Eigen Values & Eigen Vector ( problems )

Hinzugefügt: 2026-05-22

[00:00:07]hi students so the problems filter eigenvalues and eigenvectors the pro first problem is find the eigen value and eigen vector of e of a matrix a is equals to 6 minus 2 2 minus 2 3 minus 1 to minus 1 3 first you can write the given matrix what is the given matrix 6 minus 2 2 minus 2 3 minus 1 2 minus 1 3 so what is your ambition you have to find the eigen values and eigen vectors so suppose you can assume that the eigen values are lambda and eigen vectors is x let uh let lambda be the an eigen value of a and corresponding eigen vector x by the definition of an eigen vector what is the definition a minus lambda i index is equals to 0 that is 6 minus lambda minus 2 2 minus 2 3 minus lambda minus 1 2 minus 1 3 minus lambda what is exponent x y z is equals to 0 0 0 okay now so that is 6 minus lambda minus 2 2 minus 2 3 minus lambda minus 1 2 minus 1 3 minus lambda x y z is equals to 0 0 0 next you can get the characteristic equation of a is the characteristic equation of a is that of a minus lambda i equal to 0 that is the debt of a minus lambda matrix what is a minus lambda matrix this is x matrix this is the zero matrix i mean b matrix there is a debt of a minus lambda i mean 6 minus lambda minus 2 2 minus 2 3 minus lambda minus 1 2 minus 1 3 minus lambda is equals to 0 now we can find this that value by using the technique or by using the formula the formula is first you can write lambda cube that is 1 into lambda cube just minus the sum of the sum of the diagonal elements of a into lambda square plus sum of the minus of the diagonal elements of a into lambda minus that a equal to 0 that is lambda cube minus sum of the diagonal elements of a into lambda square plus sum of the minus of minus of the diagonal elements of a into lambda minus that a is equals to zero what is the lambda cube minus what's the diagonal elements in the given matrix 6 3 3 into lambda square plus into lambda square plus sum of the minus of the diagonal elements of a what is the first diagonal element that is 6 we can remove the first column and first row we get the matrix is mod that of 3 minus one minus one three plus um yes what's the second diagonal element mark three now can remove the second column and second row get the values over the mode of uh that of six two two three plus now the finally what is the third diagonal element mark 3 that is we can remove the third column on third row we get the values over that of 6 minus 2 minus 2 3 into delta into delta lambda minus that of 6 minus 2 2 minus 2 3 minus 1 2 minus 1 3 is equals to 0 that is for lambda cube minus 12 lambda square plus 3 3 is 9 minus 1 9 minus 1 6 3 is 18 minus 4 18 minus 4 plus 18 minus 4 18 minus 4 lambda this debt value is equals to 32 by using cash flow you can find this debt value that is equal to 32 is equals to zero that is one lambda cube minus 1 lambda square plus 8 plus 14 plus 40 lambda minus 32 is equals to zero okay that is equals to 1 lambda cube plus minus 12 lambda square plus 36 lambda minus 32 is equals to zero by you know solve this equation by using cashew our synthetic division method so by helping us in the reducing method what is the lambda cube coefficient the lambda cube coefficient is 1 lambda square coefficient is minus 12 lambda coefficient is 36 uh constant term is minus 32 that is 2 into 2 first you can add 0 1 plus 0 1 to 1 0 2 now minus 12 plus 2 minus 10 2 into minus 10 minus 20 36 minus 20 that is 16 2 16 32 minus 32 plus 32 0 the next one is 0 8 one plus zero one eight ones are eight minus ten plus eight minus two eight into minus two minus sixteen plus sixteen minus sixteen zero minus one is two zero one two ones are two zero now finally uh lambda is equals to 2 to 8 now we go to from verification what is the verification the sum of the three diagonal elements the sum of the eigen values is equal to the sum of the diagonal elements of a the sum of the diagonal elements of a the sum of the diagonal some of the sorry some of the eigen values eigen values are equal to the sum of the diagonal elements of a that is the sum of eigen value 2 plus 2 plus 8 is equals to the sum of the diagonal elements is 6 plus 3 plus 3 there is 2 plus 2 plus 8 is equals to 12 6 plus 3 plus 3 is equal to 12 12 is equals to 12 this is true the second condition is the product of the three eigen values are identi equal to that a that is 2 into 2 into 8 is equals to that a that is 2 2 is 4 4 8 32 that is also 32 this is also true so the finally what do you say that lambda is the eigen values of a are lambda is equals to 2 lambda is equal to 2 lambda is equals to 8 so the case 1 is to find to find the eigenvector of a corresponding lambda is equals to 2 to find the eigenvector of a corresponding lambda is equals to 2 now substitute lambda is equals to 2 in a minus lambda into x is equals to 0 that is 6 minus 2 minus 2 to minus 2 3 minus 2 minus 1 2 minus 1 3 minus 2 x y z is equals to 0 0 0 that is 4 minus 2 2 minus 2 1 minus 1 2 minus 1 1 x y z is equals to 0 0 0 so the above matrix is in the form of a x is equals to b where a is equals to 4 minus 2 2 minus 2 1 minus 1 2 minus 1 1 x is equals to x y z b is equals to 0 0 0 now can write the augmented matrix what's the augmented matrix there is 4 minus 2 2 0 minus 2 1 minus 1 0 to minus 1 1 0 but now converted the augmented matrix into equilibrium form how to convert the eclain formula so by using the first row first element the second row first element order of first elements are 0 by using row operation that is r2s gives r 2 plus 2 r 1 r 3 is gives 4 r 3 minus 2 r 1 that is 4 minus 2 2 0 minus 8 plus 8 4 minus 4 minus 4 plus 4 zero plus zero eight minus eight minus four minus half minus four four minus four zero minus zero this is identity equal to four minus two two zero zero zero zero zero zero zero zero zero so clearly this is in equilibrium pump rank a is equals to one rank a means the number of non-zero the first row is non-zero the second row is a zero row third row is also zero rank a is equals to one rank of a b is also 1 number of unknowns is 3 what do you say that rank a is equals to rank of a b is less than the number of unknowns the system gives an infinite number of solutions the number of linearly independent solutions are n minus rank a that is 3 minus 1 2 now you can solve the equations that is since ax is equals to b now what is the modified a matrix ma 4 minus 2 2 0 0 0 0 0 0 and x is equal to xyz is equals to 0 0 0 now multiply these two matrices you get the values over 4x minus 2y plus 2z is equals to 0. we get only one equation ma so from equation 1 what do you say that 4x minus 2y 4x minus 4x minus 2y plus 2z is equal to 0 what is the 4x volume of 2y minus 2z that is what is the axis depends on y and z axis depends on y and z what is the y what is the z y and z are any any para y engineer any parameters that parameter you can choose y is equals to k 1 and z is equals to k 2 where k 1 k 2 are any parameters that is 4 x is equals to 2 k 1 minus 2 k2 x is equals to 1 by 2 into k1 minus 1 by 2 into k2 so so finally the finally capped x is equals to x y z x is equals to 1 by 2 k 1 minus 1 by 2 k 2 what is the y value of 1 k 1 this 1 k 1 kelvin written as 1 k 1 plus 0 k 2 that is equals to 1 k 2 this 1 k 2 can written f zero k one plus one k two now consider the two matrices that is capital x is equals to one by two into k one one k one zero k one plus minus one by two k two zero k two one k one that is x is equals to no common the word is our 1 by 2 into k1 that is 1 by 2 into k1 1 2 0 plus 1 by 2 into k2 minus 1 0 2 so this is in the form of capital x is equals to k 1 1 by 2 into k 1 e 1 plus 1 by 2 into k 2 e 2 where e 1 is equals to e 1 means eigen vector 1 is equals to 1 2 0 and eigen vector 2 is equals to minus 1 0 2 so this is the eigen vector 1 eigen vector 2 the case 2 is to find an eigen vector of a corresponding lambda is equals to 8 now substitute lambda is equals to 8 in a minus lambda i into x is equals to 0 that is 6 minus 8 minus 2 2 minus 2 3 minus 8 minus 1 2 minus 1 3 minus 8 x y is that is equals to 0 0 0 that is 6 month a 8 means minus 2 minus 2 2 minus 2 5 minus 1 2 minus 1 5 x y z is equals to 0 0 0 the above matrix is in the form of a so the above matrix is in the form of a x is equals to b the above matrix is in the form of a x is equals to b where a is equals to what's the array matrix a is equals to minus 2 minus 2 2 minus 2 minus 5 minus 1 2 minus 1 5 x is equals to x y z b is equals to 0 0 0 now can write the augmented matrix what is the augmented matrix mark that is uh mine you can write the a matrix that is minus 2 minus 2 to minus 2 minus 5 minus 1 2 minus 1 minus 5 0 0 0 no no converted the augmented matrix into equal uniform okay equilibrium how to convert the eclain formula by using the first row first element the second row first element order of first elements are zero by using row operation that is r2 is used 2r2 minus 2r1 r3 is gives 2r3 plus 2r1 that is minus 2 minus 220 minus 4 minus of minus 4 minus 10 minus of minus 4 minus 2 minus 4 0 minus zero rp is gives 2r3 plus 2r2 now 4 minus 4 minus 2 minus 4 minus 10 plus 4 0 plus 0. now open symbol for this matrix this is identity equal to minus 2 minus 2 to 0 0 minus 6 minus 6 0 0 minus 6 0 [Music] so the first row is minus 2 minus 2 2 0 0 minus 6 minus 6 0 0 minus 6 minus 6 0 we can divide in the first row with the two minus two the second