- #1
DanielJackins
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Homework Statement
QUESTION 1:
Suppose A is a 3x3 matrix such that det(A) = 1/3.
Then det[A^3((adj(A))^-1)^2)] =
and det[3A^-1(adj(A))] =
and
QUESTION 2:
Let A and B be 3x3 invertible matrices where det(A) = 2 and det(B) = -8
Then det[A^T(adj(B))] =
and det[A^-1(adj(B))] =
The Attempt at a Solution
QUESTION 1:
For the first blank I got the correct solution but I don't know if that was just by chance. I said A^3 = 1/27 * the inverse of the adj(A) squared. I found the inverse of the adj(A) to be 3. so 1/27 * 3 * 3 = 3 which was correct.
For the second blank I said 3A^-1 is equal to 81 (because I believe you have to raise 3 to the power of the dimension of the matrix?). Then inside the brackets I tried to do the adj(A), which if I'm not mistaken is equal to det(A)*A^-1. det(A) = 1/3, which I then raised to the power of 3 as well, and multiplied by A^-1, which is 3. So all in all I had 81(1/27*3) which was not correct. Any help?
QUESTION 2:
For the first blank I said 2((-8^3)*(-1/8)) using the same logic as above and arrived at 128. (This may actually be correct as I haven't checked but I only have one attempt left and I want to be sure it's correct).
The second blank I did the same but with 1/2 on the outside as opposed to 2, and arrived at 32.
Any help would be greatly appreciated