Question

can u chick my question

Answer 1

let \[Let A= \left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]\]
determine the matrices P and Q such that PAQ is diagonal matrix

Answer 2

@ganeshie8
do u know hw to solve this ??

Answer 3

@experimentX

Answer 4

I really hate linear algebra, but seeing as you aren't having much luck at getting responses, mind if I give it a shot? ^_^

Answer 5

okay just try plz to do it
@terenzreignz

Answer 6

familiar with eigen vectors ?

Answer 7

and can you confirm if below is a typo :

\[
Let A= \left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\\color{Red}{-}2 & -1 & 3\end{matrix}\right]

\]

Answer 8

no is not typo

Answer 9

so tell me what have you tried so far

Answer 10

wolfram gives this
http://www.wolframalpha.com/input/?i=diagonalize+%7B%7B6%2C-2%2C-2%7D%2C%7B-2%2C3%2C-1%7D%2C%7B2%2C-1%2C3%7D%7D

Answer 11

nothing until now justt wanna start with u

Answer 12

are you familiar with finding eigen values and eigen vectors ?

Answer 13

no actually

Answer 14

we may work it using elementary matrices if you know row reduction

Answer 15

you need to share with us what all things you have tried and what exactly you know already to get quick help

Answer 16

@mathmale

Answer 17

i dont know anything about it but we can start from zero if u dont mind

Answer 18

Alright, lets define the zero point first :)
do you know how to multiply two matrices ?

Answer 19

yeah i know

Answer 20

@ganeshie8

Answer 21

good, lets start from there :)
lets see how to convert a matrix into triangular form :

\[
\left[
\begin{array} {ccc}
?&?&?\\
?&?&?\\
?&?&?\\
\end{array}
\right] \left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]

=

\left[
\begin{array} {ccc}
*&*&*\\
0&*&*\\
0&0&*\\
\end{array}
\right]
\]

Answer 22

any idea on what should go in those question marks in left most matrix ?

Answer 23

no

Answer 24

\[
\left[
\begin{array} {ccc}
?&?&?\\
?&?&?\\
?&?&?\\
\end{array}
\right] \left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]

=

\left[
\begin{array} {ccc}
*&*&*\\
0&*&*\\
0&0&*\\
\end{array}
\right]

\]


our goal is to change it to upper triangular matrix, notice that we need 0's below the pivot in the first column. lets fix those 0's first :


\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-1/3&0&1\\
\end{array}
\right] \left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]

=

\left[
\begin{array} {ccc}
6&-2&-2\\
0&7/3&-5/3\\
0&-1/3&11/3\\
\end{array}
\right]
\]

Answer 25

we still need to fix another 0 in the last row
but first make sure above makes sense

Answer 26

but how would we the the matrix elements

Answer 27

which matrix elements ?

Answer 28

left and right matrix

Answer 29

Okay, lets see how the row operations work

Answer 30

but we are getting left and right matrices simultaneously

Answer 31

yes :) lets first see how row operations work, it becomes clear after understanding what happens when u multiply a matrix by another matrix on left side

Answer 32

can you tell me the result of below multiplication ?

\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&0&1\\
\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]

= ?
\]

Answer 33

a1
b2
c3 in a diagonal

Answer 34

give me the full matrix

Answer 35

a1 0 0
0 b2 0
0 0 c3

Answer 36

wrong, try again

Answer 37

a1 a2 a3
b1 b2 b3
c1 c2 c3

Answer 38

BINGO! so the matrix did not change
multiplying by an identity matrix will not change the original matrix, thats the reason we call it identity matrix :

\[IA = A\]

Answer 39

next, may i know how u worked out each element in your result ?

Answer 40

didnt get u

Answer 41

you have said you got

`a1` a2 a3
b1 b2 b3
c1 c2 c3

may i know what u did to get `a1` as your first element in first row ?

Answer 42

multiply with the column

Answer 43

\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&0&1\\
\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]

=
\left[
\begin{array} {ccc}
\color{red}{a_{1}}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
\]

\(\color{red}{a_{1}} = 1*a_1 + 0*b_1 + 0*c_1\)

like this ?

