Matrices: Diagonalization of a Matrix :Best Engineering Mathematics Tips & Tricks

I am going to show you the example diagonals the matrix 1 0 0 0 3 - 1 0 - 1 3 this is the working rule to find the diagonalization of a matrix find the eigenvalues find the eigenvectors form the model matrix and find B inverse find B inverse a be the product now first thing to find the eigenvalues we need to find the characteristic equation how will you find the characteristic equation the first step as I told you earlier by using the formula lambda cube minus s1 lambda square plus s2 lambda minus s3 equal to 0 or by determinant of a minus lambda I equal to 0 now this is the given matrix so 1 minus lambda 0 0 0 3 minus lambda minus 1 0 minus 1 3 3 since we have many zeros in the matrix we can directly use the determinant of a minus lambda formula so the first step will be how will you do it to find the determinant 1 minus lambda into 3 minus lambda into 3 minus lambda which is which will give you 3 minus lambda the whole square minus 1 into minus 1 plus 1 again you will get a does minus 1 now simplify everything 1 minus lambda into 3 minus lambda the whole square so you will get it as 9 minus 6 des plus lambda square minus one now simplify it nine minus six lambda plus lambda squared minus one minus 9 lambda minus into minus faster take the first element and multiply with everything and take the next one and multiply with everything you will get it as minus 9 lambda plus 6 lambda square minus lambda cube plus lambda so when you simplify this you will be getting it as lambda cube minus 7 lambda square plus 14 lambda minus 8 therefore the characteristic equation is the characteristic equation is lambda cube minus seven lambda square plus fourteen lambda minus eight equal to zero now first step find the eigen values to find the eigen values we have found out the characteristic equation how will you do it by applying the synthetic division 1 minus 7 14 minus 8 as I told you earlier how will you find the synthetic division you will take the roots randomly that is minus 2 minus 1 and then 2 1 and then anything 3 so 1 minus 3 like that now first step I am going to solve it with 1 always put 0 for the first element you will get 1 plus 0 will be 1 and then multiply with this with the root 1 into 1 1 minus 7 plus 1 minus 6 minus 6 into 1 minus 6 again this will give you 8 8 into a 8 into 1 will give you 8 8 minus 8 is 0 are you getting the remainder 0 therefore we can take the root as 1 therefore we can say lambda is equals to 1 0 if suppose if you are not getting the remainder as 0 then go for the next one next root you can go for minus 1 even in minus 1 if you are not getting the reminder as 0 then go for the next root minus 2 and then 2 within 3 itself you will be getting the roots if not if you are going beyond the 3 or minus 4 or 4 then the characteristic equation what you're found out is wrong so once you are finding out the roots by using the synthetic division maximum you have to go for the roots - three - three if you are going beyond that then go for check for the characteristic equation you must have made mistake in the characteristic equation now you have got lambda is equals to one is the root you have got one of the road now you need to get the next two roots how by factorizing the equation lambda square minus six lambda plus 8 equal to 0 when you factorize it lambda minus 4 into lambda minus 2 which is equals to 0 therefore lambda is equals to 2 comma 4 therefore the eigen values are 1 2 1 4 therefore the eigen values are one two and four we have got the eigenvalues now first step is over now what is the step two we need to find the eigenvectors how will you find the eigenvectors by using the eigenvalues one two and four to find the eigenvector what is the condition a minus lambda I into X bar is equals to zero so this is the matrix a so how will you get it 1 minus lambda 0 0 0 3 minus lambda minus 1 0 minus 1 3 minus lambda into X 1 X 2 X 3 which is equals to 0 0 0 now what is the first case case 1 when lambda is the equals to 1 you can find out in any order you can even take when lambda is equals to 2 any order you can take but the final answer you have to get it as the diagonally laments as the roots 1 2 1 4 so we can find out the eigen vectors in any order now when you substitute lambda is equals to 1 what you get directly I will write the reduced matrix 1 minus 1 will give you 0 0 0 and then next 1 0 this will give you 2 minus 1 0 minus 1 and finally here you will get it as 2 x1 x2 x3 0 0 0 see you in the first equation you are getting it everything as 0 so you can write down the second and the third equation you can write it as 0 X 1 plus 2 X 2 minus X 3 which is equal to 0 and similarly 0 X 1 minus X 2 plus 2 X 3 is equals to zero now here the assumption method is not possible because you may get it infraction in that case I am to avoid fractions I am going to do it by rule of cross multiplication solving by Truelove cross-multiplication now to find out the role of cross multiplication I will teach you some easy method we need to take it be see a be solving the equation if suppose if you are taking this as equation one and two solving the equation one and two by the role of cross multiplication I am going to use the method BC a B that means the coefficient of x2 here you will take this as a B this as a B C now you when you take this as B what is the coefficient of x 2 here 2 and similarly what is the coefficient of x 3 there minus 1 and DA will be 0 and again B will be 2 this is for the equation 1 move on to the equation 2 same way minus 1 to 0 minus 1 now I am going to do the cross multiplication X 1 divided by 2 into 2 minus minus 1 into minus 1 this will give you 4 minus 1 which is equals to mixed 1 X 2 divided by this will be X 1 and this will be X 2 and this will be X 3 now again X 2 minus 1 into 0 minus 2 into 0 which will give you 0 minus 0 and X 3 will be again 0 into minus 1 0 into 2 so again 0 minus 0 so what are the values you are getting it X 1/3 X 2 by 0 and X 3 by 0 so therefore the eigenvector is the eigenvector for the corresponding lambda is equals to 1 is you can write this as 1 0 0 when you divide by 3 throughout you will get it as 1 0 0 so I told you in the earlier session that all the eigenvectors cannot be zeros any one value you can get it so this is correct now we will move on to the second case case 2 what is the second eigen value lambda is equals to 2 when lambda is equals to 2 what do you get directly I will substitute it and get the reduced matrix I will get it as minus 1 0 0 0 1 minus 1 0 minus 1 1 X 1 X 2 X 3 which is equals to 0 0 0 so from this we can get it as minus X 1 is equals to 0 the second equation X 1 minus X 2 is equals to 0 and the third one minus X 2 this is X 2 minus X 3 and this 1 minus X 2 plus X 3 is equals to 0 you have gotten three equations 1 2 & 3 now solving this we need to get the value of x1 x2 and x3 so from 1 we can say that X 1 is equals to 0 from Equation 2 we can write X 2 minus X 3 as X 2 Z equals 2 X 3 now assume X 2 is equals to 1 obviously X 3 will also be 1 therefore the eigenvector is therefore the eigen vector is 0 1 1 so we have got the eigenvector for two cases similarly we need to find the eigenvector for the third case similarly you will substitute lambda is equals to 3 and after simplifying everything you will get three equations after getting three equations when once you simplify it by the role of cross multiplication or by assumption method you will get the third eigenvector here for the third case I will directly write down the third eigenvector so for the third eigen vector that is for the case three that is when lambda is Z equals 2 for in that case the eigenvector is zero one minus one so we have got three eigenvectors for the corresponding three eigenvalues now step two is over now we need to find out the step three form the modal matrix what is the modal matrix columns are the eigenvectors so we have got this eigenvector 1 that is X 1 this is the second one that is X 2 this is the third one that is the X 3 now what is the modal matrix always it is denoted by B B is equal to you need to write these eigen vectors in column wise so how will you write down 1 0 0 0 1 1 0 1 minus 1 and I want to make one thing clear you can write the columns in any order you can write the eigen vectors the first eigen vector even you can write it as the third column any order but you should not change the write it in row wise always the eigen vector should be write it in column wise only so from this we have formed the modal matrix B what is the step 4 find the B inverse step four fine being buzzed how will you find V inverse what is the formula adjoint of B by determinant of B what is adjoint of B transpose of the cofactor matrix first we need to find the cofactor matrix and then we will take the transpose of them finally we will find out the determinant okay now first let me find out what is determinant of B this is the matrix B from this when you find out 1 into minus 1 you will as I told you earlier you will leave the corresponding row on the