Question

determine the matrices P and Q such that PAQ is diagonal matrix

Answer 1

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

Answer 2

@ganeshie8

Answer 3

where are we exactly

Answer 4

I think the matrix has changed
check the last element in first row

Answer 5

no its seam as before

Answer 6

it was "-2" earlier

Answer 7

ohh that one was wrong typo
sorry for that
@ganeshie8

Answer 8

since ive forgotten any quick ways to go about this

cant we just make 2 general 3x3 matrixes, work the process, and solve the system of equations that results?

Answer 9

can u try to remember any ways to solve this equation ??
@amistre64

Answer 10

let me scratch something on paper to see if my thought is workable

Answer 11

okay take ur time :)

Answer 12

D = PAQ, D being the diagonlized matrix

then P'DQ' = A

the wolf gets us a solution, but no way to actually approach it http://www.wolframalpha.com/input/?i=diagonalize%7B%7B6%2C-2%2C2%7D%2C%7B-2%2C3%2C-1%7D%2C%7B2%2C-1%2C3%7D%7D

Answer 13

find the eigenvectors of A

Answer 14

show me how plz
@Zarkon

Answer 15

\[

\left[\begin{matrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\end{matrix}\right]

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

\left[\begin{matrix}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32} & b_{33}\end{matrix}\right]

\]

\[
\left[\begin{matrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\end{matrix}\right]

\left[\begin{matrix}
6b_{11}-2b_{21}+2b_{31} & 6b_{12}-2b_{22}+2b_{32} & 6b_{13}-2b_{23}+2b_{33} \\
-2b_{11}+3b_{21}-b_{31} & -2b_{12}+3b_{22}-b_{32} & -2b_{13}+3b_{23}-b_{33}\\
2b_{11}-b_{21}+3b_{31} & 2b_{12}-b_{22}+3b_{32} & 2b_{13}-b_{23}+3b_{33}
\end{matrix}\right]

\]


\[\left[ \begin{matrix}
a_{11}(6b_{11}-2b_{21}+2b_{31}) + a_{12}(-2b_{11}+3b_{21}-b_{31}) + a_{13}(2b_{11}-b_{21}+3b_{31})=x\\
a_{21}(6b_{11}-2b_{21}+2b_{31}) + a_{22}(-2b_{11}+3b_{21}-b_{31}) + a_{23}(2b_{11}-b_{21}+3b_{31})=0\\
a_{31}(6b_{11}-2b_{21}+2b_{31}) + a_{32}(-2b_{11}+3b_{21}-b_{31}) + a_{33}(2b_{11}-b_{21}+3b_{31})=0
\end{matrix} \right.\]

\[\left. \begin{matrix}
a_{11}(6b_{12}-2b_{22}+2b_{32}) + a_{12}(-2b_{12}+3b_{22}-b_{32}) + a_{13}(2b_{12}-b_{22}+3b_{32})=0\\
a_{21}(6b_{12}-2b_{22}+2b_{32}) + a_{22}(-2b_{12}+3b_{22}-b_{32}) + a_{23}(2b_{12}-b_{22}+3b_{32})=y\\
a_{31}(6b_{12}-2b_{22}+2b_{32}) + a_{32}(-2b_{12}+3b_{22}-b_{32}) + a_{33}(2b_{12}-b_{22}+3b_{32})=0
\end{matrix} \right.\]

\[\left. \begin{matrix}
a_{11}(6b_{13}-2b_{23}+2b_{33}) + a_{12}(-2b_{13}+3b_{23}-b_{33}) + a_{13}(2b_{13}-b_{23}+3b_{33})=0\\
a_{21}(6b_{13}-2b_{23}+2b_{33}) + a_{22}(-2b_{13}+3b_{23}-b_{33}) + a_{23}(2b_{13}-b_{23}+3b_{33})=0\\
a_{31}(6b_{13}-2b_{23}+2b_{33}) + a_{32}(-2b_{13}+3b_{23}-b_{33}) + a_{33}(2b_{13}-b_{23}+3b_{33})=z
\end{matrix} \right]\]

Answer 16

eigenvalues would most likely be the shorter approach :)

Answer 17

okay
so i have to apply this and i will get the final answer

Answer 18

if we were to take a long approach, this would be how i would go about it ... but there are simpler ways that i have forgotten about over time

Answer 19

eigen vectors give you the shortest solution :
\[SAS^{-1} = \Lambda \]

that requires you to understand some theory, however elementary row operations should also give you easy solution as amistre started

Answer 20

may be manupulating a triangular matrix would be easy, so lets change the given matrix to triangular form.

fix elements below first pivot :
\[
\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 21

does that multiplication make sense ?

Answer 22

no

Answer 23

interpret row by row

Answer 24

okay i got it

Answer 25

[1 0 0]

means, "1" of first row, "0" of second row and "0" of third row :


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

\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\\
\end{array}
\right]
\]

Answer 26

[1/3, 1, 0]

means "1/3" of first row, "1" of second row and "0" of third row :
\[\left[
\begin{array} {ccc}
1/3&1&0\\

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

=

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

Answer 27

[-1/3, 0, 1]

means "-1/3" of first row, "0" of second row and "1" of third row :
\[\left[
\begin{array} {ccc}
-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}
0&-1/3&11/3\\
\end{array}
\right]\]

Answer 28

Overall :

\[\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 29

you need to look at them "row" by row

Answer 30

now im okay with it

Answer 31

then only they make sense

Answer 32

good, so is the matrix on right side triangular ?

