Question

...

Answer 1

Can you get inverse matrices?

Answer 2

What? Sorry, i dont understand.

Answer 3

I really suck at math. Just a forewarning...

Answer 4

Can you multiply matrices, though?

Answer 5

http://mathworld.wolfram.com/MatrixInverse.html

Answer 6

Oh. Im not exactly clear on how to do that, but i think so.

Answer 7

you have this
\[ AX=B\]
if A is invertible, its solution is
\[ X=A ^{-1}B \]

Answer 8

Oh?
Multiply these for me...
\[\huge \left[\begin{matrix}5 & 2 \\ 2&1\end{matrix}\right]\left[\begin{matrix}1 & -2 \\ -2&5\end{matrix}\right]\]

Answer 9

find the inverse of A

Answer 10

inverse of A not B

Answer 11

So i just do 5(1), 2(-2), 2(-2), and 1(5)?

Answer 12

I see... that's not how you multiply matrices... at all :)
It's far more complicated than that, I'm afraid...

Answer 13

Can you show me? Please?

Answer 14

Of course, but first, I need you to understand the concept of "dot product"
It's simpler than it sounds, it's a way to multiply a vector (basically a group of numbers) to another vector...
For 2x2 vectors (any dimension, it's basically the same)
<a , b> (dot) <c , d>
the answer is simply
ac + bd

for example, the dot product
<2 , 3> (dot) <1 , 5>
is equal to
(2)(1) + (3)(5) = 2 + 15 = 17

Catch me so far?

Answer 15

Oh ok so its like
[a b]
[c d]
and you multiply a and c then b and d? Then add them together?

Answer 16

Yes, something like that... so, for a small drill, what's the dot product of
<3 , -1>
and
<2 , 4>
?

Answer 17

sorry to intrupt but this might help

http://www.google.com/imgres?imgurl=http://cnx.org/content/m32159/1.4/matrixMultiply.png&imgrefurl=http://cnx.org/content/m32159/1.4/&h=1205&w=3471&sz=51&tbnid=n9VV-OF92jDwYM:&tbnh=45&tbnw=130&zoom=1&usg=__H0LfMErzJh-oknpr5VZYxwD0L4A=&docid=pGYwZla_BYnGNM&sa=X&ei=BhaNUY70H-rm4QS-xIG4BA&ved=0CGIQ9QEwBg&dur=2424

Answer 18

you have the equation
\[ \large \begin{pmatrix}
9 & 4\\ 2 & 1
\end{pmatrix}X=\begin{pmatrix}
-9 & -6\\ -1 & -8
\end{pmatrix} \]

Answer 19

or this
http://www.google.com/imgres?imgurl=http://www.mathsisfun.com/algebra/images/matrix-multiply-a.gif&imgrefurl=http://www.mathsisfun.com/algebra/matrix-multiplying.html&h=139&w=403&sz=6&tbnid=HN0U5ohaDqaPxM:&tbnh=45&tbnw=130&zoom=1&usg=__oGSpKlvFM3_c-sb5IxI4zBoNlf8=&docid=WUtotq8rktcNAM&sa=X&ei=BhaNUY70H-rm4QS-xIG4BA&ved=0CGgQ9QEwCA

Answer 20

@amir.sat Nice. I'm just telling the asker exactly what to do with the rows of the left and the columns of the right :)

Answer 21

lets find the inverse of A:
\[ \large \left(\begin{array}{cc|cc}
9 & 4 & 1 & 0\\ 2 & 1 & 0 & 1
\and{array}\right) \]

Answer 22

So i solve the [9 4] by doing 9(2) + 4(1)?
[2 1]

Answer 23

i found this sites it's pretty cool

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

Answer 24

Don't get ahead of yourself, @stop!hammertime!
we're getting there...
I asked you what the dot product of
<3 , -1>
and
<2 , 4>
is?

Answer 25

sorry

Answer 26

Oh sorry. Hold on...

Answer 27

So it would be -3 + 8 and that equals 5.

Answer 28

No.
<a , b> (dot) <c , d> = ac + bd
NOT
ab + cd

Answer 29

Product of left components plus product of right components.

Answer 30

Yeah thats what i did. I took 3(-1) + 2(4). I thought thats what i was supposed to do.

Answer 31

That's what I just said NOT to do....
Look more carefully...
\[\large <a,b> \cdot <c,d> = ac + bd\\\large <a,b> \cdot <c,d> \color{red}{\ne}ab + cd\]

Answer 32

Ohhhhh ok! Sorry. Ill try again.

Answer 33

So 3(2) + (-1(4)) = 6 + (-4) = 2

Answer 34

Okay, much better :)
Now, to get the product of matrices, remember this... in the left matrix, we deal with its ROWS, while in the right matrix, we deal with its COLUMNS.

\[\huge \left[\begin{matrix}5 & 2 \\ 2&1\end{matrix}\right]\left[\begin{matrix}1 & -2 \\ -2&5\end{matrix}\right]\]

for now I'll skip the details of matrix dimensions, but just accept that the product of two 2x2 matrices would also be a 2x2 matrix...
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}? & ? \\ ?&?\end{matrix}\right]\]

Answer 35

So, let's see what this value's gonna be...
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\boxed{\color{red}?} & ? \\ ?&?\end{matrix}\right]\]

This is in the FIRST row and the FIRST column of the resulting matrix... right?

Answer 36

Yep. So how do i find that?

Answer 37

\[ \begin{array}{cc|cc}
9 & 4 & 1 & 0\\ 2 & 1 & 0 & 1
\end{array} \]
multiply the second by -4 and add to the first row
\[ \begin{array}{cc|cc}
1 & 0 & 1 & -4\\ 2 & 1 & 0 & 1
\end{array} \]
and then multiply the first row by -2 and add to the second row
\[ \begin{array}{cc|cc}
1 & 0 & 1 & -4\\ 0 & 1 & -2 & 9
\end{array} \]

Answer 38

do u know this?

Answer 39

Well, your keywords here are "first row" and "first column"
so we get the first row of the LEFT matrix and the first column of the RIGHT matrix...
\[\huge \left[\begin{matrix}\color{red}{5} & \color{red}{2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}\color{red}{1} & {-2} \\\color{red} {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}? & ? \\ ?&?\end{matrix}\right]\]

And guess what? :P

Answer 40

We take their DOT PRODUCT.
So, what's the dot product of
<5 , 2>
and
<1 , -2>
?

Answer 41

So would i do 5(1) + 2(-2) right?

Answer 42

Yes. And...?

Answer 43

That would be 5 + (-4) = -20

Answer 44

Excuse me?
\[\huge 5 + (-4) = 20 \quad\quad\quad\color{red}{???????}\]

Answer 45

Sorry i miscalculated it = 1

Answer 46

Okay, good :)
So that upper left entry (the one on the first row and first column) would be equal to 1
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\color{blue}1 & ? \\ ?&?\end{matrix}\right]\]

Answer 47

Now what about this one?
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\color{blue}1 & \color{red}{\boxed ?} \\ ?&?\end{matrix}\right]\]

this one is in the FIRST row but in the SECOND column. So which row and column do we dot this time?

Answer 48

<5, 2>
and
<-2, 5>
?

Answer 49

YES. Exactly :)

Answer 50

So 5(-2) + 2(5) = -10 + 10 = 0

Answer 51

Mhmm :)
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\color{blue}1 & \color{blue}0 \\ ?&?\end{matrix}\right]\]
Can you do fill in the rest of the ?'s ?

Answer 52

So it would be
<2, 1>
and
<1, -2>
?

Answer 53

Yes, and you get...?

Answer 54

2(1) + 1(-2) = 2 + (-2) = 0

Answer 55

Okay...
\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\color{blue}1 & \color{blue}0 \\ \color{blue}0&?\end{matrix}\right]\]

Answer 56

Then id do
<2, 1>
and
<-2, 5>

and that =
2(1) + (-2)(5) = 2 + (-10) = -8

Answer 57

You're doing it again...
\[\huge <a,b> \cdot <c,d> \color{red}\ne ab + cd\]
dot product, please, do it correctly...

