Chapter 1 Section 1.1 Introduction to Matrices and systems of Linear Equations

Larger Systems of Equations A system of equations that has 3 or more variables can be solved by combining both the elimination method and the substitution method. 1. You eliminate variables from the other equations until you get down to one equation with one variable. 2. Solve that one equation for the variable and substitute it back into the other equations and solve for the one variable that remains. Example Solve the System of Equations: ¼ E E 1 +E 2 -3 E 1 +E 3 ⅓ E 3 - E 2. Substitute y and z into Equation 1. The solution is: x = 3, y = 2, z = -3

Subscripted Variables In systems of equations where more than 3 variables (sometime for 2 and 3 variables) are needed instead of using regular variable we use just one x but put subscripts on it x i "read x sub i ". Matrices with 1's and 0's A matrix that has 1's down the diagonal from top left to bottom right is easy to read the simultaneous solution of system of equations right from the matrix. They are the entries in the constants column. Matrix Equations Simultaneous Solution

Row Operations We will use specific row operations to change a matrix from the original equation form to the type with 1's down the diagonal and 0's everywhere else. The row operation we use keep the simultaneous solution ROW OPERATIONS For a matrix A there are 4 row operations that are allowed. The way you refer to each row operation below the second way is how the TI-83 graphing calculator refers to them. R i  R j RowSwap([A], i, j )Interchange the i th and j th rows. R i +R j Row+([A], i, j )Add the i th row to the j th row. m R i *Row( m,[A], i )Multiply the i th row by the number m. m R i +R j *Row+( m,[A], i, j )Multiply the i th row by the number m and add it to the j th row.

RowSwap Examples of row operations. R 1  R 3 RowSwap([A],1,3) Row+ R 1 +R 3 Row+([A],1,3) *Row ¼R 1 *Row(¼,[A],1) *Row+ -2R 1 +R 3 *Row+(-2,[A],1,3)

Pivoting The process of using elementary row operations to "clear out" a column and get a 1 in the diagonal position in the column and then zeros in all others is called pivoting. Only the row operations are allowed when doing this process. The diagonal entries are the pivot positions. ½R 1 *Row(½,[A],1) -3R 1 +R 2 *Row+(-3,[Ans],1,2) The first column is "cleared out"! -⅓R 2 *Row(-⅓,[Ans],2) -4R 2 +R 1 *Row+(-4,[Ans],2,1) The second column is "cleared out"! Pivot Position

Gauss-Jordan Elimination Gauss-Jordan Elimination is a process of pivoting column by column until you have all but the last column "cleared". R 1  R 2 RowSwap([A],1,2) ¼R 1 *Row(¼,[Ans],1) -½R 2 *Row(-½,[Ans],2) -½R 2 +R 1 *Row+(-½,[Ans],2,1) R 1  R 2 RowSwap([A],1,2) ½R 1 *Row(½,[Ans],1) -3R 2 +R 1 *Row+(-3,[Ans],2,1) -2R 2 +R 3 *Row+(-2,[Ans],2,3) ½R 3 *Row(½,[Ans],3) 13R 3 +R 1 *Row+(13,[Ans],3,1) -5R 3 +R 2 *Row+(-5,[Ans],3,2)

Putting it Together Lets use matrix simplification (i.e. Gauss- Jordan Elimination) to solve the system of equations below. We form the matrix. Call this A and simplify it ⅓R 1 *Row(⅓,[A],1) -2R 1 +R 2 *Row+(-2,[Ans],1,2) 3R 2 *Row(3,[Ans],2) - 7 / 3 R 2 +R 1 *Row+(- 7 / 3,[Ans],2,1) Converting back to equations gives: Check: