MATH 2000 - Gauss-Jordan Elimination and the TI-83 ~ page 1J. Ahrens 12/21/2005xy z−−−00 3 1224 2 815 1 11↔13RR−−−00 3 1224 2 815 1 11MATH 2000Gauss-Jordan Elimination and the TI-83[Underlined bold terms are defined in the glossary]A linear system such as =+−=+−=−3z 122x 4y 2z 8x5y z 11can be solved algebraically using ordinary elimination or by using an augmented matrix and elementaryrow operations. When we write a matrix, we are using only the coefficients, so it is imperative that all equations be instandard form. An augmented matrix includes the constant terms also. The augmented matrix for thissystem is given below. In larger matrices where variables may get confused, we will put the appropriatevariable at the top of each column as I have done here. In small systems this is not usually necessary.The vertical line indicates the location of the = signs.There are three elementary row operations that produce row-equivalent matrices• Two rows are interchangedijRR↔• A row is replaced by a nonzero multiple of itselfiikR R→• A row is replaced by the sum of itself and a nonzeroji ikR R R+→multiple of another row.(NOTE : means "replaces ")→You are responsible for performing all of these operations by hand, if asked, but you will be allowed to useyour calculator most of the time. Make certain you know how to do all row operations on your calculator,or you will have to do all the computations by hand! The result of a row operation is displayed on the homescreen, but it is not automatically stored (on a TI-83)! You should immediately store the result under adifferent name. It is convenient (and frequently useful) to store the results alphabetically. • Row Swap: To interchange rows 1 and 3 of matrix A: MATH C:rowSwap( NAMES 1:[A] MATRIXENTER MATRIX ENTER , 1 , NAMES 2:[B] 3)ENTER STO$ MATRIX ENTER• You must write down which row operation you are using. Please put this notation beside theresulting row in the new matrix. Note the calculator command on each screen shot.The original matrix The matrix after swapping rows 1 and 3MATH 2000 - Gauss-Jordan Elimination and the TI-83 ~ page 2J. Ahrens 12/21/2005−−−00 3 1224 2 815 1 11−+→23 31RR R2−−−−48 4 4365 132 1 12 1700 3 1224 2 815 1 11−−−• Multiplying a row by a nonzero scalar: This operation automatically replaces the old row with the new one. To multiply row 1 of matrix A by : 13 MATH E:*row( NAMES 1:[A] MATRIXENTER13, MATRIX ENTER , 1 ) NAMES 3:[C] ENTERSTO$ MATRIX ENTERThe original matrix The matrix after multiplying row 1 by 13• Adding a nonzero scalar multiple of one row to another row: Perform the multiplicationfirst, then add that result to the second equation. Replace the second equation with the result. To multiply row 2 of matrix A by and add it to row 3 12− MATH F:*row+( NAMES 1:[A] MATRIXENTER12−, MATRIX ENTER , 2 NAMES 4:[D] ,3 ) ENTER STO$ MATRIX ENTERThe original matrix The matrix after multiplying row 2 by and adding it to row 312−General directions for using Gauss-Jordan elimination:a) Write the system of equations as an augmented matrix b) Begin with the original matrix and use elementary row operations until the coefficient matrix becomes an identity matrix. c) List each row operation used to the left of the new matrix beside the new row. d) Store each result in your calculator (as a new matrix).e) Please work DOWN!Example: Solve this system of linear equations: +−=++=−−++ =−4x 8y 4z 43x 6y 5z 132x y 12x 171. Write the system as an augmented matrix and store it in your calculator as matrix [A].store as [A]→111RR3MATH 2000 - Gauss-Jordan Elimination and the TI-83 ~ page 3J. Ahrens 12/21/2005−−−−12 1 1365 132 1 12 171*row ,[E],25−−−12 1 10 5 10 1500 8 16↔23RR−−−12 1 101 2 300 8 16−−−12 1 101 2 300 1 2−−−−12 1 1008 162 1 12 17−+→12 23R R R→331RR8−−−12 1 100 8 160 5 10 15→221RR5+→13 32R R R→111RR42. Desired result: Change element to 1.11aRow op: store as [B]1*row ,[A],143. Desired result: Change element 21ato 0 by using row 1. store as [C]Row op: +−*row ( 3 ,[B],1,2) 4. Desired result: Change element 31ato 0 by using row 1 store as [D]Row op: +*row (2 ,[C],1,3)5. Desired result: Change to 122aThis is not a typical problem! We would normally multiply row 2 by the reciprocal of . 22aSince we must have a nonzero number in that location we will need to do a row swap. We do not want to swap rows 1 and 2 because we would then have a nonzero number for element . Cardinal rule: Do not undo something you just worked hard to fix!21aSwapping rows 2 and 3 is not a big deal, but it is unusual to have to do so at this point.Row op: rowSwap([D], 2, 3]store as [E]Row op: store as [F]Let’s go off on a tangent for a few moments:A. If we multiply row 3 by , we will change element to 1.1833aRow op: 1*row ,[F], 38B. Our matrix is now in row reduced echelon form. We can solve using back substitution:The third row means that 1z = !2, i.e. z = !2Back substitute into row 2: 1y + 2(!2) = !3, so y = 1Back substitute into row 1: 1x + 2(1) ! (!2) = 1, so x = !3The solution is (!3, 1, !2) Check by substituting result in all original equations..This method of solution is called Gaussian Elimination.MATH 2000 - Gauss-Jordan Elimination and the TI-83 ~ page 4J. Ahrens 12/21/2005−+→21 12R R R−−100 3010 1001 2+→32 22R R R−−−10 5701 2 300 1 2−−−100 3012 3001 2+→31 15R R R→331RR8−−−10 5 701 2 300 8 16Meanwhile, back to the original problem! 6. Desired result: Change element 12ato 0 using row 2 save as [G]Row op:+−*row ( 2,[F],2,1)7. Desired result: Change element to 133aRow op: save as [H]1*row ,[G],388. Desired result: Change element 13ato 0 using row 3 save as [I]Row op: +*row (5,[H],3,1)9. Desired result: