# Thread: Logical ways to find certain matrices...

1. ## Logical ways to find certain matrices...

I'm stuck on 2 problems. It's not the answers I care about, just the understanding of how to get to the answer.

1) Can you find a 3x2 matrix A and a 2x3 matrix B such that AB = I?

2) Find a 2x2 matrix A =/= I with A^3 = I.

How can you do these without brute forcing the solution?

2. Originally Posted by paulrb
I'm stuck on 2 problems. It's not the answers I care about, just the understanding of how to get to the answer.

1) Can you find a 3x2 matrix A and a 2x3 matrix B such that AB = I?

2) Find a 2x2 matrix A =/= I with A^3 = I.

How can you do these without brute forcing the solution?
1) For a matrix B to be the inverse of matrix A, it must satisfy the following equation:

$\displaystyle AB = I = BA$

So basically, if you can make up a random non-singular matrix and call it A, then B will be its inverse, and I'm sure you know how to find that!

2) $\displaystyle A^3 = A^2 \times A = I$

So now you have to find a matrix A whose inverse is also its square!

$\displaystyle A = \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$

$\displaystyle A^2 = \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right) \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)= \left( \begin{array}{ccc} a^2+b^2 & ab+bd \\ ac+cd & d^2+bc \end{array} \right)$

$\displaystyle A^{-1} = \frac{1}{ad-bc} \left( \begin{array}{ccc} d & -b \\ -c & a \end{array} \right)$

Hence:

$\displaystyle \frac{d}{ad-bc} = a^2+b^2$, etc. 4 equations, 4 unknowns, you do the math.

3. For #1 I'm pretty sure there's no such thing as a non-square matrix that has an inverse (such that AB = I = BA).

For #2, I know the square of the matrix is also the inverse of the matrix. But it still seems kind of unintuitive how to arrive at the solution...

4. Originally Posted by paulrb
For #1 I'm pretty sure there's no such thing as a non-square matrix that has an inverse (such that AB = I = BA).

For #2, I know the square of the matrix is also the inverse of the matrix. But it still seems kind of unintuitive how to arrive at the solution...
Ah sorry, I didn't quite read your first question as carefully as I should have!

5. Ah ok well the second one makes sense now. Thanks

6. Originally Posted by paulrb
Ah ok well the second one makes sense now. Thanks
Alternatively just find $\displaystyle A^3$ and equate it to the identity, as opposed to finding the square and the inverse and equating them...

For the first one, set up two matrices with random variables again:

$\displaystyle A = \left( \begin{array}{ccc} a & b & c \\ d & e & f \end{array} \right)$

$\displaystyle B = \left( \begin{array}{ccc} g & h \\ i & j \\ k & l \end{array} \right)$

$\displaystyle AB = \left( \begin{array}{ccc} ag+bi+ck & ah+bj+cl \\ dg+ei+fk & dh+ej+fl \end{array} \right)$

We know this must equate to:

$\displaystyle AB = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$

So, if we say that $\displaystyle b = i = c = k = 0$, and $\displaystyle a = g = 1$, then the first element is satisfied. Now look at the send element:

$\displaystyle ah + bj + cl = 1h + 0j + 0l = 0$, hence $\displaystyle h = 0$.

Now look at the 3rd element: $\displaystyle dg+ei+fk = d(1) + e(0) + f(0) = 0$, hence $\displaystyle d= 0$.

Now the last element: pmath]dh+ej+fl = 0(0) + e(0) + f(l) = 1[/tex], so just set $\displaystyle f = l = 1$, and you are sorted.

Also e = j = 0

Hence:

$\displaystyle A = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$

$\displaystyle B = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{array} \right)$