row with minus six third row with minus six that is r one is gives minus one by two into r one r two is gives minus one by six into r r3 gives minus 1 by 6 into r3 we get the new matrix is 1 1 minus 1 0 0 1 1 0 0 1 1 0 now by using the second row second element third row second element is zero by using row operation the row operation is r3s gives one r3 minus 1 r2 this is identity equal to 1 1 minus 1 0 0 1 1 0 a 1 r 3 minus 1 r 2 means 0 minus 0 1 minus 1 1 minus 1 0 minus 0 so that is 1 1 minus 1 0 0 0 1 1 0 0 0 0 zero clearly this is in equilibrium form the rank is equals to two rank a means the number of non-zero rows the first row is non-zero second row is non-zero third row is a zero row rank k is equals to two rank of a b is equals to two number of unknowns is three so what you say that rhyme k is equals to rank of a b is less than number of unknowns so the system the system gives an infinite number of solutions the number of linearly independent solutions are n minus rank a n minus rank a means 3 minus 2 that is equals to 1 since a x is equals to b what is the modified a matrix 1 1 minus 1 0 1 1 0 0 0 you can write down 1 1 minus 1 0 1 1 0 0 0 x means x y z is equals to 0 0 0 now multiply these two matrices x plus y minus z is equals to zero now call this equation one y plus z is equals to zero now call this equation two from two what do you say that from two y plus z is equal to zero means uh y is equals to minus z so why is depends on z what is any parameter so now can choose why is it that is equals to k3 where k3 is any parameter so y is equals to minus k3 so that is equals to k3 means what's the y value minus k3 now substitute z is equals to k3 y is equals to k3 in equation 1 what is the equation 1 x plus y minus z is equals to 0 that is you know x minus k3 minus k3 is equals to 0 x is equals to 2 k3 what is the caplex volume of xyz is equals to 2 k3 minus 1 k3 1 k3 no common node is over k3 capped x is equals to k3 into 2 minus 1 1 so this is in the form of capital x is equals to k3 into e3 means eigen vector 3 where eigen vector 3 is equals to 2 minus 1 1 where k 3 is any parameter the finally e 1 is equals to 1 2 0 e 2 is equals to minus 1 0 to e 3 is equals to 2 minus 1 1 our eigen vectors of a corresponding the eigen values lambda is equals to 2 lambda is equals to 2 lambda is equals to 8 this is the first problem so the second problem is find the eigen values and eigen vectors of the matrix a is equals to 1 minus 6 minus 4 0 4 to 0 minus 6 3 what is the given matrix equals to 1 minus 6 minus 4 0 4 to 0 minus 6 3 this is a given matrix in the given matrix 1 4 minus 3 is the diagonal limits 1 four minus three is the diagonal elements and that's the minor of one is minor of one so means no remove the first column and first row four two minus six three confined to that value the minor of four is remove the for second row and for second column one minus four zero two this is the minor of four what is the minor of minus three now can remove the third row and third column you get one minus six zero four we can find the debt of one minus six zero four so uh let's we can find the eigen values eigen vectors now suppose you can the eigen values is lambda and eigen vector is x so what do you say that we can assume that that lambda be the an eigen value of a and corresponding eigen vector x what is the definition of an eigen vector that is a minus lambda i into x is equals to 0 what is a minus lambda matrix 1 minus lambda minus 6 minus 4 0 4 minus lambda 2 0 minus 6 minus 3 minus lambda x y z is equals to 0 0 0 let's look at the characteristic equation of a is what is the characteristic equation that of a minus lambda is equals to zero that is that of 1 minus lambda minus 6 minus 4 0 4 minus lambda 2 0 minus 6 minus 3 minus lambda is equals to zero so now can find that this state value by using the formula what is the formula one into the lambda cube minus the sum of the diagonal elements of a the sum of the diagonal elements of a into lambda square plus the sum of the minus of the diagonal elements of a the sum of the minus minus the sum of the minus of the diagonal elements of a into lambda minus that a equal to 0 that is 11 into lambda cube the sum of the diagonal elements of a what is the diagonal elements in the matrix a that is 1 4 minus 3 1 4 minus 3 into lambda square plus the sum of the minus of the diagonal elements of a minor means the first diagonal limit is one no can remove the first column and first row the data four to minus six three this is theta four two minus six three now remember the second uh second diagonal element four you can find the minor okay remove the second row and second column the rate of one minus four zero minus three like this plus the third minor minus three you can remove the third row and third column to get the values over the debt of one minus six zero 4 1 minus 1 1 minus this is minus 6 ma 1 minus 6 0 4 into del uh lambda minus the data of a means that of 1 minus 6 minus 4 0 4 2 0 minus 6 minus 3 is equals to 0 so now solve this equation this is equals to 1 into lambda cube minus 2 lambda square minus 12 plus 12 minus 3 minus 0 plus 4 minus 0 into lambda minus 0 is equals to 0 that is lambda cube minus 2 lambda square plus lambda minus 0 equal to 0 now can using solve this equation by using cashier r the synthetic division method that is the synthetic division method is uh the lambda cube coefficient is one and lambda square coefficient is minus two lambda coefficient is one the constant term is 0 first you can last term is 0 now you can add 0 0 1 plus 0 1 0 into 1 0 minus 2 plus 0 minus 2 0 into minus 2 0 1 plus 0 1 0 into one zero yes you can use the one one into zero one plus zero one one into one one minus two plus one minus one one into minus one minus one zero so last one is one uh one one zero one plus zero one one into one one minus one plus one is zero so the finally lambda is equals to zero one one now do the verification this eigen values are right or wrong so what is the first one the sum of the eigen values is equal to the the sum of the diagonal elements of a the sum of the diagonal elements of the given matrix a that is what is the diagonal means 0 for the eigenvalues 0 1 1 what is the diagonal elements of a 1 4 minus 3 that is 2 is equals to 2 this is true the second one is the product of 3 eigen values is equal to the data of a that is 0 into 1 into 1 is equals to that a what is the 0 into 1 into 1 0 that is also 0 that is this is also true the two true conditions the two cases the two conditions are satisfied that's why lambda is equal to 0 and lambda is equals to 1 lambda is equal to 1 our eigen values of the given matrix a next one the case 1 is to find an eigenvector of a corresponding the eigenvalue lambda is equals to zero now substitute lambda is equal to zero in a minus lambda into x is equals to zero that is one minus zero minus x minus four zero four minus zero to zero minus six minus three minus zero x y z is equals to zero zero zero so the finally one minus six minus four zero four two zero minus six minus three x y z is equals to zero zero zero so the above matrix is in the form of a x is equals to b what's our a matrix 1 minus 6 minus 4 0 4 2 0 minus 6 minus 3 and x is equals to x y z b is equals to 0 0 0 now can write the augmented matrix what is the augmented matrix we can add the a matrix and b matrix of both that is 1 minus 6 minus 4 0 0 4 2 0 0 0 sorry 0 minus 6 minus 3 0 the second row is common the word is over 2 the order is also common 3 that is r 2 is gives 1 by 2 into r 2 r 3 is used minus 1 by 3 into r 3 there is 1 minus 6 minus 4 0 0 2 1 0 0 2 1 0.

[00:23:14]now convert this augmented matrix into equilion form so by using the first row first element the second row first element third row first elements are zero luckily the first row on the second row first element order of first elements are zeros that's why we go to from the second row second element the third or second element is zero that is r3 is gives 2r3 minus 2r2 there is 1 minus x minus 4 0 0 2 1 0 2 r 3 minus 2 r 2 0 minus 0 4 minus 4 2 minus 2 0 minus 0 that is 1 minus 6 minus 4 0 0 2 1 0 0 0 0 0 clearly this is in equilibrium form rank a is equals to 2 rank of a b is equals to 2 number of unknowns is 3 that is rank a is equals to rank of a b is less than n in this case the system gives an infinite number of solutions the number of linearly independent solutions are n minus rank a that is uh what is n volume of 3 rank k is equals to 2 that is 3 minus 2 means 1 we can choose one parameter that is since a x is equals to b what is the modified a matrix ma 1 minus 6 minus 4 0 2 1 0 0 0 that is 1 minus x minus 4 0 2 1 0 0 0 cap x means x y z is equals to 0 0 0 now can multiply these two matrices that is x minus 6 y minus 4 z is equals to 0 2 y plus that is equals to 0 so from 2 what do you say that 2y plus z is equal to 0 means 2y is equals to minus z what about z depends on jatna that is any parameter that is equals to k1 where k1 is any parameter there is 2y is equals to minus k1 y is equals to minus 1 by 2 into k1 so what is the jet volume no substitute that is equals to k1 y is equals to minus 1 by 2 into k1 in equation 1 what is the equation 1 x minus 6 y minus 4 z is equals to 0 in place of y you can write the values over minus 1 by 2 into k 1 in place of z you can add the values over k 1 that is x plus 3 k 1 minus 4 k 1 is equals to 0 that is x minus k 1 is equals to 0 x is equals to 1 k 1 what is the final volume capital x is equals to x y z 1 k 1 minus 1 by 2 k 1 1 k 1 what is the capital x value 1 by 2 into k 1 common voltage over 1 by 2 into k1 2 minus 1 2 so this is in the form of capital x is equals to 1 by 2 into k1 into e1 where e1 is equals to 2 minus 1 2 this is eigen vector 1 just you to find the eigen vector of a eigenvector of a corresponding the eigenvalue lambda is equals to 1 no concepts to lambda is equals to 1 in a minus lambda into x is equals to 0 that is 1 minus 1 minus x minus 4 0 4 minus 1 to 0 minus 6 minus 3 minus 1 what's x value my x y z is equals to 0 0 0 now simplify this matrix we get the values of our 0 minus 6 minus 4 0 3 to 0 minus 6 minus 4 x y z 0 0 0 the above matrix is in the form of a x is equals to b what's the array volume 0 minus 6 minus 4 0 3 to 0 minus 6 minus 4 x is equals to x y z b is equals to 0 0 0 okay now we can write the augmented matrix the augmented matrix means 0 minus six minus four zero zero three two zero zero uh minus six minus four zero in the first row the common word is over two minus two the second the second row is same the order is also common the word is over minus one by two that is r one is gives minus one by two into r two r three is also gives minus one by two into r three so this is identity equal to zero three two zero zero three two zero zero three two zero now by using the first this now convert this augmented matrix into equilibrium form neckline form what's the first condition the first row first element is non-zero element now look at the second element this is also zero element now look at the order of estimate this is also zero element in this case by using the first row second element the third second row second element third row second elements are zeros by using row operation that is r two is gives three r two minus three r one r three s gives three r3 minus the r1 the first row is same now it is 0 3 2 0 the second row means 0 minus 0 9 minus 9 6 minus 6 0 minus 0 here also 0 minus 0 rps gives 3 r3 minus 3 r1 0 minus 0 9 minus 9 6 minus 6 0 minus 0 so finally 0 3 2 0 0 0 0 0 0 0 0 clearly this is a neckline form now find rank a the first row is non-zero the second row is zero order is also zero that is rank a is equals to one rank of a b is equals to one the number of unknowns is three it is rank a is equals to rank of a b is less than number of unknowns in this case the system gives an infinite number of solutions the number of linearly independent solutions is n minus rank a it is 3 minus 2 you can choose to get 2.

[00:28:42]now this means you can choose any two parameters that is since what's the condition a x is equals to b no can write the modified a matrix 0 3 2 0 0 0 0 0 0 x y z is equals to 0 0 0 that is 3y plus 2z is equal to 0 that is 3y is equals to minus 2z y is equals to minus 2 by 3 into z so y is depends on z no can choose z is any parameter what about x x is no can choose x is also any parameter it is x is equals to k2 that is equals to k3 where k2 and k3 are any two parameters now can choose y is equals to minus 2 by 3 into k3 what is the caplex volume capital x is equals to x y z that is what is the x volume that is one k two this k two can be written as one k two plus zero k three y can be written as um zero k two minus two by three k three that can be written as zero k two plus one k three now can split this matrix is totally into two matrices that is capital x is equals to one k two zero k two zero k two plus zero k three minus two by three k three one k three that is capital x is equals to k two into 1 0 0 plus k 3 by 3 0 minus 2 3 so this is in the form of x is equals to k 2 into e 2 plus k 3 by 3 into e 3 where u 2 means the eigen vector 2 is equals to 1 0 0 and eigen vector 3 is 0 minus 2 3 the finally e 1 is equals to 2 minus 1 2 e 2 is equals to 1 0 0 e 3 is equals to 0 to 0 minus 2 3 are eigen vectors i are eigenvectors of a corresponding the eigenvalues of lambda is equals to 0 1 1 this is one of the problem so the last problem is find the eigen values and eigen vectors of the matrix 8 minus 6 2 minus 6 7 minus 4 2 minus 4 3 what is the given matrix the given matrix is a is equals to 8 minus 6 2 minus 6 7 minus 4 2 minus 4 3 so we can find this in this matrix eigen values and eigen vectors suppose eigen values is lambda and eigen eigenvectors is x we can assume that let lambda be the an eigenvalue of a and corresponding eigenvector x what is the definition of an eigenvector a minus lambda into x is equals to zero that is h minus lambda minus x2 minus 6 7 minus lambda minus 4 to minus 4 3 minus lambda x means x y z is equals to 0 0 0 now can choose the characteristic the characteristic equation of a is