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6.4.2 Solutions of Linear Systems Using Matrices - Part II In 6.4.1, we wrote the system: 2x + 4y + 4z = 4 x + 3y + z = 4 -x + 3y + 2z = -1 as the augmented matrix: 2 4 4 4 1 3 1 4 1 3 2 1 and then demonstrated (forward) Gaussian elimination, transforming the augmented matrix into row-echelon form: 1 2 2 2 0 1 1 2 0 0 1 1 We then rewrote the system as: x + 2y + 2z = 2 y - z = 2 z = -1 and used back-substitution to finish the solution. Instead of back-substitution, we can complete the process on the system in matrix form using backward elimination. 6.4.2-1 Backward elimination start from the bottom, and work on one row at a time to the top (that’s why it’s called backward elimination) turning the entries above the leading 1's to 0's (row op 3) The resulting matrix will be: 1 0 0 2 0 1 0 1 0 0 1 1 This matrix is in reduced row-echelon form (RREF): every non-zero row has a leading 1 below and above the leading 1’s we find only 0’s the leading 1's "march" from left to right rows of all zeros are at the bottom (not shown here) Rewriting the matrix as a system, the solution is apparent: x y z = 2 = 1 = -1 I will illustrate backward elimination with our example system at the board. 6.4.2-2 What else can “go wrong”? A matrix in RREF form may have a row of 0’s: 1 0 5 5 0 1 2 3 0 0 0 0 this indicates a system with many solutions a dependent system Q: How do we state "the solution" when there are infinitely many solutions? A: We do so parametrically. The above reduced form corresponds to the system: x y + 5z + 2z =5 =3 Here’s how to state the solution parametrically: 1. Solve each equation for its first variable: x = 5 - 5z y = 3 - 2z The variables x and y do not appear on the right. z can be chosen arbitrarily to be any real number (z is referred to as a parameter), and will determine the corresponding numerical values for x and y. 2. State your (infinitely many) solutions parametrically: (5 - 5z, 3 - 2z, z) z = 0 (5, 3, 0) is a solution z = 1 (0, 1, 1) is a solution, etc. for many solutions 6.4.2-3