1. Proving Matrices

Hi all.
I'm very much stuck up on these three problems (and they're the only problems I haven't been able to solve). Anyways, here they are:

1.) If A and B are nonsingular N x N matrices and C = AB, show that C^-1 = (A^-1)(B^-1).
Hint: You must use the associative property of matrix multiplication.
PS: For reference, the associative property of matrix multiplication is A(BC) = (AB)C

2.) Let A be an M x N matrix and B be an N x P matrix.
(a) How many multiplications are needed to calculate AB?
(b) How many additions are needed to calculate AB?

3.) Find (X)(X^T) and (X^T)(X), where X = (1,-1,2).
Note: X^T is the transpose of X.

Any help would be very very very much appreciated. I'm so desperate!

2. Originally Posted by zeugma
Hi all.
I'm very much stuck up on these three problems (and they're the only problems I haven't been able to solve). Anyways, here they are:

1.) If A and B are nonsingular N x N matrices and C = AB, show that C^-1 = (A^-1)(B^-1). I think there is an error here. I think it should be that we are to show that C^{-1}=B^{-1}A^{-1}...
Hint: You must use the associative property of matrix multiplication.
PS: For reference, the associative property of matrix multiplication is A(BC) = (AB)C
Since $C=AB$, and $C^{-1}=B^{-1}A^{-1}$, all you need to do is see if $CC^{-1}=I$. The proof will then be complete.

2.) Let A be an M x N matrix and B be an N x P matrix.
(a) How many multiplications are needed to calculate AB?
Disregard my previous answer...see Iso's post below...XD

(b) How many additions are needed to calculate AB?
I'm not sure what is to be done here...

3.) Find (X)(X^T) and (X^T)(X), where X = (1,-1,2).
Note: X^T is the transpose of X.

Any help would be very very very much appreciated. I'm so desperate!
Hint: the dimension of $XX^T$ is 1 x 1 and the dimension of $X^TX$ is 3 x 3.

Can you try to take it from here?

3. Originally Posted by zeugma
Hi all.
I'm very much stuck up on these three problems (and they're the only problems I haven't been able to solve). Anyways, here they are:

1.) If A and B are nonsingular N x N matrices and C = AB, show that C^-1 = (A^-1)(B^-1).
Hint: You must use the associative property of matrix multiplication.
PS: For reference, the associative property of matrix multiplication is A(BC) = (AB)C
$CC^{-1} = ((AB)(B^{-1}) A^{-1}) = (A(B B^{-1})) A^{-1} = (AI)A^{-1} = AA^{-1} = I$

2.) Let A be an M x N matrix and B be an N x P matrix.
(a) How many multiplications are needed to calculate AB?
(b) How many additions are needed to calculate AB?
(a) do you realize that every entry in AB is made up of a dot product of two n length vectors? And there are MP entries in AB.

(b) Again each entry in AB is made up of a dot product of two n length vectors. And we need n-1 additions for that (per entry)

3.) Find (X)(X^T) and (X^T)(X), where X = (1,-1,2).
Note: X^T is the transpose of X.
This is just applying the definition... Exactly on which part are you stuck? Show us your work .... we will comment on it.

4. Originally Posted by isomorphism

this is just applying the definition... Exactly on which part are you stuck? Show us your work .... We will comment on it.
$XX^{T}=(1)(1) + (-1)(-1) + (-2)(-2) = 1 + 1 + 4 = 6$

and

$X^{T}X =(1)(1) + (-1)(-1) + (-2)(-2) = 1 + 1 + 4 = 6$

I know there's something wrong with my work.

5. XX^T is a row vector times a column vector... So what you have done is right. However X^T X is a 3 x 1 column vector times a 1 x 3 row vector. The dimension of the product matrix X^T X, should be 3 x 3, isnt it?

Follow the rules of matrix multiplication and carefully multiply X^T X, you will get a 3 x 3 matrix as the answer. It generally called the outer product.

6. 3.) Find (X)(X^T) and (X^T)(X), where X = (1,-1,2).
Note: X^T is the transpose of X.
Dude, here, two matrices are multiplied
so the answer will be in matrix form

(X)(X^T)=(1)(1) + (-1)(-1) + (-2)(-2) = 1 + 1 + 4 = (6)

(X^T)(X)=( 1 -1 2)
(-1 1 -2)
( 2 -2 4)

7. Originally Posted by Isomorphism
XX^T is a row vector times a column vector... So what you have done is right. However X^T X is a 3 x 1 column vector times a 1 x 3 row vector. The dimension of the product matrix X^T X, should be 3 x 3, isnt it?

Follow the rules of matrix multiplication and carefully multiply X^T X, you will get a 3 x 3 matrix as the answer. It generally called the outer product.
Oh I guess I overlooked the book. If you don't mind, would you check my answer below for the $X^{T} X$:

$X^{T} X$= [(1)(1) + (1)(-1) + (1)(2)] + [(-1)(1) + (-1)(-1) + (-1)(2)] + [(2)(1) + (2)(-1) + (2)(2)]
= (1 -1 + 2) + (-1 + 1 -2) + (2 - 2 + 4)
= 2 - 2 + 4
$X^{T} X = 0$

8. Originally Posted by zeugma
Oh I guess I overlooked the book. If you don't mind, would you check my answer below for the $X^{T} X$:

$X^{T} X$= [(1)(1) + (1)(-1) + (1)(2)] + [(-1)(1) + (-1)(-1) + (-1)(2)] + [(2)(1) + (2)(-1) + (2)(2)]
= (1 -1 + 2) + (-1 + 1 -2) + (2 - 2 + 4)
= 2 - 2 + 4
$X^{T} X = 0$
Sigh... I already told you that X^T X is a 3 x 3 matrix assuming X is a row vector and yet you are working out the same old dot product. Is the answer 0 a 3 x 3 matrix?

I want you to get this answer on your own, thats why we give hints. So I will show an example...

If X = (1,2), what is X^T X?
$
X^T X =\begin{pmatrix} 1 \\ 2 \end{pmatrix} (1\, 2)$
$=\begin{pmatrix} 1.1 & 2.1 \\ 1.2 & 2.2 \end{pmatrix} = \begin{pmatrix}1 & 2 \\ 2 & 4 \end{pmatrix}$

Try the problem again...

9. Oh I guess I overlooked the book. If you don't mind, would you check my answer below for the :

= [(1)(1) + (1)(-1) + (1)(2)] + [(-1)(1) + (-1)(-1) + (-1)(2)] + [(2)(1) + (2)(-1) + (2)(2)]
= (1 -1 + 2) + (-1 + 1 -2) + (2 - 2 + 4)
= 2 - 2 + 4
hey I gave u the answer

(X^T)(X)=
( 1 -1 2):row1
(-1 1 -2):row2
( 2 -2 4):row3
its a 3x3 matrix
got it....

10. Ok got it ... thanks.