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Mathematical Investigations II Name Mathematical Investigations II Linear Thinking Gaussian Elimination or Row Reduction Part I. Read the example CAREFULLY Consider how we might solve the following system of equations: x y z 5 x 3y 2z 2 2x y z 1 Add equation 1 (or row 1, r1) to equation 2 (or row 2, r2) and copy over the other equations: [We will denote this by r1+ r2 r2.] r1 r1 r1 r2 r2 r3 r3 x y z 5 4y 3z 7 2x y z 1 Next, multiply equation 1 (r1) by -2 and add it to equation 3 (r3), copying over the other equations: r1 r1 r2 r2 2r1 r3 r3 x y z 5 4y 3z 7 y z 9 Next, swap equation 2 and equation 3: r1 r1 r3 r2 r2 r3 x y z 5 y z 9 4y 3z 7 Finally, multiply (equation 2) by (–1): r1 r1 r2 r2 r3 r3 x y z 5 y z 9 4y 3z 7 LinearThinking 10.1 Rev. S06 Mathematical Investigations II Name We have used equation 1 (r1), in conjuction with equations 2 and 3 to obtain two new equations that contain only 2 variables (y and z). Let us now use equation 2 in a similar fashion; that is, to remove the y variable from equations 1 and 3. Replace equation 1 with –(equation 2) + (equation 1) and replace (equation 3) with -4 • (equation 2) + (equation 3) to obtain: r2 r1 r1 r2 r2 4r2 r3 r3 x 4 y z 9 z 29 Finally replace equation 2 with – (equation 3) + (equation 2) to obtain: 4 x y 20 z 29 Which is the solution to the system? Describe the general process! LinearThinking 10.2 Rev. S06 Mathematical Investigations II Name Part II. If you stop to think for a minute, the essential information used to solve the system in Activity I is contained in the following augmented matrix. 1 1 1 5 1 3 2 2 2 1 1 1 This matrix is obtained from augmenting the matrix of the coefficients with the matrix of the constant terms. Repeat the operations from the system of equations above to transform that matrix to the one containing the solution. Remember that in this situation, you must use a ‘0’ wherever the coefficient of a variable is 0. 1 1 1 5 1 3 2 2 2 1 1 1 1 0 0 4 0 1 0 20 0 0 1 29 This process is called "row reduction" or "Gaussian Elimination". How can you tell whether or not row reduction is finished? LinearThinking 10.3 Rev. S06 Mathematical Investigations II Name Part III. Solve the following systems using Gaussian Elimination on the appropriate augmented matrices. Use the notation described on the previous pages to indicate how you solved each system (e.g., -r1 + r2 r2). a) x y z 1 x 2z 3 x y 3z 1 b) x 3y 2z 8 4x 11y 6z 34 3x 2y 5z 13 c) x y 2z w 6 x y w 12 x 3y z 6 24 2x y z LinearThinking 10.4 Rev. S06 Mathematical Investigations II Name d) 2x y z 3 4x 3y 2z 1 6x 4y z 2 f) What happens when you try to use Gaussian Elimination on the last two sets of systems? e) 2x y z 3 4x 3y 2z 1 6x 5y 5z 1 Use matrix algebra to solve these systems. What happens? 2 0 5 8 In problem d), you should have ended up with something like 0 1 4 5; 0 0 0 5 2 0 5 8 for e) your matrix should look something like 0 1 4 5. 0 0 0 0 The last equation for d) states that 0x + 0y + 0z = -5, which is impossible. There is no solution to the system of equations. The system is called inconsistent. The last equation for e) states that 0x + 0y + 0z = 0, which is true for all (x, y, z). The system has been reduced from 3 equations in 3 variables to 2 equations in 3 variables. For this system, there are infinitely many solutions. The system is called dependent. LinearThinking 10.5 Rev. S06 Mathematical Investigations II Name Inconsistent System of Equations A system of equations that has no solutions. A system of equations that has at least one solution is called consistent. Dependent System of Equations A system of equations that has infinitely many solutions. A system of equations that has at most one solution is called independent. Thus for any system of linear equations there are three possibilities for the answer: or inconsistent (no solution) = independent usually just called inconsistent. consistent (1 unique solution) = independent usually just called independent consistent (infinitely many solutions) = dependent usually just called dependent Part IV. There is another viable use for Gaussian Elimination. Recall that the identity matrix for the n x n square matrices is the matrix with ones in the main diagonal and zeroes everywhere else. For example the 3 x 3 identity matrix is 1 0 0 0 1 0 0 0 1 The matrix B is called the inverse of A if AB = BA = I (the identity matrix) and it is denoted by A-1. Let us explore how we can use Gaussian Elimination to derive the inverse of a 3x3 (or any square) matrix. LinearThinking 10.6 Rev. S06 Mathematical Investigations II Name Consider the matrix of the coefficients from problem (b) above: 1 3 2 Let A = 4 11 6 3 2 5 Now we want to augment matrix A with the identity matrix to get: 1 3 2 1 0 0 [A|I] = 4 11 6 0 1 0 3 2 5 0 0 1 Now row reduce the left side (matrix A) of the above 3x6 matrix: 1 3 2 1 0 0 4 11 6 0 1 0 3 2 5 0 0 1 Locate A–1. (You can check this with your calculator.) LinearThinking 10.7 Rev. S06 Mathematical Investigations II Name Use this same process to find the inverse of the following matrices: b. 1 1 2 1 1 2 1 1 0 1 1 1 0 3 1 0 d. 2 1 1 4 3 2 6 5 5 a. 1 1 1 1 0 2 . 1 1 3 c. 2 1 1 4 3 2 6 4 1 e. What happens in the last two cases? How does this relate to your work above? LinearThinking 10.8 Rev. S06