Answer 58

uck sorry.
2(-2) + 1(5) = -4 + 5 = 1

Answer 59

Better.\[\huge \left[\begin{matrix}{5} & {2} \\ {2}&{1}\end{matrix}\right]\left[\begin{matrix}{1} & {-2} \\ {-2}&{5}\end{matrix}\right]=\left[\begin{matrix}\color{blue}1 & \color{blue}0 \\ \color{blue}0&\color{blue}1\end{matrix}\right]\]

Answer 60

Okay, now that you have matrix multiplication down, let's get to your original problem...
You have to find the inverse of
\[\huge \left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]\]
Finding inverses is generally a very tasky endeavour, but fortunately, for 2x2 matrices, there is a shortcut.
\[\huge \left[\begin{matrix}{a} & {b} \\ {c}&{d}\end{matrix}\right]^{-1} =\frac1{ad-bc} \left[\begin{matrix}{d} & {-b} \\ {-c}&{a}\end{matrix}\right]\]

Answer 61

Using that formula, find
\[\huge \left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]=\color{red}?\]

Answer 62

Sorry...
\[\huge \left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]^{-1} = \color{red}?\]

Answer 63

So do i do 9(2) + 4(1)?

Answer 64

This particular bit has nothing to do with dot products. Use this formula...
\[\huge \left[\begin{matrix}{a} & {b} \\ {c}&{d}\end{matrix}\right]^{-1} =\frac1{ad-bc} \left[\begin{matrix}{d} & {-b} \\ {-c}&{a}\end{matrix}\right]\]

Answer 65

Ok give me a couple minutes.

Answer 66

1/9 - 8 = 1/1 or 1

[1 -4]
[-2 9]

Answer 67

Okay, so that's the inverse of this matrix
\[\huge\left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]\]
So back to your original question \[\huge \left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]\color{green}X=\left[\begin{matrix}{-9} & {-6} \\ {-1}&{-8}\end{matrix}\right]\]

Answer 68

What we're going to do is multiply both sides of this equation by the inverse of \[\huge\left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]\]
Which is
\[\huge\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\]

Answer 69

\[\huge\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]\color{green}X=\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\left[\begin{matrix}{-9} & {-6} \\ {-1}&{-8}\end{matrix}\right]\]
Catch me so far?

Answer 70

I think so.

Answer 71

Be sure, before we continue.
(You also said "I think so" when I asked you if you could multiply matrices, and it turned out, you didn't. )

Answer 72

haha good point. Ill look it over here a sec.

Answer 73

Ok yeah i get it.

Answer 74

Keep in mind, this is only a *crash course* I'm cramming so many lessons into one, so this is by no means a complete reference. However, it may serve as a good guideline in the future.

(You're not supposed to be doing questions like this if you don't know how to multiply matrices...)

Answer 75

Anyway, since these two are inverses of each other...
\[\huge\color{blue}{\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\left[\begin{matrix}{9} & {4} \\ {2}&{1}\end{matrix}\right]}\color{green}X=\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\left[\begin{matrix}{-9} & {-6} \\ {-1}&{-8}\end{matrix}\right]\]

They just cancel out...
\[\huge \color{green}X=\left[\begin{matrix}{1} & {-4} \\ {-2}&{9}\end{matrix}\right]\left[\begin{matrix}{-9} & {-6} \\ {-1}&{-8}\end{matrix}\right]\]
And now you have X.
Just multiply these two matrices, and you'll have your answer.

Answer 76

So first row first column = 1(-4) + (-9)(-1) = -4 + 9 = 5

Answer 77

I'll tell you this one last time...
\[\Huge \bf <a,b>\cdot <c,d> \color{red} \ne ab + cd \]

I have now told you this three times, if you make that same mistake, then ...

Answer 78

Ahhh! Sorry! I just cant seem to get that right can i? Hold on...

Answer 79

<1,-4>
and
<-9,-1>

1(9) + (-4)(-1) = 9 + 4 = 13. But thats not one of my answer choices...

Answer 80

Where did i screw up this time?

Answer 81

Of course. The answer is a matrix. What you're trying to solve for now is the upper left entry of the matrix.
Wrong, by the way, it's -9, and not 9


Why are you doing this sort of problem if you haven't been taught matrix multiplication yet?

Answer 82

I have, its just im not good at it. At all! As you can probably see. Its so confusing.

Answer 83

Well, anyway... if all else fails, use another formula... a direct one
\[\huge \left[\begin{matrix}{a} & {b} \\ {c}&{d}\end{matrix}\right]\left[\begin{matrix}{p} & {q} \\ {r}&{s}\end{matrix}\right]=\left[\begin{matrix}\color{red}{ap+br} & \color{blue}{aq+bs} \\ \color{green}{cp+dr}&\color{orange}{cq+ds}\end{matrix}\right]\]