Matrices Solution Of Matrix :Best Engineering Mathematics Tips (AU,JNTU,GATE,DU)

now I am going to show you the system of a equations example first thing what is given your test to fault consistency and if it is consistent then find the solution for the system of equations they will give you the three system of equations or four system of equations they can give you any number of simultaneous equation what is the first step you need to write it in the matrix form the given system of equations you have to write it in the matrix form so let a is equals to what is the coefficient of x here one coefficient of Y will be 1 and the coefficient of Z is 1 similarly for the next equation 2 1 minus 1 you need to take the coefficients of the variables XY and Z you will write it in the matrix form and what about for the last equation 4 minus 1 2 so I have taken the given system of linear equations as a matrix form now the argumented matrix is the argument ER matrix a comma B is equals to you will write the matrix along with the right-hand side of the linear equations so what is it 1 1 1 and for the first equation what is the right-hand side 3 and what about the next equation 2 1 minus 1 and what is the right-hand side of the equation two on the last one 4 minus 1/2 and what does the right-hand side of last equation five so what is our aim to check for consistency we need to check whether the rank of a and the rank of the argumented matrix a comma B is equal or not if it is equal then we say that it is consistent if it is not equal then we say that it is not inconsistent it is inconsistent so first I am going to find the rank of a matrix for the argument matrix a comma B find rank of a comma B so again step one what is step one as I told you while explaining the rank of a matrix you need to make the below the diagonal element should be zero so what is below the diagonal elements of the matrix a 2 4 minus 1 you need to make these three elements as zeros only the three elements should be zero we call this as below the diagonal elements so how will you make it a zero by using the first row on the second rope first I am going to write it as R 3 is equals to R 3 minus 4 R 1 so I am going to make changes only in the last row the first row and the second row should be written as it is 1 1 1 3 2 1 minus 1 2 now 4 minus 4 4 minus 4 will give you 0 and then again for the next 1 minus 1 minus 4 will give you minus 5 and for the next 1 2 and 1 for this you will get it - - and for the last one 5-12 which will give you minus 7 so in the first step we have got the first element of the third man third row is zero now step two first element of the second row has to be made 0 so R 2 is equals to R 2 minus 2 R 1 so from this what do you get first row will be written as it is 1 1 1 3 and the next row will be 2 minus 2 will give you 0 and for the next one 1 minus 2 which will give you minus 1 and for the next one minus 1 minus 2 will give you minus 3 and for the last one 2 minus 6 will give you minus 4 and the third row will be written as it is no more no changes should be made for the first and the third only for the second row we have made the changes so 0 minus 5 minus 2 minus 7 now you can see that we we have made the first second and the third row as zero students will always make a mistake that after finding out the third row immediately they move on to the second element of the third of no you will not get the answer when you go for the same row step by step you have to do it first the third row to be made 0 on the second row to be made 0 then when he again you will move on to the third row so the step 3 will be second element of the third row what is the second element minus 5 has to be made 0 so how will you make it by using the second row not by using the first row you by using the second row you will make it as 0 so R 3 is equals to R 3 minus 5 R 2 this will be 1 1 1 3 0 minus 1 minus 3 minus 4 and the last row will be 0 you can see 5 minus 5 will give you 0 and again when you simplify this you will get it as 10 -15 will give you 13 and then here also you will get it as 13 now how we got the below the diagonal elements are 0 you can see this by upper triangular matrix 0 0 0 now we have to write the rank of the argumented matrix rank of argumented matrix that is a comma B what is the rank of argumented matrix whatever we have done all these steps that is the rank of a argumented matrix now from this you can see how many nonzero rows are here 1 2 3 3 nonzero rows therefore we can say that the rank is 3 here what is it since the nonzero ROS are three therefore the rank of the argumented matrix a comma B is equals to three now we need to find the rank of a how will you find the rank of a by deleting the last column of the argumented matrix what is it that is first I will write the matrix 1 1 1 0 minus 1 minus 3 0 0 13 this is the transformed matrix from this we need to write the rank of a therefore here how many nonzero rows are here again 3 you can see 3 is the rank here therefore R of a is equals to 3 so from this we can say that rank of a comma B which is equal to rank of a therefore the solution is consistent when will you say that the solution is inconsistent when this is not equal when the rank of argumented matrix is not equal to rank of a then we say that the solution is inconsistent here this is both are equal so we can say that the solution is consistent so what is the rank of the argumented matrix you have got it here nonzero rows has 3 and here also the rank is 3 so if both are equal then we say that the solution is in if it is not inconsistent you will leave it as it is but here it is consistent in that case you need to find the solution how do you find the solution for the system of linear equation from the transformed matrix what is the transform the matrix here this is the transformed matrix from the last step whatever we have got it as the matrix that is the reduced matrix which is nothing but the transformed matrix from this we can say that from last row from last row we can write 13 Z is equals to 13 which is nothing but say from the last row we can take this as 13 0 is equals to 13 so that is equals to 13 by 13 which will give you 1 so therefore the Z value is 1 here and similarly what is the next one from the second row from the second row from the second row we can say that minus y minus 3 Z which is equals to minus 4 now substituting the value of substituting the value of z in the above equation we can get the value of y so what do you get minus y -3 in to substitute the value of Z that is 1 which is equals to minus 4 so minus y -3 when you bring it to the other side you will get it as minus 4 plus 3 which is nothing but minus 1 minus an minus gets canceled therefore Y is equals to 1 now we have got the value for y and z now finally you need to get the value for X from the first row so how will you get it from the first row we can write it as X plus y plus Z which is equals to 3 now substitute the value of y unzip so from this we will get it as expressed what is the Y value 1 here and what is the z value 1 so 1 plus 1 which is equals to 3 so X is equals to 3 minus 2 which will give you 1 therefore the solution is the solution is X is equals to 1 y is equals to 1 and Z is equals to 1 so when will you get these solutions only when the system of equations is consistent when it is not consistent that is when it is inconsistent then you will not get even a single solution

Channel: Btechguru BodhBridge ESPL

Views: 28,303

Rating: 4.8947368 out of 5

Keywords: engineering, mathematics, maths tricks, maths formulas, math questions, Nptel, GRE, GMAT, IBPS, BANK, BANK PO, Gate, Matrix, matrices, learning matrix, application of matrices, algebra, solution of matrix, matrix solve, matrix equation, solve matrix equation, how to solve matrices, system o flinear equations, math, maths, jntu

Id: 2UedS9YSm1Q

Length: 13min 38sec (818 seconds)

Published: Mon Oct 03 2016