Matrices- I : Best Engineering Mathematics Tips (Anna University ,JNTU ,GATE,Delhi University)

engineering mathematics one first let me tell you what are the topics covered in engineering mathematics one mattresss differential calculus analytic geometry functions of several variables and multiple integrals so in this I am going to teach you the matrices what does mattress is a rectangular arrangement of elements is called as mattresses it can be a two by two matrix or three by three matrix before I move on to mattresses I just want to tell you the basic definitions that is diagonals of a matrix and the kingís of a matrix diagonal of a matrix your square matrix in which all the elements except the diagonal elements are zeros is called the diagonal of a matrix what is it if suppose if you take a matrix a is equals to 1 0 0 0 2 0 0 0 3 you can see in this these are the diagonal elements except these diagonal elements all the other elements are 0 so we call this as a diagonals of a matrix next inverse of a matrix what is inverse of a matrix inverse of a matrix suppose let a be the given matrix let a be the given matrix then a inverse the inverse of the matrix is defined by a inverse which is equals to adjoint of a by determinant of a adjoint of a by determinant of a what is our joint of a it is the transpose of the cofactor matrix so these two are the important definitions which we are going to apply in the matrices that is the first thing let me tell you what are the topics subtopics in the mattresses first one eigen values and eigenvectors eigenvalues and eigenvectors number two number two Calley Hamilton theorem Cali Hamilton theorem number three diagonal ization of a matrix diagonalization of the matrix so in this the first topic that is eigenvalues and eigenvectors which i am going to teach you now what is the eigen values what do you mean by that i ghen values the equation the equation obtained by the equation obtained by [Music] expanding the determinant expanding the determinant and equating to 0 expanding the determinant and equating to 0 so what do you mean by eigenvalues and eigenvectors so first thing the equation obtained by expanding the determinant and equating to 0 so in this this is the formula that is determinant of a minus lambda equal to 0 is the formula what will be them a is a given matrix and lambda is the scalar and I will be the unit matrix so the unit matrix will be taken according to the given matrix it can be a 2 by 2 matrix on 3 by 3 matrix if suppose if it is a 2 by 2 matrix then the unit matrix will be 1 0 0 1 if it is a 3 by 3 matrix then the unit matrix will be I is equals to 1 0 0 0 1 0 0 0 1 so this is the eigenvalues characteristic equation before we move on to the eigenvalues you should know how to find the characteristic equation and then the roots of the characteristic equation will give you the eigenvalues let me show you through some example so they will ask you they will ask you the question find the eigenvalues and eigenvectors of the matrix a so the first step you need to find the characteristic equation under roots suppose this is the question find the eigenvalues find the eigenvalues and eigenvectors of the matrix of the matrix 3 2 1 4 if this is the matrix given what is it it is a two-by-two matrix if this is given and asked to find out the eigenvalues and eigenvectors what did I tell you first thing you need to find the characteristic equation using the formula determinant of a minus lambda I equal to 0 determinant of a minus lambda equals to 0 what is a a will be the given matrix and lambda will be the scalar I will be the unit matrix now from this first let me take let a is equals to 3 2 1 4 and what is lambda lambda is a scalar that is lambda I is equals to lambda 0 0 lambda as I told you earlier if it is the two-by-two unit matrix then only the diagonal elements will be the unit 1 and the other elements will be 0 now we need to substitute in this formula what is that formula a minus lambda I equal to 0 what is a is the given matrix that is 3 2 1 4 and minus lambda I that is lambda 0 0 lambda which is equals to 0 now we need to simplify this and get the characteristic equation how will you simplify it 3 minus lambda 2 minus 0 will give you 2 1 minus 0 will give you 1 again 4 minus lambda will be written as it is now usually how do you find the determinant 3 minus lambda into 4 minus lambda minus of 2 into 1 so you need to cross multiply to find the determinant how will you cross multiply three minus lambda into four minus lambda minus two equal to zero now when you multiply this 3 into 4 which will give you 12 minus lambda into 4 will give you minus 4 lambda again 3 into minus lambda will give you minus 7 lambda minus lambda into minus lambda plus lambda square minus 2 equal to 0 so we say this is the characteristic equation when you simplify this again you will get this as lambda square minus 7 lambda 12 minus 2 will give you 10 which is equals to 0 is the characteristic equation now to find the