column to find the determinant 1 into minus 1 minus 1 so you will get it as 1 minus 1 which is equal to minus 2 therefore determinant of B is minus 2 now to find the adjoint of B you need to find the cofactor matrix cofactor matrix B which is nothing but V 1 1 minus b12 b13 minus B 2 1 B 2 2 minus B 2 3 B 3 1 minus B 3 2 B 3 3 so I will show you some examples what is how to find out B 1 1 and B 1 2 so on the minor of each element that is the cofactor matrix first one B 1 1 what is B 1 1 this element you will leave the corresponding row and column what is left out 1 1 1 minus 1 so you will get it as minus 1 minus 1 which is nothing but minus two similarly b12 for b12 you will leave the corresponding row and column this is b12 row and column you will get a does 0 0 so 0 1 0 minus 1 0 minus 0 you will get it as 0 B 1 3 again for this element you will leave this grow on the column you will get it as 0 1 0 1 which is now again nothing but 0 so for the rest you can do it similarly for this one the rows and the columns are left out 0 0 1 minus 1 so directly I will write down B 1 2 will be 0 and then B 2 2 will be 1 and then B 2 3 will be 1 for the next one B 3 1 again for the third row for this element you will leave the corresponding row on the column what is left out 0 into 1 minus 1 into 0 again which will give you 0 and then B 3 2 will give you 1 B 3 3 will give you minus 1 after finding out all these you have to write the inverse of the matrix so we have found out the cofactor matrix now what is the transpose of this transpose of the cofactor matrix which is equals to this is what is given you can directly write the transpose of this so what is it - - and then next one b12 is zero anyway it is going to be zero zero zero next one is B - one for B - 1 it is zero and then next element is one and then here it is minus one and for the final one zero minus 1 and minus one so finally therefore B inverse is equals to adjoint of B by determinant of B so determinant of B already we have found out s minus 2 so minus 1 by 2 into minus 2 0 0 0 1 minus 1 0 minus 1 minus 1 we have got B inverse now what is the final one last step find the product B inverse a B which will give you the diagonal form of the matrix we have found out B inverse minus 1 by 2 minus 2 0 0 0 1 minus 1 0 minus 1 minus 1 and then a that is the given matrix a 1 0 0 and 0 3 minus 1 0 minus 1 3 next last matrix the matrix B that is minus 2 the matrix B is 1 0 0 0 1 1 0 1 minus 1 so when you multiply all these you will get the diagonals form of the matrix that is 1 0 0 0 2 0 0 0 4 so from this you can see that the diagonal elements are all the eigenvalues the rest of the elements are all zeros this is what the diagonal form of the matrix so here no formulas nothing you can just go according to the working road that is the procedure what is the first step you will find the eigenvalues step 2 you will find the eigenvectors and then step 3 form the model matrix what is the model matrix the whose columns are the eigenvectors three eigenvalues you have found out corresponding eigenvectors you have found out those eigenvectors you will write it in column wise and also you can change the order accordingly you will get the final answer the eigenvalues will be changed here so any order you can write the eigenvectors and that will be the model matrix step 4 find the B inverse what is the formula for the inverse adjoint of a B by determinant of B whatever you are taking matrix B or anything so B inverse is equals to adjoint of B by determinant of B what is our joint of the transpose of the cofactor matrix and finally you need to find the product B inverse a B will give you the diagonal form of the matrix I'll teach you one simple way to find out the product form first to always go for the B inverse a you first multiply with these two and then whatever the resulting matrix you are getting it and from that you multiply the last one this will give you the easy way for finding out the diagonals are three eigenvalues

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Keywords: diagonalize a, engineering, mathematics, maths tricks, maths formulas, math questions, Nptel, GRE, GMAT, IBPS, BANK, BANK PO, Gate, Matrix, matrices, learning matrix, application of matrices, algebra, diagonalizable, diagonalmatrix, similar martices, symmetric matrix, orthogonal matrix, skew symmetric, orthogonally diagonalize, properties of diagonal matrices, properties of symmetric matrices., sat, math, maths

Id: ERB5GY1P-Ns

Length: 25min 36sec (1536 seconds)

Published: Mon Oct 03 2016