899 9.2 Matrix Solution of Linear Systems y - 2112 = -2 Substitute 1 for z in the second equation. y = 0 Add 2. x - 2102 + 3112 = 0 Substitute 0 for y and 1 for z in the first equation. x = -3 Subtract 3. The solution set is 51-3, 0, 126. Check the solution in the original system. S Now Try Exercise 55. The ref (row-echelon form) command from the Matrix menu performs the Gaussian elimination method on the TI-84 Plus graphing calculator. Unlike the reduced row-echelon form, there can be more than one row-echelon form corresponding to a matrix. Figure 8 shows the matrices that result when the ref command is applied to the linear system from Example 5. Although the matrix in Figure 8(c) is different from the matrix found in Example 5, back-substitution results in the same solution set. (b) (c) (a) Figure 8 7 The Gauss-Jordan method may be preferred for solving systems of linear equations with few variables, and it is essential for solving linear systems using matrix inverses, as discussed later in the text. The Gaussian elimination method may be preferred in the following situations. 1. When we want only to determine the number of solutions (that is, one, none, or infinitely many solutions), Gaussian elimination is quicker. 2. When the system is very large (say, thousands of variables) and will be solved using a computer, Gaussian elimination executes significantly faster than Gauss-Jordan elimination. 3. When numerical instability is a concern, Gaussian elimination is less likely to produce inexact results. For example, if the leading coefficient of one of the rows is very close to zero, dividing by that number and proceeding with decimal computations results in a row that should be used as little as possible during the elimination process. 9.2 Exercises CONCEPT PREVIEW Answer each question. 1. How many rows and how many columns does this matrix have? What is its dimension? c -2 5 8 0 1 13 -6 9d 2. What is the element in the second row, first column of the matrix in Exercise 1?
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