( 1/3*-2^2/3* 1/1/2)
= 17/9

Is this also correct?

Answer 4

sorry answer should be 27/8

Answer 5

A= 3 , B= -⅔, C= ½ <--oops, C is ½ not 2

⅓ * (-⅔ * -⅔) * 2 = 8/27
invert the whole thing to get 27/8

Answer 6

Thanks :)

Answer 7

yw