Simpler way to factorize - Instead of expanding the brackets to get a polynomial to the 3rd power you could just leave the quadratic and manipulate the other terms to get (let h = lambda): (2-h)[h^2 - 22h + 40 - 9] - 11 + h + 9 - 18 +9h (2-h)[h^2 - 22h + 40 - 9] + 10h - 20 < manipluate this term (2-h)[h^2 - 22h + 31] - (20 - 10h) (2-h)[h^2 - 22h + 31] - 10 (2-h) h = 2 [h^2 - 22h + 21] can be solved using the quadratic formula [B+- squaroot (b^2 - 4ac)]/2 This leaves you with: h=[22+20]/2 & [22-20]/2 which gives you h = 21 & 1
THANK YOU SO MUCH. I am finally capable of doing this right cuz our prof could only send hand written notes for the lesson due these horrible covid hazard. Plus we didn't have online lectures. So thx again
Very easy to understand, but…you get the roots from looking at a graph you can’t generate during an exam so I’m guessing the graph/ intercepts will be provided?
Why do we need to change the cofactor to negative when you add the number "1" at 3:28 min into the video? I don't understand why we need to multiply with a number, in this situation we multiply with (-1).
why did you subtract the (3R2-R1) in the end, do we really need to do the last step?,however we have already fulfill the conditions of an echelon form, I guess
I don't know if this is helpful still, but he did it because a row that is all zeros has to be at the bottom of the matrix. I believe it is an issue about ordering the matrix when you have a row of zeros.
To solve the system of equations, swapping rows is not necessary. To find the row equivalent matrix that is in echelon form, all rows of zero need to be at the bottom of the matrix.
+Maheen Hashmi You just have to test the value of the polynomial function for every integer root + or - 1,2,3,5,7,14,21,42 (you have to test them until you find 3 roots because a polynomial has n roots, where n is the degree of the polynomial). It is simple but it will take time. If you find a root, you can divide the original polynomial by x - root and continue finding roots for the "new polynomial"
Maheen Hashmi, Its very simple you just apply the factor theorem on that equation (polynomial of degree 3 equation), After that you will get a factor of that then divide the equation with that factor, so you will get a polynomial of degree 2. then simply factorize, that's all I hope you understand.
Thanks..that was very helpful..but I wanna say that you would have used the remainder theorem and the factor theorem by picking a value that makes the polynomial =0..then you will have one factor and the other two you can easily find them since you got a quadratic equation..
if you multiplied the vector X*a, "a" a scalar equal to 6, shouldn't you multiply the original matrix by that scalar also? a*L*A=a*L*X=L(a*A)=L(a*X) *=multiplication a=scalar L=lamda eigenvalue A=original matrix X=eigenvector
this is another faster method and easy method ...try this ...............lambda cube+ (lambda square)sum of the diagnols (2+2+20)+(lambda)sum of coefficient of diagnols+determinant=0
Too long. You can quickly notice that one eigenvalue is 1. The the trace = 24, so the other λ's add to 23. The determinant, after R3-R1-R2, is 14*(4-1)=42, so their product is 42. That gives 2 and 21.
hell no u teaching we guys the incorrect .. u got my mind cleared and i forgot whatever i had learned till date about maths ,,lol ,,so many errors..seriously
![Diagonalizing 3x3 Matrix - Full Process [Passing Linear Algebra]](/img/n.gif)
Finally a clear concise step by step showing the details of how to solve the matrix using lamda to produce eigenvalue and eigenvector.
Simpler way to factorize - Instead of expanding the brackets to get a polynomial to the 3rd power you could just leave the quadratic and manipulate the other terms to get (let h = lambda):
(2-h)[h^2 - 22h + 40 - 9] - 11 + h + 9 - 18 +9h
(2-h)[h^2 - 22h + 40 - 9] + 10h - 20 < manipluate this term
(2-h)[h^2 - 22h + 31] - (20 - 10h)
(2-h)[h^2 - 22h + 31] - 10 (2-h) h = 2
[h^2 - 22h + 21] can be solved using the quadratic formula [B+- squaroot (b^2 - 4ac)]/2
This leaves you with:
h=[22+20]/2 & [22-20]/2 which gives you h = 21 & 1
Thanks for this quick hint.
You are a live saver bro 🙏🙏🙏
Thanks DrBrainWalton, many thumbs up for explanation !
This video is a saving grace!
Great video. Learned it in 15 mins with your help
Thank you for your straightforward approach. My god I lost count of the number of explanations that over complicate this.
You literally saved me before my engineering mathematics quiz
you deserve a nobel...
w h Y
So clarity in your explanation. No words to praise you. Thank you.
..please keep doing on other topics
At 3:20 why is the co-factor -1???
BECAUSE OF THIS MAN, I EXCELLED ON MY EC QUIZ AND GOT TO OPT OUT OF MY FINAL!!!
K
after 6 years, still this video is useful
thanks a lot
its very clear explaination for beginners.Thank you.
The explanation is very great it has really given me deep understanding
Thanks a lot for such a detailed explanation of the problem!! Thanks for every step..
cheers :D
+Raghi Pandit can i add u in insta??
Thank very much for making this topic easy to understand
so good and excellent
thanks Dr
I liked this before watching it, it better be good!
w h Y
Selamlar. Çok güzel bir ders. Ağzına sağlık kardeşim. Tam aradığım dersti. Teşekkürler.
work of a genius!
Perfectly explained :) Thank you bro.
w h Y
THANK YOU SO MUCH. I am finally capable of doing this right cuz our prof could only send hand written notes for the lesson due these horrible covid hazard. Plus we didn't have online lectures. So thx again
I usually leave the (2-x) out so I can get a quadratic in the middle so I can factor later to find the roots.
Thanks! You really helped me understand my tutorial
thank you very much sir, i finally can catch up with my classTT
It's very clear. Thank you for making this video Sir.
Very easy to understand, but…you get the roots from looking at a graph you can’t generate during an exam so I’m guessing the graph/ intercepts will be provided?
Great explanation
w h Y
It's helpful and understandable.
Thanks
Why do we need to change the cofactor to negative when you add the number "1" at 3:28 min into the video? I don't understand why we need to multiply with a number, in this situation we multiply with (-1).
Thank you..I had no idea what was going before I saw video.
w h Y
At 9:19, how does the graph not go to minus infinity?
EDIT NM I figured it out, it crosses the X axis again at 21 and goes to minus infinity.
How do i find the eigenvector if when I reduce the nullspace I get the vector [100, 010, 001] instead of [100,010,000]?
Damn this is a huge process
Thanks again for your effort....
I don't understand why when eigenvalue=21, the eigenvector needs to multipled by 6. Is it a must for the eigenvector to be integers?
why did you subtract the (3R2-R1) in the end, do we really need to do the last step?,however we have already fulfill the conditions of an echelon form, I guess
Excellent video thanks alot!
w h Y
why do you swap row2 and row3? do we need to let row3=0?
I don't know if this is helpful still, but he did it because a row that is all zeros has to be at the bottom of the matrix. I believe it is an issue about ordering the matrix when you have a row of zeros.
To solve the system of equations, swapping rows is not necessary. To find the row equivalent matrix that is in echelon form, all rows of zero need to be at the bottom of the matrix.
how are you finding the elements of matrix as free and not free
in which condition are you multiply wuth x,y,z
Can u plz do this question without the calculator?
+Maheen Hashmi You just have to test the value of the polynomial function for every integer root + or - 1,2,3,5,7,14,21,42 (you have to test them until you find 3 roots because a polynomial has n roots, where n is the degree of the polynomial). It is simple but it will take time. If you find a root, you can divide the original polynomial by x - root and continue finding roots for the "new polynomial"
Maheen Hashmi, Its very simple you just apply the factor theorem on that equation (polynomial of degree 3 equation), After that you will get a factor of that then divide the equation with that factor, so you will get a polynomial of degree 2. then simply factorize, that's all I hope you understand.
but 4 can also be an answer
I get -lambda^3 + 4lambda^2 - lambda - 18 (this is another example from the book) and it does not crosses with x, what should I do
i have one confusion ain't homogeneous system of equation solved by echelon form ? where on the basis of rank it gives us the solution.
GREAT EXPLANATION. THANK YOU (:
w h Y
In getting eigenvector in each eigenvalues, can i use gaussian-elimination??
thank you very much sir .. ^_^ you r a life saver ..
thanks ooh this helps me alot God bless you
VERY EXUISITE METHOD
Thanks..that was very helpful..but I wanna say that you would have used the remainder theorem and the factor theorem by picking a value that makes the polynomial =0..then you will have one factor and the other two you can easily find them since you got a quadratic equation..
thanks a looooooottttt man..... hats off may god bless ;u
Look at the coefficient of each term , the sum of them be equal to zero, so the first solution of this equation = 1.
It was good! Very good!
w h Y
could you please explain (80-44+2lamda^2) how did you get this?
Mehrab Fatima yuko vizur
if you multiplied the vector X*a, "a" a scalar equal to 6, shouldn't you multiply the original matrix by that scalar also?
a*L*A=a*L*X=L(a*A)=L(a*X)
*=multiplication
a=scalar
L=lamda eigenvalue
A=original matrix
X=eigenvector
love the video... i'm shocked that i got the answer right !
Why didnt we say the z is free when we were solving for lamda=1 ? :/
i know, right? it must have been.
It was great help!!!
7:20 .....how come 80-18-20-9+9-18= 42?????
Thanks Sujoy
very nice
All I need now is a method to create the cubic equation of lamdas in a programing routine.
this is another faster method and easy method ...try this ...............lambda cube+ (lambda square)sum of the diagnols (2+2+20)+(lambda)sum of coefficient of diagnols+determinant=0
this shit was money
w h Y
Can someone explain to me how 3 times 3 became -9?
I m having a trouble in solving step 2
thank you so much..
w h Y
Thank you so much. :)
will Z be allways the free value?????
Is it A_lambda or lambda _A
Good
thank you for this
holly shit its so easy thank u so much
thanks.
nice
w h Y
Thank you!
i guess there is a mistake when you calculate the det in 3rd row you should do 3(1st row((2-y)(1)) 2nd row((1)(2-y))
👍
preach!!!
I have never been able to factorise.
thanx
perfect
I think that he made a mistake in finding the determinants because it is -1 if it is because of, when finding the determinants
Thaaaaaaanks
Can my eigen vector be 0, 0, 0?
I believe that it would not be considered an eigenvector because it is a trivial solution to (A-lamda*I)x=0
good video but wish would stop jumping steps
The explanation was good, but the scrawl was nearly illegible.
@ Dr brain Walton,I totally doubt u here,ur wrong
I have a confused in last step
now instead of just doing it explain the steps we are watching the video to learn not to see if you can solve it or not
Too long. You can quickly notice that one eigenvalue is 1. The the trace = 24, so the other λ's add to 23. The determinant, after R3-R1-R2, is 14*(4-1)=42, so their product is 42. That gives 2 and 21.
that lamda calculation is so long and boring
nice explanation, crappy writing.
hell no u teaching we guys the incorrect .. u got my mind cleared and i forgot whatever i had learned till date about maths ,,lol ,,so many errors..seriously
+Utkrisht Sharma you are so dumb
+hassan alrehamy WTF shut your hole ,, pisssing oveeer myyy post
Thank you!