Please Help! V. Hard Integration/Matrix Questions (1 Viewer)

[pink_rabbit_<3]

Hey guys.
Im stuck on these difficult questions and maybe u can help...Thanks Heaps.

1. Find a function f and a number a so that

486 + 5
S x
a f(t)t7/2 dt = 2x5/2.

2. Write the following system of equations in the form Ax = b, find the inverse of A, and use it to solve the
system. Without further calculation, what can you say about the determinant of A?
− x2 − 2x3 = 0
x1 + x2 + 4x3 = 7
x1 + 3x2 + 7x3 = 4

3. Suppose detA = 3, detB = −2/3 , detC = 1/2 . Calculate
(a) det(A(^2)B(^−1)C) (b) det(A(^−1)B(^2)C(^−1))(^−1)

4. Suppose A and B and invertible n × n matrices. Show that
(I + AB)(^−1)A = A(I + BA)(^−1)

 

[ronnknee]

I want to help but I don't understand the notation of the conditions for the first question

 

[Iruka]

For Q2, remember that if a matrix is invertible then it has a non-zero determinant.

For Q3, det(AB) = det(A) det(B), det(A^(-1)) = 1/det(A).

For Q4, remember that A=[A^(-1)]^(-1), and [AB]^(-1) = B^(-1)A^(-1)

 

[leoyh]

wth is with all this weird terminology

 

[pLuvia]

The last three questions are taught at undergrad maths. Unless you are in melb

 

[ronnknee]

Mm yea I thought so, I've never heard of matrices before. Isn't this in the wrong forum then?

 

[pLuvia]

The matrices questions are but not sure about the first one. I'll move it

 

[sicmacao]

Hey guys.
Im stuck on these difficult questions and maybe u can help...Thanks Heaps.

1. Find a function f and a number a so that

486 + 5
S x
a f(t)t7/2 dt = 2x5/2.

2. Write the following system of equations in the form Ax = b, find the inverse of A, and use it to solve the
system. Without further calculation, what can you say about the determinant of A?
− x2 − 2x3 = 0
x1 + x2 + 4x3 = 7
x1 + 3x2 + 7x3 = 4

3. Suppose detA = 3, detB = −2/3 , detC = 1/2 . Calculate
(a) det(A(^2)B(^−1)C) (b) det(A(^−1)B(^2)C(^−1))(^−1)

4. Suppose A and B and invertible n × n matrices. Show that
(I + AB)(^−1)A = A(I + BA)(^−1)

for
4. Suppose A and B and invertible n × n matrices. Show that
(I + AB)(^−1)A = A(I + BA)(^−1)


Assuming (I + AB) and (I + BA) are invertible.

Proving (I + AB)(^−1)A = A(I + BA)(^−1)

is equivalent to proving
A(I + BA) = (I + AB)A

So LHS = A(I + BA)
= A + ABA
= (I+AB)A
= RHS.

Therefore A(I + BA) = (I + AB)A
Hence (I + AB)(^−1)A = A(I + BA)(^−1)
Q.E.D.