MAT 122 Lecture 14: Systems of Linear Equations and Matrices
Document Summary
Mat 122- lecture 14- systems of linear equations and matrices. Echelon method: triangular form, when each equation starts further to the right than the one that proceeds it, desirable, when equations have a coef cient of 1 (this will be easy to solve using back substitution) 2x + 11y + 8z= 19: start by getting the rst row to have a coef cient of 1 for the x" variable. = x + 4y + 5z= 2 new rst row. Row2 + 3*row1 (-3x - 3y - 12z= 66) = 0 + 9y + 3z= 72 new second row. Row3 - 2*row1 (2x + 11y + 8z= 19) = 0 + 3y - 2z= 15 new third row. The rst row is not permanent, it will not change anymore: next try to get a 1 coef cient for the y" variable in the second equation. = 0 + y + 1/3z = 8.
