Logical ways to find certain matrices...

**paulrb** · April 2nd 2009, 05:54 PM

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?

**Mush** · April 2nd 2009, 05:56 PM

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:

$AB = I = BA$

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

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

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

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

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

Hence:

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

**paulrb** · April 2nd 2009, 06:03 PM

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...

**Mush** · April 2nd 2009, 06:09 PM

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!

**paulrb** · April 2nd 2009, 06:11 PM

Ah ok well the second one makes sense now. Thanks

**Mush** · April 2nd 2009, 06:14 PM

Originally Posted by paulrb

Ah ok well the second one makes sense now. Thanks

Alternatively just find $A^3$ 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:

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

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

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

We know this must equate to:

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

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

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

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

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

Also e = j = 0

Hence:

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

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

Thread: Logical ways to find certain matrices...

LinkBack

Thread Tools

Display

Logical ways to find certain matrices...

Similar Math Help Forum Discussions

Find ways to invest money

two ways to find hypotenuse

Prove/disprove using logical using logical arguments

Using truth tables to show logical consistancy between logical expressions

find the number of ways

Search Tags