x y z 0 0 3 4 5 MATH 000 Gauss-Jordan Elimination and the TI-3 [Underlined bold terms are defined in the glossary] 3z = A linear system such as x + 4y z = x + 5y z = can be solved algebraically using ordinary elimination or by using an augmented matrix and elementary row operations. When we write a matrix, we are using only the coefficients, so it is imperative that all equations be in standard form. An augmented matrix includes the constant terms also. The augmented matrix for this system is given below. In larger matrices where variables may get confused, we will put the appropriate variable 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 interchanged Ri Rj A row is replaced by a nonzero multiple of itself kri Ri A row is replaced by the sum of itself and a nonzero kr j + Ri Ri 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 use your 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 home screen, but it is not automatically stored (on a TI-3)! You should immediately store the result under a different name. It is convenient (and frequently useful) to store the results alphabetically. Row Swap: To interchange rows and 3 of matrix A: MATRIX MATH C:rowSwap( ENTER MATRIX NAMES :[A] ENTER,, 3 ) ENTER STO$ MATRIX NAMES :[B] ENTER You must write down which row operation you are using. Please put this notation beside the resulting row in the new matrix. Note the calculator command on each screen shot. The original matrix The matrix after swapping rows and 3 0 0 3 4 5 R R 3 MATH 000 - Gauss-Jordan Elimination and the TI-3 ~ page J. Ahrens //005
Multiplying a row by a nonzero scalar: This operation automatically replaces the old row with the new one. To multiply row of matrix A by : 3 MATRIX MATH E:*row( ENTER 3, MATRIX NAMES :[A] ENTER, ) ENTER STO$ MATRIX NAMES 3:[C] ENTER The original matrix The matrix after multiplying row by 3 0 0 3 4 5 R R 3 Adding a nonzero scalar multiple of one row to another row: Perform the multiplication first, then add that result to the second equation. Replace the second equation with the result. To multiply row of matrix A by and add it to row 3 MATRIX MATH F:*row+( ENTER, MATRIX NAMES :[A] ENTER,, 3 ) ENTER STO$ MATRIX NAMES 4:[D] ENTER The original matrix The matrix after multiplying row by and adding it to row 3 0 0 3 4 5 R + R R 3 3 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! 4x + y 4z = 4 Example: Solve this system of linear equations: 3x + 6y + 5z = 3 x + y + x = 7. Write the system as an augmented matrix and store it in your calculator as matrix [A]. 4 4 4 3 6 5 3 7 store as [A] MATH 000 - Gauss-Jordan Elimination and the TI-3 ~ page J. Ahrens //005
. Desired result: Change element a to. Row op: *row,[a], 4 R R 4 3 6 5 3 7 store as [B] 3. Desired result: Change element a to 0 by using row. Row op: *row + ( 3,[B],,) 3R + R R 0 0 6 7 store as [C] 4. Desired result: Change element a3 to 0 by using row Row op: *row + (,[C],,3) R + R3 R3 0 0 6 0 5 0 5 store as [D] 5. Desired result: Change a to This is not a typical problem! We would normally multiply row by the reciprocal of a. Since we must have a nonzero number in that location we will need to do a row swap. We do not want to swap rows and because we would then have a nonzero number for element a. Cardinal rule: Do not undo something you just worked hard to fix! Swapping rows and 3 is not a big deal, but it is unusual to have to do so at this point. Row op: rowswap([d],, 3] R R 3 0 5 0 5 0 0 6 store as [E] Row op: *row,[e], 5 R R 5 0 3 0 0 6 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 a 33 to. Row op: *row,[f],3 R 3 R 3 0 3 0 0 B. Our matrix is now in row echelon form. We can solve using back substitution:. The third row means that z =!, i.e. z =! Back substitute into row : y + (!) =!3, so y = Back substitute into row : x + ()! (!) =, so x =!3 The solution is (!3,,!) Check by substituting result in all original equations. This method of solution is called Gaussian Elimination. MATH 000 - Gauss-Jordan Elimination and the TI-3 ~ page 3 J. Ahrens //005
Meanwhile, back to the original problem! 6. Desired result: Change element a to 0 using row Row op: *row + (,[F],,) R + R R 0 5 7 0 3 0 0 6 save as [G] 7. Desired result: Change element a 33 to Row op: *row,[g],3 R 3 R 3 0 5 7 0 3 0 0 save as [H]. Desired result: Change element a3 5R3 + R R 0 0 3 to 0 using row 3 0 3 save as [I] Row op: *row + (5,[H],3,) 0 0 9. Desired result: Change element a3 to 0 using row 3 Row op: *row + (,[I],3,) R3 + R R 0 0 3 0 0 0 0 The solution can now be read directly from the matrix: x =!3 y = z =!, in other words, (!3,,!) This method is called the Gauss-Jordan Elimination Method. Its algorithm is ideal for use with calculator and/or computer programs and is the basis for linear programming problems. Summary of the Gauss-Jordan Elimination Method: Get a " in position,. Then use row to get 0"s in the rest of column. Get a " in position,. Then use row to get 0"s in the rest of column. Get a " in position 3,3. Then use row 3 to get 0"s in the rest of column 3. You will be required to solve one problem on the test by showing the individual row operations as demonstrated above. You may use your calculator to do the actual calculations on each step. Study your owners manual to see if your calculator has a row reduced echelon form command. If so, it will save you a lot of time and effort! If not, practice until you can do the necessary steps quickly and accurately. On the TI-3, matrix A can be changed from augmented form to row reduced echelon form in one step using: MATRIX MATH B:rref( ENTER MATRIX NAMES :[A] ENTER! MATH 000 - Gauss-Jordan Elimination and the TI-3 ~ page 4 J. Ahrens //005
Not all linear systems have solutions The augmented matrix is Solve: 6 3 3 x + 6x = 3 x + 3x = 3 The row reduced echelon form is 0 0 7 The last row of the row reduced echelon form means that 0x + 0x = 7, i.e. 0 = 7. This system has no solution. Such a system is said to be inconsistent. Some linear systems have multiple solutions The augmented matrix is 4 6 3 Solve x x = 4 6x + 3x = The row reduced echelon form is 0 0 0 The last row of the row reduced echelon form means that 0x + 0x = 0, which is true regardless of the values of the variables. Since the bottom row is always true, we must determine when the first row is true. Introduce a parameter t, and let x = t. [t is a real number] Then x x = x = t + All solutions to this system have the form t, t t + R Examples of solutions are (, 0), (!,!6), and (4, 4). This is a consistent system. MATH 000 - Gauss-Jordan Elimination and the TI-3 ~ page 5 J. Ahrens //005