Determinant Prob. (1 Viewer)

[Yip]

This is kinda off-syllabus, but i was wondering how one would go about this problem:

(a) Let A be a 3x3 non-singular matrix, and I be the 3x3 identity matrix. Show that:

det(A^-1-xI)=[-x³/det(A)][det(A-x^-1I)]

(b) Let A=
0 1 0
0 0 1
4 -17 8
(i) Show that 4 is a root of det(A-xI)=0 and hence find the other roots in surd form
(ii) Solve det(A^-1-xI)=0

I can only work out (b)(i) ~.~
thx in advance ^^



 

[KeypadSDM]

Don't you just replace x with 1/x from part i and use part a) to solve?

 

[Yip]

oh ic....~.~ that means that the roots are simply the reciprocals of the roots of (b)(i) eh...i have no clue about proving part a though ~.~ im rather new to this matrix stuff...

 

[ngai]

properties of determinant that u need:
det(AB) = det(A)det(B)
det(A^-1) = 1/det(A)
det(cA) = cⁿdet(A) for A in Rⁿ
hope uve seen those before =)

so then det(A-x^-1I) / det(A)
= det((A-x^-1I)A^-1)
= det(I - x^-1A^-1)
= (-1)³det(A^-1x^-1 - I)
= - det(A^-1 - xI) / x³

 

[Templar]

Hmm...you returning for good, ngai, or just a temporary visit?