Show that the determinant of the matrix C does not depend on x (Linear Algebra)?

Determinant of a 3*3 matrix

a,b,c

d,e,f

g,h,i is

(aci + bfg + cdh) - (gec + hfa + idb)

aci +bfg + cdh = sin^2(x) + 0 + 0 = sin^2(x)

gec + hfa + idb = 0 + 0 - cos^2(x)

so det is sin^2(x) - (-cos^2(x)) = sin^2(x) + cos^2(x) = 1

Calculate the determinant and you will find come cancellation and some expressions like cos^2(x) + sin^2(x) which is equal

to 1.

Very simple, you can explicitly calculate it, by cofactor expansion on the last column:

|C| = |sin(x) cos(x); -cos(x) sin(x)| * 1 + |.| * 0 + |.| * 0 = sin^2 + cos^2 = 1

To evaluate a determinant you can expand by

minors along any row or column. In this case it seems

simplest to take the last column.

Indeed

det(C) = 0*det(C_13) - 0*det(C_23) + 1*det(C_33)

Here C_13 means remove the first row and 3rd column.

This means det(C) = det(C_33) = det(sin(x)*sin(x) - (-cosx)*cos(x)) = sin^2(x) + cos^(x) = 1

so det(C) = 1 which is independent of x

permit A_lambda denote the lambda-eigenspace of A. (c) follows right this moment from (b): if dim A_lambda = ok, meaning there's a foundation (v_1, ..., v_k) for A_lambda; if so (P^(-a million)v_1, ..., P^(-a million)v_k) is a foundation for B_lambda, so dim B_lambda = ok. To practice (b), use the undeniable fact that P is invertible. If (v_1, ..., v_k) is a foundation for A_lambda, then you definately comprehend that (P^(-a million)v_1, ..., P^(-a million)v_k) is a linearly self reliant checklist of vectors (in any different case the v_j might themselves be linearly based), and you comprehend that each and all of the P^(-a million)v_j lie in B_lambda. So this is a minimum of a partial foundation. think it weren't an entire foundation for B_lambda; permit w in B_lambda be linearly self reliant from each and all of the P^(-a million)v_j. yet then, via (a), Pw is in A_lambda, and linearly self reliant from each and all of the v_j, contradicting the determination of (v_1, ..., v_k) as a foundation for A_lambda. This completes the evidence.

the determinant is 1 and is thus independent of x...

0[(-cos)(sin+cos)-(sin)(sin-cos)] - 0[(sin)-0(sin-cos)] + 1[(sin)(sin)-(cos)(-cos)]

=sin^2+cos^2=1