Diagonalization of matrix | In English | With Easy Tips

hello everybody today we are going to see about diagnostician of a matrix diagnostician of a matrix involves simple steps look at here this is the first step find the eigenvalues then next step we are going to find the eigenvectors then next we are going to form the modal matrix using the eigenvectors from the mat modal matrix IAM in the next step we are going to find a mailers by a factor method the next step is to find m inverse am this then we will get the diagonalized matrix let us consider an example diagonally is the matrix a is equal to 2 2 minus 7 2 1 2 0 1 minus 3 first we are going to find the eigenvalues how can we find the eigen values eigen values can be found by the characteristic equation the characteristic equation of the matrix is given by the formulae characteristic equation of the matrix is given by the formula determinant of a minus lambda I which is equal to 0 this is the equation to find the characteristic equation is the given matrix lambda is a scalar I is the identity matrix I am going to substitute in this equation a 2 2 minus 7 2 1 2 0 1 minus 3 then next minus lambda into identity matrix all the diagonal entries will be 1 and the rest of the elements will be 0 it is equal to zero here look at it all the diagonal enter miss of one so I am going to substitute directly lambda inside this lambda 1 into lambda lambda 0 into lambda will be 0 so we are leaving it 1 into lambda lambda the 1 into lambda lambda next step we are going to fourth sub sub tract lambda I from a 2 minus lambda we get 2 minus lambda 2 minus 0 2 minus 7 minus 0 - 7 2 - 0 - 1 minus lambda 1 minus lambda 2 - 0 - 0 - 0 0 1 minus 0 1 minus 3 minus lambda minus 3 minus lambda this is a - sorry determinant this is a minus I a minus lambda I am it is equation 1 from this we are going to find the characteristic equation from the characteristic equation we are going to find the eigen values first take the first element 2 minus lambda I to minus lambda leave the first row and the first column remaining elements 1 minus lambda n 2 minus 3 minus lambda minus 1 into 2 - next - take the second element 2 2 into leave the first row and the second column 2 into minus 3 minus lambda 2 into minus 3 minus lambda here 0 into 2 0 so I am leaving as it the next element minus 7 minus 7 into 3 the first row and the third column remaining elements 2 into 1 - 0 into 1 minus lambda Z which is equal to zero now we are going to substitute this values inside right to minus lambda as it is we are going to multiply 1 minus lambda into -3 minus lambda 1 into minus 3 minus 3 1 into minus lambda minus lambda minus lambda into minus 3 we get plus 3 lambda minus lambda into minus lambda plus lambda square right the same - - as it is minus 2 into I am going to multiply this 2 inside 2 into minus 3 we get minus 6 2 into minus lambda minus 2 lambda here I'm going to multiply the 7 inside - 7 into 2 we get 14 which is equal to 0 so I - putting a 0 now we are going to simplify again 2 minus lambda into minus 3 minus 2 wicked -5 minus lambda plus 3 lambda we get plus 2 lambda less the lambda square then you're multiplying is - 2 inside we get less to me into minus 2 into minus 6 we get plus 12 minus 2 into minus 2 lambda we get plus 4 lambda I am writing this - 14 isit is which is equal to 0 next time I'm going to minus 2 lambda X 2 minus lambda inside this 2 into minus 5 we get minus 10 2 into 2 lambda we get 4 lambda 2 into 2 plus lambda square we get 2 lambda square minus lambda into minus y we get plus Phi lambda minus lambda into plus 2 lambda we get minus 2 lambda square minus lambda into plus lambda square we get minus lambda cube see here plus 1 into 1 plus 12 minus 14 we can write as - 2 + 4 lambda writing as it is it is equal to 0 next I am going to the rearranged rearrange and simplify - lambda Q M I am writing first then next I am going to simplify lambda square values look at your list 2 lambda square minus 2 lambda square will get cancelled then and let's go for 4 lambda values plus 4 lambda plus 5 lambda 9 lambda plus 4 lambda we get 13 lambda minus 10 minus 12 minus 2 we get minus 12 which is equal to 0 changing the sign lambda cube minus 13 lambda plus lambda is plus 12 which is equal to 0 this is the characteristic equation from this character the roots of this characteristic equation will give you the eigenvalues how can we find the roots if there is a square we can factorize and easily write the eigenvalues but there is a cube so we have to go for synthetic division or the Oder synthetic division right the coefficient is off this values coefficient of lambda cube is 1 so I am writing 1 there is no coefficient for lambda squared so I am writing 0 lamb coefficient for lambda is minus 13 the constant down this list 1 to find the roots first substitute 0 1 2 any values in this lambda and if you get 0 not it will that is the root for the lambda here 3 will be the root first put 0 here 1 plus 0 will be 1 1 into 3 will be 3 0 plus 3 is 3 3 into 3 will be 9 minus 3 plus 9 will be minus 4 minus 4 into 3 will give minus 12 plus 12 minus 1 is 0 this is the reduced equation and this is the route with this boat we are going to find the eigenvalues lambda minus 3 into reduced equation is lambda square plus 3 lambda minus 4 lambda square plus 3 lambda minus 4 which is equal to 0 now easily we can factorize and find the roots in order to get minus 4 4 into minus 1 we'll get minus 4 4 minus 1 we get less 3 so I am substituting here lambda minus 3 into lambda plus 4 into lambda minus 1 which is equal to 0 the roots are lambda is equal to 3 comma minus 4 comma 1 these are the roots and these roots are the eigen values eigen vector can be found from this eigen values the form to find eigen vectors a minus lambda I into X bar which is equal to 0 this is the farm to find the eigen vector we are going to substitute the values of lambda in this equation and we are going to find the three eigen vectors K can take guess 1 lambda is equal to 1 I am going to take this value or you can take in any order that doesn't matter lambda is equal to 1 substitute here this is the a minus lambda egg we have found which here we have formed already 2 minus 1 we get 1 then write the 2 as it is there is no lambda here then minus 7 as it is the next 2 1 minus 1 we get 0 2 0 1 minus 3 minus 1 make it minus 4 in the form to find eigen vector there is an X bar X bar is nothing but x1 x2 and x3 which is equal to zero to find the values of x1 x2 x3 I'm going to do my cross multiplication and I'd x1 x2 and x3 is equal to minus my I had devices to find the to find the values of x1 x2 3 using cross multiplication we must conserve only two equations any two equations you can take I'm just going to consider the first two equations to find the X 1 values first hide this X 1/4 mile cross multiply the remaining values 2 into 2 4 minus 0 into 7 0 then next to find the X 2 values hi the x2 but cross multiply the remaining 1 into 2 2 2 into minus 7 minus 14 already there will be 4 minus sign so I'm writing plus 14 next you to find the x3 value hide x3 and cross multiply the remaining 4200 minus 2 into 2 4 then I am simplifying it x1 by 4 which is equal to minus x2 by 16 is equal to x3 by minus 4 then the eigen vectors are X 1 X 2 X 3 which corresponds to lambda 1 eigen values are 4 minus 16 minus 4 these are the device these number these vectors are divisible by 4 so I am reducing it 1 minus 4 and minus 1 this is the eigenvector which corresponds to I can value 1 take it as 2 then next we are going to fight find eigen vector using the eigenvalue three three I am going to substitute the velum de is equal to three in the first equation and I am going to find the eigen picture two minus three minus one the two are as it is minus 7 2 1 minus 3 we get -2 + 2 0 1 minus 3 minus 3 we get minus 6 X 1 X 2 and X 3 which is equal to 0 I'm going to find the eigen vectors of corresponding lambda lambda is equal to 3 using cross multiplication method by cross multiplication method as