1. ## matrices help!

I'm teaching myself matrices because it could prove useful in the future.

I'll just run through a few things I've done to check if I'm correct:

i'm confident in adding/subtracting and multiplying.

What I'm having trouble getting my mind around is identities, inverses and determinants for 2x2 matrices.

In my book it says that for matrices of a given dimension (2x2) a identity I exists...

$\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

What does this I mean?

Then after this it says

If $\displaystyle A =\left(\begin{array}{cc}2&3\\1&4\end{array}\right) , AI = IA = A$

I have no idea in the world what this means. Could someone clarify it for me please =]

2. The matrix $\displaystyle I$ has the same function to matrices as the number $\displaystyle 1$ has to numbers.

If you multiply any number by one and you will get the same number back.

If you multiply any matrix by $\displaystyle I$ and you will get the same matrix back.

Nothing more to it!

3. Oh okay! Cool stuff.

Now... there's an example question here which I don't really get...

For the Matrix A = $\displaystyle \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$ find: $\displaystyle A^-1$

For the solution they wrote:
$\displaystyle A^-1 = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$

how does that work?

4. Originally Posted by jgv115
Oh okay! Cool stuff.

Now... there's an example question here which I don't really get...

For the Matrix A = $\displaystyle \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$ find: $\displaystyle A^-1$

For the solution they wrote:
$\displaystyle A^{-1} = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$

how does that work?
(Put the -1 in { } to get the whole thing as the exponent.)
If you multiply [tex]AA^{-1}= \left(\begin{array}{cc}3&2\\1&6\end{array}\right) \left(\begin{array}{cc}3&2\\1&6\end{array}\right)= \left(\begin{array}{cc}16 & 0 \\ 0 & 16\end{array}\right)[/quote]
The "1/16" in front gives the indentity matrix- the determinant of A is 16 and that is one over the determinant. In other words, "I" is like "1" in the real numbers and "$\displaystyle A^{-1}$ is like 1/A.

There are a variety of different ways to find the inverse of a matrix (provided it has one- many matrices do not have inverses) and I don't know which your text gives.

My preference is: Write A and the identity matrix side by side. Use a series of "row operations" to reduce A to the indentity matrix while simultaneously applying those same row operations to the indentity matrix. When A has been reduced to the indentity matrix, the identity matrix will have been change to $\displaystyle A^{-1}$.

5. See attachment

6. Originally Posted by jgv115
Oh okay! Cool stuff.

Now... there's an example question here which I don't really get...

For the Matrix A = $\displaystyle \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$ find: $\displaystyle A^-1$

For the solution they wrote:
$\displaystyle A^-1 = \frac{1}{16} \left(\begin{array}{cc}3&2\\1&6\end{array}\right)$

how does that work?
Use the formula and work out on your own instead, the solution provided is wrong.