Question

inverse of matrices

a)21
52

b) 101
023
121

Answer 1

Know how to find the inverse of a matrix? There are a few ways.

Answer 2

it just says find the inverse of the matrices idk any way honestly

Answer 3

Well, for the 2X2 there is an easy one:

\(\begin{bmatrix}
a&b\\c&d
\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\-c&a
\end{bmatrix}\)

That is a stock formula that always works but is based on something else.

Answer 4

If A is a matrix that is invertible, then \(A^{-1} = \dfrac{1}{det(A)}(adj(A))\) where det means determinant and adj means adjoint.

There is another one as well, called adding on the identity matrix and using row reduced echelon form.

Any of these sound familiar from your class?

Answer 5

pretty much all have been mentioned as far as i remember i just don't understand how to solve the problem i don't think the method matters if that's what your asking

Answer 6

Reduced row echelon form, or RREF, or Gauss-Jordan elimination is mathematically written as:
\([A|I] \leftarrow [I|A^{-1}]\)

That might be in your book. What is means is taking on an identity after your matrix:

\(\begin{bmatrix}
a&b\\c&d
\end{bmatrix}\) becomes \(\left[\begin{array}{cc|cc}
a&b&1&0\\c&d&0&1
\end{array}\right]\)

Then you solve it so that the \(\left[\begin{smallmatrix}
a&b\\c&d
\end{smallmatrix}\right]\) becomes \(\left[\begin{smallmatrix}
1&0\\0&1
\end{smallmatrix}\right]\) and the new right hand side is the inverse.

Well, I don't want to work with you on something your class is not doing. Might be a bit confusing if I do!

Answer 7

Oops... put an arrow in their wrong. Meant a right arrow on the AI to IA-1

Answer 8

Well, your first one is: \(\begin{bmatrix}
2&1\\5&2
\end{bmatrix}\)

So how about putting that into the formula?

\(\begin{bmatrix}
a&b\\c&d
\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}
d&-b\\-c&a
\end{bmatrix}\)

Oh, and the determinant of a 2X2 matrix is handy to remember. It comes in very useful at times.

Answer 9

yeah i did that one just like u said earlier now if thats how u do that one i just dont know how to do the next part like what formula to use or anything

Answer 10

\(\begin{bmatrix}
1&0&1\\ 0&2&3\\ 1&2&1
\end{bmatrix}\)

Well, do you want to use the determinant and adjoint, or do you want to use rref?

Answer 11

To be honest, it is set up pretty well for either.

Answer 12

actually the next problem is asking same numbers for determinants and this one says inverses of the matrices

Answer 13

Well, then if you find the determinants then you have a step in answering both.

Answer 14

i dont understand the difference honestly so the first answer u gave me would be the answer for the determinants one right?

Answer 15

For a 2X2 the determinant is ad-bc

Answer 16

so i did that and got that answer now for the 2x2 inverse the answer is what u showed me before?

Answer 17

oh wow i finally get what ur saying

Answer 18

When you multiply the inverse of the determinant through he special form, you get that.

Answer 19

Yah, linear algebra is like that. Really confusing until you finally connect with it. Then suddenly it makes sense.

Answer 20

yeah i got that now so whats the formula for the 3x3 determinant and inverse

Answer 21

did not mean to sound rude sorry ur a great help...im just anxious to understand

Answer 22

You did not sound at all rude! It is great you finally get part of it.

Answer 23

For a 3X3 there is trick to finding the determinant. You tack on a couple columns, then properly sum and difference the products of the diagonals. That is really a pain in the anatomy to describe, but the good news is that someone else did the work for me!

http://www.purplemath.com/modules/determs2.htm

Answer 24

For anything larger than a 3X3, there are no tricks. But for a 2X2 and 3X3 the determinant is not too hard to get.

Answer 25

oh ok thank you i appreciate all of the help

Answer 26

\(\begin{bmatrix}
1&0&1\\ 0&2&3\\ 1&2&1
\end{bmatrix}\) becomes \(\begin{bmatrix}
1&0&1&1&0\\ 0&2&3&0&2\\ 1&2&1&1&2
\end{bmatrix}\)

Now, you would do the diagonals on that just like on that site. The good news is, all the 0s make a lot of it simple!

Answer 27

thank i see that

Answer 28

OK, what did you get for the det? (I did it here... hehe)

Answer 29

for the 3x3 -2 is that right?

Answer 30

im doing it over tho to make sure

Answer 31

I got a little lower number.

Answer 32

-6?

Answer 33

=)

Answer 34

is the inverse
-4 2 -2
3 0 -3
-2 -2 2

Answer 35

Don't know. I did not find it yet.

Answer 36

oh ok thought u did it already lol

Answer 37

Now, you are looking to fill in the parts of this formula:

\(A^{-1} = \dfrac{1}{det(A)}(adj(A))\)

Well, you just found the det(A) part:

\(A^{-1} = \dfrac{1}{-6}(adj(A))\)

What is needed is the adj(A) part. Now that takes a little work.

Answer 38

And it looks like this one will have fractions in it.

Answer 39

The adj(A) is a done this way:

Find the cofactor matrix (which is where ad-bc come in handy)
Transpose the cofactor matrix.

Answer 40

In a 2X2 you can do this all in one nice step! That is the formula earlier.

In a 3X3, they make you sweat it more.

Don't ask about 4X4 until they tell you to.

Answer 41

ok

Answer 42

i think i got it all at least what i need

Answer 43

I am going to show you what is sometimes called the coverup method. Did they talk about that?

Answer 44

honestly the reason im so confused is my professor speaks barely any english and what he can speak sounds like gibberish i think he has a speech issue and he never told us what book to use

Answer 45

so i cant even read it myself

Answer 46

Hehe. OK.

Answer 47

What class is it in?

Answer 48

finite mathematics i would probably have to research a good book to use

Answer 49

Finite? OK. Then you can take a reasonably strong book on it. You already had the lower stuff.

Here is a good thing for a 3X3, and I'll show you some linear and finite books.
http://www.wikihow.com/Inverse-a-3X3-Matrix

Answer 50

Oh, and my answer had a lot of sixths, thirds, and halfs. So if you got something like that, you are probably good.

Answer 51

http://www.math.dartmouth.edu/~doyle/docs/finite/cover/

Answer 52

And here are a bunch of free, peer reviewed textbooks.
http://www.aimath.org/textbooks/textbooklist.html

Matrices are covered in Linear. You will probably run into any of the non-calculus topics in finite.

Answer 53

ok thanks and yeah im getting it now i appreciate all the help really

Answer 54

And here are some reviews of open books, including ones on finite:
http://www.collegeopentextbooks.org/opentextbookcontent/thereviews/mathematics

Answer 55

I am looking at the finite it links and it has lots of example problems with solutions available. Might show you the bits your lecture is missing!

Answer 56

Oh, and you can download that collection as a PDF, making it into a book you can view on your machine or print to have.

Answer 57

At 374 pages, it is a tad small for a college book! LOL