eigenvalues we have got the characteristic equation so how will you get the eigenvalues when you solve this characteristic equation and get those roots those roots will give you the eigenvalues let me tell you how will you get the roots so this is the characteristic equation lambda square minus 7 lambda plus 10 equal to 0 how will you get the roots only when you factorize this if this is the equation you need to check what is the constant here what is the constant 10 what are the possible ways of multiplying 10 five times two and ten times one when you add this or subtract this you should get the middle value which one will you take so when you add these two you will get seven lambda therefore we will take this as the roots so now split the roots lambda minus five into lambda minus two equal to zero so this will give you lambda as equals to 2 comma 5 so these roots are called the eigenvalues or characteristic roots so we will call lambda is equals to 2 comma 5 are the eigenvalues so always to find the eigenvalues the first step you need to find the characteristic equation after getting the characteristic equation when you solve it you will get the roots those roots are called as I can values and eigenvectors now this is over for to find the eigenvalues we have found out this much now we need to find the eigenvectors how many roots we have got two roots lambda is equals to two and five so we need to find two cases if suppose if they have given three by three matrix obviously we will be getting three roots so how many cases we need to find out four eigenvectors three cases that is lambda is equal Sleeps suppose for example lambda is equals to 1 2 5 so three cases we need to find for the eigenvectors now let me show you how to find the eigenvectors now case one to find the eigenvectors to find the eigenvectors case 1 when lambda is equals to two what is the formula to find the eigen vector a minus lambda I into X bar is equals to zero this is the formula to find the eigen vector a minus lambda I into X bar is equals to zero we know already what is a minus lambda what is rate the given matrix and lambda is the scalar and I is a unit matrix so directly I am going to substitute in this and according to the given matrix I am going to write the eigen vectors so a minus lambda I means 3 minus lambda 2 1 4 minus lambda into X bar as I told you according to the given matrix we will take X bar that is X 1 and X 2 suppose for example if it is a 3 by 3 matrix then we will take the X bar as x1 x2 x3 now for this one only x1 and x2 which is equals to 0 now look at the case when lambda is equals to 2 substitute this lambda value here and you will get the simplified matrix what is that simplified matrix when we substitute the lambda you will get it as 3 minus 2 will give you 1 and here 2 you will write it as it is and again in the next row you will write 1 as it is and 4 minus 2 will give you 2 into X 1 X 2 which is equals to 0 now you can look at this now in this matrix I am going to multiply that X bar what is it 1 into X 1 plus 2 into - that is rows into columns rows into columns we'll give you 2 equations 2 into X 2 which is R equals to 0 now again the next row X 1 + 2 + 2 X 2 now we have got two equations 1 & 2 I am going to solve these two equations and get the values for X 1 and X 2 that will give you the eigenvectors now what is the equation first and second you can look at the equation both the equations are same so we cannot go by simultaneous equation I am going to do it by assumption method from 1 X 1 plus X 2 2 X 2 is equals to 0 from this I am going to write X 1 is equals to minus 2 X 2 now shall I assume the value X 2 s 1 when I assume it X 2 s 1 assume X 2 is equals to 1 then what will be the X 1 value X 1 is equals to minus 2 so we have got two values X 1 and X 2 therefore the eigen vector corresponding to the value therefore the eigen vector corresponding to the value lambda is equals to 2 is X 1 value and X 2 value that is minus 2 and 1 this is the eigenvector for the case 1 similarly we need to find the eigenvector for the case 2 what is the case to the case 2 value is when lambda is the equals to 5 same way using the same formula M - lambda into X bar equals to zero I am going to find the eigenvector for the next case that is three minus five which will give you minus 2 and here - one and again four minus five will give you minus five and this one into x1 x2 which is equals to zero as we did for the case one previous case the same way I am going to multiply rows into columns so this will give you 2 - 2 X 1 plus 2 X 2 which is equals to 0 and X 1 minus X 2 is equals to 0 so from this this is 1/2 so from this I am going to pick any one of the equation if I pick the second one you should check among these two which one is the easier to way to solve the eigen vector from this I am going to take the second equation that is x1 minus x2 is equal to 0 I am going to rewrite this as x1 is equal to x2 because to assume the value for any one of the variable