that of a minus lambda is equals to 0 that is 8 minus lambda minus 6 2 minus x 7 minus lambda minus 4 2 minus 4 3 minus lambda is equals to 0 we can find this z value by using the formula the formula means uh lambda cube 1 lambda cube minus the sum of the diagonal elements of a the sum of the diagonal elements of a into lambda square plus the sum of the minus of the diagonal elements of a into lambda minus that a equal to zero minus that a equal to zero just listen what is the formula one into lambda cube minus the sum of the the sum of the diagonal elements of a into lambda square plus the sum of the diagonal elements of a sum of the minus of the diagonal elements of a into lambda minus delta a equal to zero that is uh 1 into lambda cube what is the diagonal means minus 8 plus 1 plus 3 into lambda square what is the first diagonal element 8 we can remove the first row and first column we get the values over 7 minus minus four three the data of seven minus four minus four three plus now okay remember the second diagonal element that is seven now remove the second column and second row we get eight two two three you can find the debt plus the third diagonal moment is 3 you can find the minor of 3 the rate of 8 minus 6 minus 6 7 how to find this value we can remove the third column and third row into lambda minus that a that is my debt of 8 minus 6 2 minus 6 7 minus 4 2 minus 4 3 is equals to 0 now can solve simplify this equation 1 lambda cube minus 18 lambda square 7 3 is 21 minus 16 plus 8 is 24 minus 4 plus 8 756 minus 36 into lambda and this that value is equals to 0 the finally 1 lambda cube minus 18 lambda square plus 5 plus 20 plus 20 into lambda minus 0 is equals to 0 that is 1 lambda cube minus 18 lambda square plus 45 lambda minus 0 is equals to 0 by using this equation and solving this equation by using cashew as synthetic division method what is the synthetic division method the lambda cube coefficient is 1 lambda square coefficient is minus 18.

[00:33:59]lambda coefficient is 45 constant term is 0. first you can last term is 0 by using 0 zero one plus zero one zero into one zero minus eighteen plus zero minus eighteen zero into minus eighteen zero uh for forty five plus zero forty five zero into forty five zero and by using three so uh first term is zero one plus zero one three ones are three minus eighteen plus three minus fifteen three into minus fifteen minus forty five plus forty five minus forty five zero fifteen first word is zero one plus zero one fifteen ones are fifteen the minus fifteen plus fifteen is zero so how to solve this equation by using cashio first you can press the mode button is three times first you can press the mode button is three times first time second time totem you get the equation matrix vectors what is the equation one so how the unknowns so there is no need from unknowns this is the home button in home button you can press this side arrow what is the degree degree is uh what is the ice power of lambda 3 you can choose 3 okay what's the a a means lambda cube coefficient that is 1 is equals to the second b lambda square coefficient minus 18 is equals to c i mean lambda coefficient 45 is equals to the dm is a constant term that is zero so x one means uh first first eigen vector 15 0 3 r 3 0 15 okay so lambda is equal to 0 3 15.

[00:35:46]now go to from verification ma so this eigenvector is right or run now can check it that is the sum of the eigen values is equal to the the diagonal elements of the given matrix a okay the sum of the eigen values are equal to the the sum of the diagonal elements of a 0 plus 3 plus 15 is equals to 8 plus 7 plus 3 that is 18 is equal to 18 this is true the next one is case 2 is the product of the three eigenvalues uh eigenvalues are equal to the that of a that is 0 into 3 into 15 is equals to that a so 0 is equal to 0 okay this is also true so the two cases are true the given uh lambda values is equals to zero lambda is equals to three lambdas equals 15 are eigen values of the given matrix a so first one is to find the eigenvector of a corresponding lambda is equal to zero substitute lambda is equal to zero in a minus lambda into x is equals to zero that is eight minus zero minus six to minus six seven minus zero minus four two minus four three minus 0 x y z is equals to 0 0 0 that is 8 minus 6 2 minus 6 7 minus 4 2 minus 4 3 x y z is equals to 0 0 0 the above matrix is in the form of a x is equals to b what's your a matrix ma 8 minus 6 2 minus 6 7 minus 4 2 minus 4 3 x is equals to x y z b is equals to 0 0 0 look at the augmented matrix what's argumentative matrix ma we can write a matrix and b matrix that is 8 minus 6 2 minus 6 7 minus 4 0 to minus 4 3 0 so now converted this argumentative