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Section 8.1/8.2 Matrix Solutions to Linear Systems Inconsistent and Dependent Systems

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Overview When solving systems of linear equations in two variables, we utilized the following techniques: 1.Substitution 2.Elimination 3.Graphing In this section we will develop techniques that can be used for larger systems.

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Three-variable systems A system of linear equations in three variables is in the form The solution to a three-variable system is an ordered triple (x,y,z).

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Solving a three-variable system We will utilize matrix techniques. One method involves using row operations, and is done by hand. The other method involves using your graphing calculator.

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The augmented matrix If is a system of linear equations in three variables, then is the augmented matrix for the system.

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Row Operations When you have a matrix, there are operations you can perform on the rows in that matrix: 1.You can swap rows:

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Row Operations 2.You can multiply (or divide) a row by a constant (and put the result in the place of the original):

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Row Operations 3.You can add two rows together and put the result in the place of one of the rows you added (the other row is unchanged):

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Row Operations 4.You can do combinations:

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The Goal Of Row Operations Use row operations to convert your augmented matrix to row-echelon form: Let’s see why this form is useful.

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Solve The Following System:

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Getting There Is Half The Fun How to convert an Augmented Matrix into row-echelon form: 1.Start working with the first column. If there is a “1” at the top of that column, go to Step 2. If there is not a “1” at the top of that column, swap that row with a row that has a “1” in the first position.

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2.The next objective is get a “0” in the remaining positions of the first column. Multiply Row 1 by the opposite of the coefficient of the first entry in the second row, then add to Row 2 and put the result in Row 2. 3.Repeat the process with Row 3, again using Row 1.

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4.Now move on to the second column. You want to get a “1” in the second position of the second column. If there is already a “1” there, go on to Step 5. If not, divide Row 2 by the entry in the second position of the second column and put the result back into Row 2. 5.Now, you want to get a zero in the last position of the second column. Multiply Row 2 by the opposite of the entry in the second position of Row 3, then add to Row 3 and put the result in Row 3.

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6.Finally, divide Row 3 by the entry in the third position of the third row (it is also the third position of the third column) and put the result in Row 3.

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Example 1

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Example 2

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Using The Graphing Calculator* 1.Press 2 nd MATRIX to enter matrix mode. 2.Arrow over to EDIT, then choose the matrix you want to use, then press ENTER. 3.Input the number of columns, press ENTER, then the number of rows, press ENTER. The cursor should move to the first entry in the matrix. *these directions are for a TI-83 and TI-84 only.

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4.Input the entries row by row, from left to right, pressing ENTER between entries. 5.When you finish, press 2 nd QUIT to exit matrix mode. 6.Now, go back into matrix mode by pressing 2 nd MATRIX. 7.Arrow over to MATH, then down to “rref(“. Press ENTER.

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8.Press 2 nd MATRIX, arrow down to the matrix you used, then press ENTER. 9.The resulting matrix will be in reduced row echelon form.

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Inconsistent and Dependent Systems If the bottom row of your rref matrix has all 0’s, then the system is dependent (has an infinite number of solutions). Write both x and y in terms of z. If the bottom row of your rref matrix has all 0’s and a 1, then the system is inconsistent (has no solution).

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Example 3 (Dependent)

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Example 4 (Inconsistent)

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