assume X 1 is equals to 1 when I am assuming X 1 is equals to 1 obviously X 2 will also be 1 why is it because the equation is X 1 is equals to X 2 so similarly therefore you will get X 2 is equals to 1 you can write down the eigen vector corresponding to lambda is equals to 5 is 1 and 1 so this is how you need to find the eigenvalues and eigenvectors of a 2x2 matrix now you can look at this what are the procedures first thing you need to find the characteristic equation the next you need to find the eigenvalues how will you find the eigenvalues when you solve it when you factorize it you will get those roots those roots will give you the eigenvalues and finally according to the roots you need to find the eigenvectors now how will you check whether this eigenvector is correct or not you need to substitute this value in any one of the equation now suppose for example this is the eigenvector you got it for the value lambda is equals to 5 what is the eigenvector 1 and 1 these are the equations we have got it substitute in this when you take the second equation this is to check suppose if I am taking the second equation X 1 minus X 2 is equals to 0 when you substitute this 1 and 1 here 1 minus 1 which will give you 0 0 is equals to 0 hence it satisfies this equation so we call the eigenvector what we have done this correct so this is how you need to check the eigenvector for the corresponding eigen values so far we have done the matrix that is 2 by 2 matrix eigen values and eigen vectors now I am going to tell you for the 3 by 3 matrix same way for 3x3 matrix to find the characteristic equation this is the formula a minus lambda I equal to zero we can also find by using the formula lambda cubed minus es 1 lambda square plus s 2 lambda minus s 3 is equals to 0 this is the formula to find the 3 by 3 matrix where s 1 is equals to sum of the diagonal elements s 1 is equal to sum of the diagonal elements yes 2 is equals to some of the minus of the diagonal elements some of the minor of the diagonal elements X 3 is equals to determinant of a so this is the easy method to find out the 3x3 matrix eigenvalues and eigenvectors you can see lambda cube minus s1 lambda square plus s 2 lambda minus s3 is equals to 0 so what is s 1 s 2 s 3 s 1 some of the diagonal elements what is the sum of the diagonal elements whatever suppose if they are giving you the matrix 1 0 2 & 3 1 4 & 1 2 3 what is the diagonal element here 1 1 3 so the sum will give you the S 1 1 plus 1 plus 3 which will give you 5 so s 1 is equals to 5 you will substitute it here similarly what is yes to some of the minors of the diagonal element what do you mean by the minor if it is a three by three matrix you need to find the minor for each diagonal element what are the diagonal elements 1 1 3 when you take the first element you have to leave the corresponding row and the column this will give you the minor and similarly you need to find for all the diagonal elements on determinants you know how to find out one into the cross multiplication of this and again for the next and for the third element this is what s1 s2 and s3 I will show it through example No find the eigenvalues and eigenvectors of the matrix 3 -1 0 and similarly here minus 1 2 minus 1 0 minus 1 3 if this is the 3x3 matrix the same procedure which I am going to follow it now that is characteristic equation I ghen values and eigen vectors first characteristic equation how we are supposed to find out characteristic equation by using the formula lambda cube minus s1 lambda square minus s 2 lambda minus s 3 equal to 0 so from this what is yes 1 sum of the diagonal elements what are the diagonal elements here 2 3 2 3 that is 3 plus 2 plus 3 which will give you 8 similarly yes - what is yes to the sum of the minors of the diagonal elements when you are taking 3 first diagonal element as 3 leave the corresponding row on the column so what is the left of numbers 2 minus 1 minus 1 3 2 minus 1 minus 1 3 plus what is the next diagonal element - same way leave the corresponding column on the row what is left out 3 0 0 3 what is the next last diagonal element three again you will leave the corresponding row on the column what is left out three minus one minus one two now when you simplify all these you will get some value those value we call it as s2 so when you simplify it what do you get you can look at this how will you simplify it the determinant

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Keywords: Engineering (Industry), Engineering Mathematics, Mathematics (Field Of Study), Matrix (Literature Subject), Identity Matrix, matrices, engineering, mathematics, maths tricks, maths formulas, math questions, calculus, Nptel, GRE, GMAT, IBPS, BANK, BANK PO, Gate, matrice, learning matrix, matrices problems, matrix order, matrix 4, matrix meaning, definintion of matrix, matrix explained, matrix questions, application of matrices, sat, math, maths, tips, tricks

Id: EJ8Ptnk7dzU

Length: 26min 13sec (1573 seconds)

Published: Wed Nov 25 2015