 how to find the eigenvalue and the corresponding eigenvector by RALJ power method learning outcomes at the end of this session the student will be able to find the largest eigenvalue and the corresponding eigenvector using power method pause the video and answer the question using the power method find the dominant eigenvalue and the corresponding eigenvector of the matrix the five four one two perform one iteration i hope all of you answer the result solution let a is equal to the given square matrix that is five four one two and let the we consider the initial eigenvector x naught as an the column matrix one zero that is then the by the power method now take the multiplication between the square matrix a and the initial eigenvector that is x naught a into x naught is equal to matrix five four one two into column matrix one zero now taking the matrix multiplication that is first to the column matrix that is five now multiply the second row to the column matrix that is one into one one plus two into zero zero that is one plus zero is one now in this column matrix take the largest element as a common factor and simplify that becomes five into the matrix the one zero point two which is equal to lambda one into x one now the first iteration of eigenvalue is lambda is equal to five and the corresponding eigenvector is x is equal to the matrix one zero point two now come to an example find the largest eigenvalue and the corresponding eigenvector of the matrix five one two first row one three zero second row two zero minus four using the power method solution let a is equal to a square matrix that is twenty five one two first row one three zero second row two zero minus four is the third row and let now consider the initial eigenvector x naught is equal to a column matrix one zero zero now by the power method you take the multiplication of the square matrix and the initial eigenvector that is a into x naught is equal to the matrix twenty five one two first row one three zero second row two zero minus four the third row into x naught that is a column matrix one zero zero now take the multiplication between the square matrix and the eigenvector that is first you have to multiply the first row to the column matrix that is twenty five into one twenty five plus one into zero zero plus two into zero zero that is twenty five plus zero plus zero that is twenty five now multiply the second row to the column matrix that is one into one one plus three into zero zero plus zero into zero zero that is one plus zero plus zero which becomes one now multiply the third row to the column matrix that is two into one two plus zero into zero zero minus four into zero that is zero that please become that two here now in the column matrix now take the largest element as a common factor and simplifying we will get the twenty five into the matrix one zero point zero four zero point zero eight which is equal to lambda one into x one which is the first approximation where the lambda one is equal to twenty five and x one is equal to matrix one zero point zero four zero point zero eight now come to the second iteration that is a into x one which is equal to matrix twenty five one one one three zero two zero minus four into the x one that is the column matrix one zero point zero four zero point zero eight now take the now the second iteration is a into x one is equal to the matrix twenty five one two one three zero two zero minus four into the x one that is the column matrix one zero point zero four zero point zero eight now take the multiplication that is a first two to the column matrix that twenty five into one plus one into zero point zero four plus two into zero point zero eight that becomes the twenty five point two and multiply the second row to the column matrix that is one into one plus three into zero నినిల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల్ల సిటోఔ్వండివదiqué మిరోటునTony Nord, led película sign చంపంఴ్టణ� brown పికాసురా మిిఖండ్రటోం. న్ఘంపి దింపీ passat�రా. మేన్భ రింరరా ముకాతీమ్ప్చి చురోటిందా నిమైల్గేటితీరిందిల్చిదుదా ఇ వపారాన్స్మా మునుధానుమా తీడినింది పెటిళినినేటిానేటిక� now taking the multiplication between these two matrix and we get the matrix the 25.183 and 1.135, 1.724 చాల్లీనినికిందాకింద్ల౔యరినిక్భందిమాత్రి. నాలాల్దాక్ని గానుల్ల్లందింద్విల్లులేలుని causaలునిలిందిందిందాలు. పెటాలికోగధరి� లృవింింింింింింి. క్నివింాదింసిక్ంి఍కి. డ్రింత్స్రుర్రంలుచి. అ౻త్నురూత్ధింిలి. ప్రూతింిన్సింికితేన్. కిచిఆలికిధోత్ని 3 5 1.724. In this matrix, take the largest element as a common factor and we will and simplify we get the 25.181 into matrix 1 0.045 0.068 which is equal to lambda 5 into x 5. Now we observe that the x 4 is equal to x 5 hence we conclude that the largest eigen value of a is lambda is equal to 25.181 and the corresponding eigen vector x is equal to matrix 1 0.045 0.068 transpose