Answer 44

yup

Answer 45

there is another powerful way to multiply, and thats the key in understanding row operations

Answer 46

let me show you the method with an example :


\[
\left[
\begin{array} {ccc}
1&0&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=?
\]

Answer 47

what the left matrix says is this :
take the combination : "1" of first row, "0" of second row and "0" of third row

\[
1*\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\

\end{array}
\right]

+
0*\left[
\begin{array} {ccc}
b_{1}&b_2&b_3\\

\end{array}
\right]
+
0*\left[
\begin{array} {ccc}
c_{1}&c_2&c_3\\

\end{array}
\right]

\]

you get :

\[
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\

\end{array}
\right]
\]

Answer 48

you should think matrix multipilication as : `combination of rows`

Answer 49

not as some weird meaningless arithmetic with elements

Answer 50

In light of above information, can you guess the result of below multiplicaiton ?

\[
\left[
\begin{array} {ccc}
0&1&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=?
\]

Answer 51

b1 b2 b3

Answer 52

Exactly !

so you have taken : 0 of first row, 1 of second row and 0 of last row

right ?

Answer 53

right

Answer 54

what about below :


\[
\left[
\begin{array} {ccc}
0&-2&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=?
\]

Answer 55

-2b1 -2 b2 -2b3

Answer 56

Excellent, last example on row muultiplication :

what about below :


\[
\left[
\begin{array} {ccc}
1&-2&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=?
\]

Answer 57

just think of it as `combination of rows` :

```
1 of first row

+

-2 of second row

+

0 of third row
```

Answer 58

a1 a2 a3
-2b1 -2b2 -2b3
0 0 0

Answer 59

careful, multiplying by a row gives you a row
not a 3x3 matrix.

Answer 60

here is the answer :


\[
\left[
\begin{array} {ccc}
1&-2&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=

\left[
\begin{array} {ccc}
a_1-2b_1 &a_2-2b_2&a_3-2b_3\\

\end{array}
\right]
\]

Answer 61

i see

Answer 62

great ! lets work another final example

Answer 63

\[
\left[
\begin{array} {ccc}
1&-2&3\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=

?
\]

Answer 64

a1-2b1+3c1 a2-2b2+3c2 a3-2b3+3c3

Answer 65

good, so can you tell me now what u need to multiply on left side to get second row as result ?

Answer 66

yeah i guess

Answer 67

good, then tell me :

\[
\left[
\begin{array} {ccc}
?&?&?\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
b_1&b_2&b_3\\

\end{array}
\right]

\]

Answer 68

what goes in those question marks ?

Answer 69

0 1 0

Answer 70

\[
\left[
\begin{array} {ccc}
?&?&?\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
a_1-b_1&a_2-b_2&a_3-b_3\\

\end{array}
\right]

\]

Answer 71

what about this ?

Answer 72

0 -1 0

Answer 73

try again

Answer 74

you're taking "1" of first row and "-1" of second row right ?

Answer 75

so you will get


1 -1 0

Answer 76

\[
\left[
\begin{array} {ccc}
1&-1&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
a_1-b_1&a_2-b_2&a_3-b_3\\

\end{array}
\right]

\]

Answer 77

its just say how you combine `rows`

Answer 78

lets see how to interpret multiplication of a full matrix :



\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&0&1\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
?

\]

Answer 79

its exactly same as before,
you just need to repeat the process two more times thats all

Answer 80

you can get the first row of result by interpreting the multiplication by first row :

lets see how to interpret multiplication of a full matrix :



\[
\left[
\begin{array} {ccc}
1&0&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
a_1&a_2&a_3\\

\end{array}
\right]
\]

Answer 81

similarly, you can work second and third rows :

\[
\left[
\begin{array} {ccc}
0&1&0\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
b_1&b_2&b_3\\

\end{array}
\right]
\]


\[
\left[
\begin{array} {ccc}
0&0&1\\

\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
c_1&c_2&c_3\\

\end{array}
\right]
\]

Answer 82

Overall :

\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&0&1\\
\end{array}
\right]
\left[
\begin{array} {ccc}
a_{1}&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\
\end{array}
\right]
=
\left[
\begin{array} {ccc}
a_1&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\\

\end{array}
\right]
\]

Answer 83

see if that makes sense...

Answer 84

yeah its seam

Answer 85

good, lets go back to orignal problem

Answer 86

okay

Answer 87

this thread is lagging because of huge latex code
can u start a new post ?

Answer 88

okay