question help on 7-10 7. For two n x r matrices A and B, we define the commutato

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question help on 7-10

7. For two n x r matrices A and B, we define the commutator, den oted [A,B], by [A, B):= AB-BA. a) what is [4,1], where l is the n × n identity nnatrix? b) What is [A, A] for any n x n matrix A ? c) Give an example of two nonzero 2 x 2 matrices A and B such that [A, B] is not the zero matrix d) Fix an n x n matrix C. Show that [A +B, C] = [A,C] + |B,C] e) [Optional - but fun!] Compute out this sum [A, [B,C]]+IB, [C, A+IC, IA, B]. 8. Use matrix algebra to show that if A is invertible and D satisfies AD 1, then 9. Suppose A is an n × n matrix with entries a,i and all the rest zero. Determine a 10. Suppose and n × n matrix A has det(A-1. If det(2A-16, what is the size, n, of A. D = A-1 formula for A-1

Explanation / Answer

7)

For the problem, A is a 2x2 matrix (n=2) =

PART a) I is the identity matrix. By definition, AI = IA = A. It's the matrix equivalent of 1 where any matrix multiplied by I gives back the matrix itself. Like 5x1 = 5 or ax1 = a ----> AI = IA = I

(2x2) The matrix I =  

So, [A,I] = AI - IA

By definition, AI = IA = A

----> [A,I] = AI - IA = A - A = 0.

Thus, we would get the 2x2 matrix with all entries = 0.

PART b) [A,A] = AA - AA

For clarity,

AA =

X

=

Anyway, that's how AxA would look like.

But [A, A] would cancel both the terms out ----> [A,A] = AA - AA = 0

This would be the 2x2 matrix with all entries = 0.

PART c) An example where [A,B] = AB - BA is not equal to the zero matrix -->

Okay, so let A =

and B =

PART d) [A+B , C] = (A+B)(C) - (C)(A+B)

[A+B , C] = (AC+BC) - (CA+CB)

[A+B , C] = (AC-CA) + (BC-CB)

[A+B , C] = [A,C] + [B,C]

8) PROOF:

Since AB=I, we have

det(A)det(B)=det(AB)=det(I)=1

This implies that the determinants det(A) and det(B) are not zero.
Hence A and B are invertible matrices: A1,B1 exist.

Now we compute

I=BB1=BIB1=B(AB)B1=BAI=BA , since AB=I

Hence we obtain BA=I
Since AB=I and BA=I, we conclude that B=A1

Hence Proved.

9) Okay, so for a 2x2 matrix:

Let A =

Det(A) = |A| = 2

A-1 =

For a 3x3 matrix,

A =

Det(A) = |A| = (1)(6) + 2(3) + 3(2) = 18

A-1  =

10)

If A is a square n matrix, det(x.A)=xn.det(A)

Therefore, as det(A) = 1 and det(2A) = 16 --->

det(2A) = 2n.det(A)

16 = 2n(1)

n = 4

Cheers!

a b c d

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