matrix into equilibrium form how to convert how to convert ma by using the first row first element the second row first element third row first elements are zero that is r2s gives eight r2 plus six r1 r3s gives eight r3 minus two r1 that is 8 minus 6 2 0 what is r 2 8 r 2 plus 6 r 1 minus 48 plus 48 58 plus minus 36 minus 32 plus 12 0 plus 0 the third row means rpgs gives 8 r 3 minus 2 r1 16 minus 16 minus 32 minus half minus 12 24 minus 4 0 minus 0 it is 8 minus 6 2 0 0 20 minus 20 0 0 minus 28 20 is zero so the first second row is the second row is common load is 20 the order is also common what is the word 20 that is r2s gives 1 by 20 into r2 r3 also gives 1 by 20 into r3 we get the matrix is 8 minus 6 to 0 0 1 minus 1 0 0 minus 1 1 0 again by using the second row second element the third row second element is zero r3s gives one r three plus one r one eight minus six two zero zero one minus one zero one r three plus one r two zero plus zero minus one plus one one minus 1 0 plus 0 this is identity equal to 8 minus 6 2 0 0 1 minus 1 0 0 0 0 0 clearly this is in a clan form rank a means the number of non-zero rows the first row is non-zero the second row is non-zero third row is a zero row rank a is equals to two rank of a b is equals to two number of unknowns is three there is rank a is equals to rank of a b is less than number of unknowns the system gives an infinite number of solutions the number of linearly independent solutions are n minus rank a that is 3 minus 2 1 since ax is equals to b what is a modified a matrix 8 minus 6 to 0 1 minus 1 0 0 0 x is equal to x y z is equals to 0 0 0 that is 8 x minus 6 y plus 2 z is equal to 0 y minus z is equal to 0 so from 2 what is say that y minus z is equal to 0 means y is equals to z y is depends on z that is what about z that is any parameter so you can choose that is equals to k1 where k1 is any parameter so that is equals to k1 means what's the y is equal to y is also k1 now can substitute that is equals to k1 and y is equals to k1 in equation 1 there is 8x minus 6 k1 plus 2 k1 is equals to 0 8x minus 6 k1 plus 2 k1 that is 4 k1 is equals to 0 that is x is equals to four by eight into k one x is equals to one by two into k one so finally capital x is equals to x y z one by two k one one k one one k one capped x is equals to one by two into k one one two two this is in the form of a x is equals to 1 by 2 into k1 e1 where e1 is equals to 1 2 2 and the k1 is any parameter k1 is any parameter unless we go to from condition 2 means case 2 to find the eigenvector of a corresponding the eigenvalue lambda is equals to 3 to find an eigenvector of a corresponding the eigenvalue lambda is equals to 3 so substitute lambda is equals to 3 in a minus lambda into x is equals to 0 it is 8 minus 3 minus 6 to minus 6 7 minus 3 minus 4 2 minus 4 3 minus 3 x y z is equals to 0 0 0 now simplify this matrix that is 5 minus 6 2 minus 6 4 minus 4 2 minus 4 0 x y z 0 0 0 again the above matrix is in the form of a x is equals to b that is what is the array matrix map 5 minus 6 2 minus 6 4 minus 4 2 minus 4 0 x is equals to x y z b is equals to 0 0 0 now conduct the augmented matrix what's the augmented matrix my a b is equals to 5 minus 6 to 0 minus 6 4 minus 4 0 to minus 4 0 0. now convert this accommodation matrix into a clone form how to convert ma by using the first row first element the second row first element order of first elements are zero that is r two is gives five r two plus six r one r three s gives five r three minus two r one that is five minus six two zero what is r two m five r two plus six r one that is 5 into minus 6 minus 30 plus 30 20 minus 636 minus 20 plus 12 0 plus 0 rps gives 5 r3 minus 2 r1 10 minus 10 minus 20 minus half minus 12 0 minus 4 0 minus 0 it is 5 minus 6 to 0 0 minus 6 minus 8 0 0 minus 8 minus 4 zero the second row is common the word is over eight the third row common is our minus one by four that is five minus six two zero zero two one zero two one again by using the second row second element third row second element is zero that is r3 is gives 2 r3 minus 2 r2 5 minus 6 2 0 0 2 1 0 rps gives 2 r3 minus 2 r2 0 minus 0 4 minus 4 2 minus 2 0 minus 0 that is 5 minus 6 2 0 0 2 1 0 0 0 0 0 clearly this is in equilibrium form the rank is equals to 2 rank of a b is equals to 2 number of unknowns is 3. that is rank a is equals to rank of a b is less than n the system gives an infinite number of solutions the number of linearly independent solutions are n minus tank a 3 minus 2 1 that is you can choose one parameter since a x is equal to b what is the modified a matrix 5 minus 6 to 0 to 1 0 0 0 and x is x y z is equals to 0 0 0 that is 5 5x minus 65 plus 2z is equals to 0 2y plus that is equals to 0.

