Question

A =
\[\left[\begin{matrix}1 & 2 & 4 \\ -4 & 1 & 3 \\ 2 & 4 & -2 \end{matrix}\right]\]

Without determining A^-1 or A^T evaluate, if possible, with reasons,

a) det(4A^-1)
b) det ((A^-1)^T)

Answer 1

The determinant of this matrix is -90. Don't know how to approach it further...

Answer 2

the determinant of a matrix is the same as that of its transpose, but how to get around finding the inverse...

Answer 3

Should I calculate the inverse of this to answer part(a) ?

Answer 4

it says not to, so let's try to avoid it
oh wait, I think i found something....

Answer 5

so\[\det(A^T)=\det(A)\]now we also have that\[I=AA^{-1}\]\[\det I=\det(AA^{-1})\]\[1=\det A\cdot\det(A^{-1})\implies\det(A^{-1})=\frac1{\det A}\]

Answer 6

oh, and for an \(n\times n\) matrix we also have that\[\det(cA)=c^n\det A\]

Answer 7

so\[\det(4A^{-1})=4^3\det(A^{-1})=\frac{4^3}{\det A}\]and\[\det((A^{-1})^T)=\det(A^{-1})=\frac1{\det A}\]now it should be a piece of cake

Answer 8

This really help me a lot! I have part a as \[-\frac{ 32 }{ 45 }\] and part b as \[-\frac{ 1 }{ 90 }\] Is this correct?

Answer 9

oh man, you're gonna make me check the arithmetic? let me see if I can wolfram it to avoid silly mistakes...

Answer 10

yeah that's right :)

Answer 11

You are a life saver! Thank you so much!!!!!!!

Answer 12

anytime !