Answer 33

looks we need to fix another element in the bottom row, eh ?

Answer 34

\[
\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&\color{red}{-1/3}&11/3\\
\end{array}
\right]

\]

Answer 35

if we make that 0, it becomes triangular, right ?

Answer 36

i think yeah

Answer 37

how to make it 0 ?

Answer 38

tell me a row to left multiply

Answer 39

will this work ?

[0 1/7 1 ]

Answer 40

this one with which row ??

Answer 41

\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&1/7&1\\
\end{array}
\right]

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

\]

Answer 42

okay

Answer 43

multiply the right hand side and see what you get

Answer 44

\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&1/7&1\\
\end{array}
\right]

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

\]




becomes



\[
\left[
\begin{array} {ccc}
1&0&0\\
0&1&0\\
0&1/7&1\\
\end{array}
\right]

\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&\color{red}{0}&72/21\\
\end{array}
\right]

\]

Answer 45

we're mostly done as we reached a triangular matrix.
lets just do the multiplication on left hand side also

Answer 46

after multiplying left side matrices, you get :

\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&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&\color{red}{0}&72/21\\
\end{array}
\right]

\]

Answer 47

still wid me ? :)

Answer 48

yeah im with u

Answer 49

but im not getting seam value but still okay later i will see y

Answer 50

no, first make sure you're also getting same value... otherwise we won't get right result in the end

fyi : i do lot of arithmetic mistakes, so watch out >.<

Answer 51

okay now im getting seam answer

Answer 52

good, two more steps and we will be done!

Answer 53

so we're here :


\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&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&\color{red}{0}&72/21\\
\end{array}
\right]
\]

Answer 54

lets split the right side matrix into "diagonal" and "triangular"

Answer 55

okay

Answer 56

\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&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&\color{red}{0}&72/21\\
\end{array}
\right]
\]

becomes :


\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&1\\
\end{array}
\right]

\left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]
\\~\\
=
\left[
\begin{array} {ccc}
6&0&0\\
0&7/3&0\\
0&\color{red}{0}&72/21\\
\end{array}
\right]
\left[
\begin{array} {ccc}
1&-2/6&-2/6\\
0&1&-5/7\\
0&\color{red}{0}&1\\
\end{array}
\right]
\]

Answer 57

we finally go the diagonal matrix !
just make sure you're 100% at peace with the splitting, one last step is there to get it to the required form : `PAQ = diagonal matrix`

Answer 58

okay sure

Answer 59

last step is to simply find the inverse of right most matrix and multply both sides :




\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&1\\
\end{array}
\right]

\left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]
\left[
\begin{array} {ccc}
1&-2/6&-2/6\\
0&1&-5/7\\
0&\color{red}{0}&1\\
\end{array}
\right]^{\color{Red}{-1}}
\\~\\
=
\left[
\begin{array} {ccc}
6&0&0\\
0&7/3&0\\
0&\color{red}{0}&72/21\\
\end{array}
\right]

\]

Answer 60

we're done !


thats the required `PAQ = diagonal` form

Answer 61

ofcourse you need to find the inverse and plug there to be called officially done :)

Answer 62

so now we r done ??

Answer 63

yes, just find the inverse and plugin

Answer 64

wolfram says inverse is
http://www.wolframalpha.com/input/?i=inverse%7B%7B1%2C-2%2F6%2C-2%2F6%7D%2C%7B0%2C1%2C-5%2F7%7D%2C%7B0%2C0%2C1%7D%7D

Answer 65

\[\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&1\\
\end{array}
\right]

\left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]
\left[
\begin{array} {ccc}
1&-2/6&-2/6\\
0&1&-5/7\\
0&\color{red}{0}&1\\
\end{array}
\right]^{\color{Red}{-1}}
\\~\\
=
\left[
\begin{array} {ccc}
6&0&0\\
0&7/3&0\\
0&\color{red}{0}&72/21\\
\end{array}
\right]\]

plugging in the inverse you get :


\[
\left[
\begin{array} {ccc}
1&0&0\\
1/3&1&0\\
-6/21&1/7&1\\
\end{array}
\right]

\left[\begin{matrix}6 & -2 & -2 \\-2 & 3 & -1 \\2 & -1 & 3\end{matrix}\right]
\left[
\begin{array} {ccc}
1&1/3&4/7\\
0&1&5/7\\
0&\color{red}{0}&1\\
\end{array}
\right]
\\~\\
=
\left[
\begin{array} {ccc}
6&0&0\\
0&7/3&0\\
0&\color{red}{0}&72/21\\
\end{array}
\right]
\]

Answer 66

thats it !

Answer 67

okay thank u so much man

Answer 68

np :)