Question

Linear Algebra question:

Suppose A is a 3 x 3 matrix such that det (A) = 4. Then det (4(A^-1)^T) is equal to?

Answer 1

\[\int\limits_{?}^{?}\]

Answer 2

^T at the end means transpose correct? It's been quite a while since my linear algebra days.

Answer 3

Well here is my reasoning ok so determinant of A =4 then

\[\left| A^{-1} \right| = 1/\left| A \right|\]

therefore if det(A)=4 then det(A^-1)=1/4.

since we are multiplying by 4 then determinant is now 1

finally if i remember correctly...

\[\left| A \right|=\left| A^T \right|\]

there fore

there answer should just be 1 i think. i would love for someone to double check my logic though if anyone else remembers this stuff.

Answer 4

looks good to me
i have forgotten all the determinant properties

Answer 5

Ok good thank you =). I had to reach WAY back for those ha ha

Answer 6

i know the properties are right i was just unsure in the logic and how i followed them.

Answer 7

yeah i just looked up some determinant info
you are correct :)

Answer 8

NOOO i did it wrong. i got the right answer but i did my order of operations wrong i should of transposed first before i multiplied.

so it should be

det(A^-1)= 1/4
transpose = 1/4
mupltipy by 4 = 1

Answer 9

thanks for the answer. So if i understand you first have to do the inverse A^-1 which gives you 1/A and then you have to get the transpose which gives?

Answer 10

det(A)=det(A^T) so determinant of A is equal to the transposed determinant of A

Answer 11

Just follow order of operations and use each property i gave where necessary

Answer 12

Thanks, I have one more question. Can I post it here? I'm new to this.

Answer 13

just post new thread its easier

Answer 14

the answer is \[4^3\cdot\frac{1}{4}=16\]

Answer 15

I see you beat me to it. But I made up this example in MATLAB, so I'll post it.

Answer 16

thanks guys for your input