multiplication consider this two equation to find the values for three variables X 1 divided by minus X 2 divided by which is equal to X 3 divided by 2 to find the values for X 1 hide the X 1 values cross multiply the remaining 2 into 2 we get 4 that will be minus minus 2 into minus 7 nigut plus 14 then minus X to hide this values and prosper together a many minus 2 into minus 1 into 2 minus 2 minus 2 into minus 7 we get minus 14 there's a minus sign before so we write 14 the next three values hide the X 3 and then cross multiply the remaining minus 1 into minus 2 I get 2 2 into 2 4 there is a minus sign already next way is we can simplify this tip X 1 by 4 minus 14 we get minus 10 which is equal to minus X 2 divided by minus 2 plus 14 we get 12 is equal to X 3 2 9 6 3 2 minus 4 minus 2 these are eigen vectors X 1 X 2 X 3 which is equal to minus 10 minus 12 minus 2 look at here all the all the vectors have the same sign so they neglected simply write 5 6 1 this is the second eigen vector which corresponds to the eigen value 3 and name it has name this equation as 3 case 3 lambda the remaining eigen value is minus 4 substitute the lambda is equal to minus 4 in the a minus lambda equation and we can find the matrix 2 minus of minus 4 picot 6 2 minus 7 to 1 minus minus 4 we get 5 to 0 1 minus 3 minus of minus 4 we get 1 into X 1 X 2 and X 3 which is equal to 0 we are going to find the eigen vectors using the cross multiplication method consider I'm going to consider only this two equations to find the eigen value sorry for considering these two equations to find the x1 x2 x3 values takes one you got an right which is equal to minus x2 divided by it is equal to X 3/2 find the X 1 values hide this spot and cross mentally the remaining 4 in to it of 2 4 minus sign is there 5 into minus 7 we get minus 35 am writing is plus 35 minus X to hide this part the remaining cross multiply 6 into 2 we get 12 stew into say minus 7 we get minus 14 under already there will be minus sign so we write plus 14 X 3 value hide this part and cross multiply the remaining 6 into 5 we get 32 into 2 we get 4 minus 4 simplify me in this next step X 1 you would have a 4 plus 35 39 minus X 2 divided by 12 plus 14 with you 26 X 3 divided by 30 minus 4 we get 26 this is the eigen vector X 1 X 2 X 3 which corresponds to the lambda is equal to minus 4 39 minus 26 this 26 I'm writing it here these numbers are divisible by 13 notes dividing it 3-2 to this is the equation for from the equations 2 3 & 4 we get the eigen vectors I am writing the eigen vectors along X 1 X 2 X 3 to this one this eigen vector we got from the lambda is equal to 1 eigen value 1 minus 4 minus 1 found in the intuition to then equation 3 5 6 1 this one we got from the eigen value corresponding to 3 next we are going to write die again with the corresponding today lambda is good - for fourth equation these are the eigen vectors which we got from the eigen values next we are going to form the model matrix M model matrix M model matrix M we can form it easy from the Seiken vector simply we are going to write this three eigen vectors as it is 1 minus 4 minus 1 then 5 6 1 3 minus 2 2 this is the model matrix M you can rearrange you can write in any order either you can write 5 6 1 at first and next 1 minus 4 minus 1 this will not affect your matrix so you can write as you wish but must you must write only in the column vector you should not write this eigen vectors in row vector you must write only in column vector next we are going to find the air members how can we find a members a members can be found by the formulae are GM divided by determinant of M first we must take the determinant of M whether it is equal to or not equal to 0 if it is not equal to 0 only not you will M inverse will be exist first right the first element 1 into 6 into 2 they get 12 minus 1 into minus 2 plus 2 the next element 5 only the first row and second column ma cross multiply remaining element minus 4 into 2 minus 8 minus 1 into minus 2 plus sorry minus 2 plus third element 3 leave the first row and third column minus 4 into 1 minus 4 minus 1 into my 6 minus 1 into 6 plus 6 I am going to simplify in the next system determinant of M is equal to 1 into 12 plus 2 we get 14 minus 5 minus 8 minus 2 we get minus stay plus 3 into minus 4 minus plus 6 we get to again simplifying pointing to 14 14 minus 5 minus 10 we get my in plus 15 plus 3 into 6 2 we get 6 14 plus 50 plus 6 may give you 70 this is the determinant of him which is not equal to 0 so mmm us exists we have found determinant of M next we are going to find the axiom what is our gym our GM is the transpose of cofactor M transpose of cofactor iam cofactor matrix M can be written by a11 minus a12 a13 - a21 a.22 a23 a31 sorry the little - yeah - a32 a33 this is the cofactor meum they are going to find the - a 1 1 and a 1 2 for the corresponding values and we are going to substitute here from the M minus of M - of M how can we find the - of him a 1 1 leave the first row and first column and consider that minor element 6 - 2 1 - 6 - 2 1 2 cross multiply 6 into 2 we get 12 minus 1 into 2 - my 1 into minus 2 minus 2 so plus 2 we get 14 then next a 1 2 minor of a 1 to leave the first row and the second column the remaining elements minus 4 minus 2 - one two cross multiply minus 4 into 2 minus 8 minus minus 1 minus 2 we get 2 minus 10 next we are going to find the minor a 1 3 to find a 1 3 delete the first row and the third column write the remaining elements - or 6 - one word - 4 6 - 1 1 cross multiply minus 4 into 1 minus 4 minus minus 1 into minus 6 plus 6 simplify it we get to now be aware define indicate what minor to find out - what really the second row and the first column write the remaining elements 5 3 1 2 cross 1 to play 5 into 10 minus 1 into 3 3 is equal to 7 next to minor element a2 to define this there is the second row and the second column write the remaining elements won 3-1 to cross multiply 1 into 2 2 minus minus 1 into minus a ticket + 2 + 3 5 a 2 3 a23 delete the front second row and the third column remaining won 5-1 1 cross multiplying 1 into 1 1 minus minus 1 into 5 we get plus 5 X equal to 6 8 3 1 this is equal to delete the third row and the first column remaining elements 5 3 1 sorry hi 3 6-2 I went to - -10 60 degree - AJ we get minus twenty eight eight three two minor there is the third row under second column 1 3 - 4 - 2 1 3 - 4 - 2 cross multiply 1 into minus 2 minus 2 minus 3 into minus 4 minus 12 they get minus 2 plus 12 we get it a 3 3 minor there is a 0 at the third column by winning 1 5 - 4 6 1 5 - 4 & 6 cross multiply a 1 into 6 6 minus minus 4 into my minus 4 into 5 we get minus 20 now begins bliss 23 26 now we are going to substitute all this - in the poor fracture in matrix what is the a11 value I am directly writing it a 100 are you 14 then age to k12 value is - them already that is - so we hiking acid Tim take a 1/3 valueless to the next eight to one value already there is a minus sign I hate to value with seven so it becomes minus seven next 8 to 2 X 8 to 12 is 5 8 to 3 value is 6 the right of ready to 3 is 6 all right in it - 6 the value of 8 3 1 s - ready H value of a 3-2 value that 3 to this till alikum a 3 3 is 26 this is that over this is the four factor matrix here the transpose of this cofactor matrix will give you the idea so we are going to write the idea what this idea what is transpose interchanging the rows into columns I am going to write a first row and first column 14 tin to the interchanging of rows and columns is the transfers second row second column - 7 5 - 6 the third row has third column to the - radiate - ding and 26 now we are going to write in English I click with this M inverse matrix r GM divided by determinant of M determined or determine value is 71 by there's some Indian writing here are Jimmy jockey of 1410 - -7 5 - 6 - variates - ding time trains to die L is a matrix we must find the product value mmm us a visit this will give you the diagonal instamatics first we are going to find a in junior that is simple not gives multiplication this a matrix and my guess we are going to learn it in the next step what does he market - 2 - 7 - 1 2 - 3 into em FS here ordered - 1 5 3 - 4 6 - 2 - 1 1 2 we are going to look a simple markets multiplication the seventh method first row and the first column 2 into 1 - 2 into minus 4 minus 8 minus 7 2 minus 1 plus 7 the next first row at the second column 2 into 5 10 2 into 6 plus 12 minus 7 into 1 minus 7 the next first row and the third column 2 into 3 6 2 into minus 2 minus 4 minus 7 into 2 minus 40 next we are going for next arrow and the next Apsara 2 into 1 2 1 into minus 4 minus 2 into minus 1 minus 2 so into five 10 1 into 6 6 2 into 1 2 2 3 6 1 into minus 2 minus 2 24 plus 4 next we are going for the third row third all the first column 0 into 1 0 1 into minus 4 is for I honestly do minus 1 we get plus 3 next zeroing we are going for the second fire 0 into Phi 0 22 6 6 minus 3 into 1 minus 3 then cut our know and the third color 0 in 3 0 1 into minus 2 minus 2 minus 3 into 2 minus 6 in the next step we are going to simplify this area markets and what I did here which is equal to root 2 minus 8 7 will give you 110 plus 12 22 22 minus 7 15 6 minus 4 minus 14 will get minus 12 2 minus 4 minus 2 we get minus 4 10 plus 6 plus 12 plus 2 we get 18 6 minus 2 plus 4 it is 8 minus 4 plus 3 you get minus 1 6 minus 3 3 minus 2 minus 6 you get - please this is the matrix a now we are going to multiply the MLS air we are going to multiply this matrix which am him unless we are going to find Emma and the CM he solves a simple Martin's multiplication we are going to multiply and when her smartbox into this also simple matrix multiplication we are going to multiply F inverse minus 6 EI EI m inverse minus 1 by 70 Udo 14 minus 7 minus 1 th anyway - thing but don't - 626 and the product am mark exist yet 115 -12 -4 18 gauge minus 1 3 minus 8 they're going to multiply this both might assign a value right which is equal to 1 by 17 to 14 in 214 minus 72 minus 4 we get plus 28 - 28 - - o- + 28 and next coding through in delegates to 110 - 72 18 with it - 126 - maybe unity you will get P for next asteroid typewriter - 14 into - 12 they get 1 6 8 - 7 into a 2 million - 56 - - 28 - a period plus 220 for the next row and the first column I need 110 5 in the pole you get - wait wait - to me until - one may take less step then Express when you call up then in the 50 I get 1/2 feet filing to a 5 into 18 you will get 90 10 into 3 will get minus 30 the next second or third color you need to elevated - 155 until you will get 14 minus 10 into minus 8 will get mg the next row and the first column 2 into 100 under 6 into - all related this ready for work 6:26 into -1 will get minus 26 our rocks leaks the third row and the second color to even stay we will get 13 - its into 18 you will get - go on a date when the 6 into 3 you will get 7 th row has a third color to meet you - no you will get - ready for - ECT you will get minus 48 when mrs. new I SAT with it - 2 8 6 again simplify one 572 pudding please forgive me sweetie dexterity 21 - 106 - calculated 0 minus 1 68 minus 6 plus 2 order it 8-0 10-20 western-oriented 0 150 plus 90 minus that we will go over well then 2 minus 120 + 40 verse 8 you get 0 2 plus 24 minus 96 you will get 0 30-100 let's go zero - 24 - for the - - not a giving it - - 8 multiplying with 70 inside dividing its audience dividing the 17 side we get 1 0 0 0 3 0 0 0 -4 look at here 1 3 minus 1 is the eigen value which we fought for this matrix yep thanks the markets diminished thank you

Channel: MECH Tech.

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Keywords: engineering mathematics, anna university, indian university, iit, iit maths, iit jee, jee, maths, engineering, how to, shortcut, neet, gate, neet exam, gate exam, easy, eigen, value, eigenvector, eigenvalue, matrix, vector, syllabus, find, adjoint, ad joint, cofactor, co factor, orthogonal matrix, derivation, theorem, matrices

Id: cx2fbYcIVfw

Length: 37min 33sec (2253 seconds)

Published: Mon Apr 01 2019