# Using Matrix Multiplication to Obtain Another Matrix

1. Mar 1, 2013

### annpaulveal

1. The problem statement, all variables and given/known data

I need to use matrix multiplication of matrices A-D to obtain matrix E. I also need to set a equal to some value that would allow me to perform this multiplication.

2. Relevant equations

The matrixes I need to multiply:
A = [1, 1; 0, 1]
B = [1, 0; 1, 1]
C = [a, 0; 0, 1]
D = [1, 0; 0, a]

By multiplying the above matrices in some order, I need to obtain the following matrix:

E = [0, 1; 1, 0]

3. The attempt at a solution

I have tried setting a = 0, 1, -1, 2, and -2 (since those seemed like the most logical choices) and tried all combinations of multiplication. For the life of me, I can't figure out the order the multiplication should go in and which of the four matrices to multiply together to get matrix E. There has to be a simpler way of approaching the problem, right? I just need some help getting onto the right path.

2. Mar 1, 2013

### jbunniii

Note that $CD = DC = aI$ (i.e., the scalar $a$ times the identity matrix), and $AB(aI) = A(aI)B = (aI)AB$. None of these will work, because $AB$ is not a multiple of $E$. Similarly, $BA$ is not a multiple of $E$. So you know $C$ and $D$ can't be adjacent.

Maybe you can make some other observations which will help you reduce the possibilities further. Anything has to be better than just randomly trying different arrangements, because there are a lot of them ($4! = 24$).

Another observation is that $E^2 = I$, not sure if that will be helpful.

Last edited: Mar 1, 2013
3. Mar 1, 2013

### annpaulveal

Thanks for your help, but I'm still struggling tremendously. I understand that a will most likely be 0 or -1 (or maybe 1, but that would make C and D the same identity matrix). Is there any other guidance I could get?

4. Mar 1, 2013

### Ray Vickson

It might be easier to look at the transformations the matrices induce. If
$$x= \pmatrix{x_1\\x_2}$$ then
$$A: x \longrightarrow \pmatrix{x_1+x_2\\x_2}\\ B: x \longrightarrow \pmatrix{x_1\\x_1+x_2}\\ C: x \longrightarrow \pmatrix{ax_1\\x_2}\\ D: x \longrightarrow \pmatrix{x_1\\ax_2}$$ and you want to find a sequences of such transformations (and a value of 'a') that gives
$$\pmatrix{x_1\\x_2} \: \longrightarrow \pmatrix{x_2\\x_1}$$

Last edited: Mar 1, 2013
5. Mar 1, 2013

### jbunniii

You can use determinants to find the possible value(s) for $a$.

6. Mar 1, 2013

### annpaulveal

Sorry, I still haven't been able to make any progress at all. Thanks anyway.

7. Mar 1, 2013

### jbunniii

Not even in finding the possible value(s) for $a$? Did you try taking determinants of both sides as I suggested? The answer might surprise you: $a$ can't even be a real number, assuming I did the calculation correctly.

8. Mar 1, 2013

### annpaulveal

I haven't been taught how to use determinants like you said, and it wouldn't be possible for a to be an imaginary number for this solution. If I wasn't clear, I have to figure out some kind of combination of multiplying matrixes A-D in order to get E as the result. First, I have to figure out what a should be. So, for example, a solution would be (A*B)*(C*D), or some similar combination.

I've tried everything and I've been working on this for hours. I just want to cry.

9. Mar 1, 2013

### jbunniii

I don't think a solution exists. Here is how you use determinants to find $a$. Regardless of the order in which you multiply $A$, $B$, $C$, and $D$, the resulting product will have the same determinant, namely $\det(A)\det(B)\det(C)\det(D)$, where for a general matrix $[a\textrm{ }b; c\textrm{ }d]$, the determinant is $ad - bc$. Using this formula, we get $\det(A) = \det(B) = 1$, $\det(C) = \det(D) = a$, and $\det(E) = -1$. As $\det(A)\det(B)\det(C)\det(D) = \det(E)$, we have $a^2 = -1$, so $a = i$ or $a = -i$.

Out of curiosity, I wrote a brute-force Matlab script which computes the products of all 24 possible permutations, assuming either $a = i$ or $a = -i$. No permutation even resulted in a real-valued matrix, let alone $E$. So either I made a coding error or there is no solution.

10. Mar 2, 2013

### Ray Vickson

You can do it, but you need more than 4 multiplications. Recall how the matrices transform vectors:
$$A:\pmatrix{x\\y}\rightarrow\pmatrix{x+y\\y}\\ B: \pmatrix{x\\y} \rightarrow \pmatrix{x\\x+y}\\ C: \pmatrix{x\\y} \rightarrow \pmatrix{ax\\y}\\ D: \pmatrix{x\\y} \rightarrow \pmatrix{x\\ay}$$
If we set $a = -1$ we can transform
$$\pmatrix{x\\y} \rightarrow \pmatrix{y\\x}$$
through a sequence of such moves:
$$\pmatrix{x\\y} \rightarrow \pmatrix{-x\\y} \rightarrow \pmatrix{-x\\y-x} \rightarrow \pmatrix{x\\y-x} \rightarrow \pmatrix{y\\y-x} \rightarrow \pmatrix{-y\\y-x} \rightarrow \pmatrix{-y\\-x} \rightarrow \pmatrix{y\\-x} \rightarrow \pmatrix{y\\x}.$$