[00:44:00]that is from 2 what do you say that 2y plus z is equal to 0 or y is equals to minus 1 by 2 into z so y is depends on z what about z that is any parameter that is equals to k2 where k2 is any parameter so that is equals to k2 means what's the y value minus 1 by 2 into k1 now substitute that is equals to k2 y is equals to 1 by 2 into k2 in equation 1 what is the equation 1 ma 5x minus 65 plus 2 z is equal to 0 in plus of 5 you can write the values over minus 1 by 2 into k2 in plus object we can write the values over k2 that is 5x minus 6 into minus 1 by 2 into k2 plus 2 k2 is equals to 0 that is 5x plus 3 k2 plus 2 k2 is equal to 0 5x is equals to minus k2 the positive 55 is cancelled x is equals to 1 into k2 that is minus 1 into k2 cap x is equals to x y z minus 1 k 2 minus 1 by 2 k 2 1 k 2 what's the capital x value now common what is over k 2 by 2 minus 2 minus 1 2 this is in the form of x capital x is equals to k 2 by 2 into e 2 where e 2 is eigen vector 2 there is eigen vector 2 means minus 2 minus 1 2 and k 2 is any parameter so the next one is so the next one is the last case is case 3 is to find an eigen vector of a corresponding lambda is equals to 15.

[00:45:29]now substitute lambda is equal to 15 in a minus lambda index is equals to zero that is eight minus fifteen minus six two minus six seven minus fifteen minus four two minus four three minus fifteen xa means x y z zero zero zero now simplify this matrix minus 7 minus 6 2 minus 6 minus 8 minus 4 2 minus 4 minus 12 x is equals to x y z is equals to 0 0 0 so this is in the form of a x is equal to b where a is equals to minus 7 this is minus 7 minus 7 minus 6 2 minus 6 minus 8 minus 4 2 minus 4 minus 12 capital x is x y z b is equals to zero zero zero now can write the augmented matrix what is the augmented matrix a b is equals to minus seven minus six two zero minus six minus eight minus four zero two minus four minus twelve zero now convert this accommodating matrix into a clan form how to convert mum by using first row first element the second row first element third row first elements are zero by using row operation what's the row operation ma seven r two minus six r one r three s gives 7r3 plus 2r1 that is it is -7 minus 620 minus 42 minus of minus 42 minus 56 minus half minus 36 minus 28 minus 12 equal 0 minus 0 14 minus 14 28 minus 28 minus 12 minus 84 plus 4 0 minus 0 minus 7 minus 6 2 0 0 minus 20 minus 40 0 0 this is minus 40 minus 80 0 the second row common the order is over minus 20 the order is also common the reservoir minus 40 r 2s gives minus 1 by 20 into r2 r3 squares minus 1 by 40 into r3 this is identity equal to minus 7 minus 6 2 0 0 1 2 0 0 1 2 0 by using the second row second element third row second element is zero r3s gives one r3 minus one r2 that is minus seven minus six two zero zero one two zero zero minus zero one minus one two minus two uh zero minus zero now simplify this word that is minus seven minus six two zero zero one two zero zero zero zero zero clearly this is in equilibrium form rank a is equals to number of non-zero rows the number of non-zero rows is two non-zero rows rank of a b is also two number of unknowns is three it is rank as equals to rank of a b is less than number of unknowns the system gives an infinite number of solutions the number the linearly independent solutions are n minus tank a there is 3 minus 2 means 1 now you can choose one parameter since a x is equal to b it is minus 7 minus 6 2 0 1 2 0 0 0 x y z is equals to 0 0 0 now multiply these two matrices ma we get the values over so minus 7x minus 6y plus 2z is equals to 0 y plus 2z is equals to 0 from 2 we have y y plus 2 z equals 0 means y is equals to minus 2z what about z that is any parameter that parameter is k3 so z is equals to k3 where k3 is any parameter what is the y value minus 2 k3 now substitute z value y value in equation 1 that is minus 2 7 x minus 6 into minus 2 k 3 plus 2 k 3 it is minus 7 x plus 2 k 12 k 3 plus 2 k 3 equal to 0 that is minus 7 x is equals to minus 14 k 3 x is equals to 2 k 3. our capital x is equals to x y z that is 2 k 3 minus 2 k 3 1 k 3 no common divorce over k 3 2 minus 2 1 so this is in the form of capital x is equals to k 3 into e 3 where i 3 means i can vector 3 2 minus 2 1 k 3 is any parameter so finally e 1 is equals to 1 2 2 e 2 is equals to minus 2 minus 1 2 and e 3 is equals to 2 minus 2 1 our eigenvalue eigenvector of a and corresponding the eigenvalues 0 3 15. thank you thank you everyone