Example:

1. Linear systems were solved using substitution and elimination in the two previous section. This section describes how to solve these systems by a matrix approach. Using matrices are particularly suitable for large systems having many unknowns. A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns and enclosed by brackets. Each number in the array is an element of the matrix. Rows Example: Columns This matrix has 2 rows and three columns. It is called a 2 x 3 matrix (“2 by 3”). A matrix with m rows and n columns is an m x n matrix. The number of rows is always given first. Examples of matrices: 3 x 2 matrix 3 x 3 matrix 1 x 3 matrix Next Slide

2. With every system of linear equations, we can associate a matrix that consists of the coefficients and constant terms. This is called the augmented matrix of the system. The goal is to transform the augmented matrix into one in which the variables will be easy to see. Since each column in the matrix represents the coefficients of one variable, it should be transformed so that it is of the form Examples: The rows of the augmented matrix can be treated just like the equations of a system of linear equations. Since the augmented matrix is nothing more than a short form of the system, any transformation of the matrix that results in an equivalent system can be performed. Elementary row operations. • Any two rows may be interchanged. • Any row of the matrix can be multiplied by a nonzero real number. • Any row of the matrix can be replaced by the sum of a nonzero multiple of another row plus that row. for real numbers a and b. Once the augmented matrix is in this form, the matrix can be rewritten as a linear system to get x = a and y = b (two equations with two unknowns). Next Slide

3. If the system is three equations with three unknowns, the augmented matrix should be transformed so that it is of the form to get x = a and y = b and z = c. These matrices are called reduced echelon form. The dashed vertical line is optional. Steps to obtaining reduced echelon form of a 2 x 2 augmented matrix using elementary row operations. *Note: getting a 1 in the upper left hand column can be done by multiplying by the reciprocal of that entry. This often involves fractions and can get messy. Using row operations is often easier. Next Slide 1. Get a 1 in the upper left hand column. * 2. Get a 0 in the lower left hand column. 3. Get a 1 in the lower middle column. 4. Get a 0 in the upper middle column. This is the reduced echelon form of the augmented matrix.

4. Example 1. Solve the system using a matrix approach: Solution: 0 9 -18 Answer: 1 0 4 Your Turn Problem #1 Solve the system by a matrix approach: 1. Get a 1 in the upper left hand column by interchanging columns. 2. Get a 0 in the lower left hand column. (multiply row 1 by -2 and add it to row 2) 3. Get a 1 in the lower middle column. (multiply row 2 by 1/9) 4. Get a 0 in the upper middle column. (multiply row 2 by 3 and add it to row 1.) x=4 and y=-2

5. Example 2. Solve the system using a matrix approach: Solution: 1 -10 -17 Solve the system by a matrix approach: Answer: 0 52 104 1 0 3 Your Turn Problem #2 1. Get a 1 in the upper left hand column. (multiplying by 1/3 will result in fractions, easier to multiply row 1 by 2 and subtract row 2) 2. Get a 0 in the lower left hand column. (multiply row 1 by -5 and add it to row 2) 3. Get a 1 in the lower middle column. (multiply row 2 by 1/52) 4. Get a 0 in the upper middle column. (multiply row 2 by 10 and add it to row 1.) x=3 and y=2

6. Then we have the solution x=a, y=b, and z=c. Next Slide Steps to obtaining reduced echelon form of a 3 x 3 augmented matrix. 1. Get a 1 in the upper left hand column. 2. Get a 0’s in the first column beneath the 1. 3. Get a 1 in the second row, second column. 6. Get 0’s above the 1 in the third column. 4. Get 0’s above and below the 1 in the second column. 5. Get a 1 in the third row, third column.

7. Example 3. Solve the system using a matrix approach: Solution: Solve the system by a matrix approach: -2R1+R2 Answer: -5R1+R3 -2R3+R2 -1R2+R1 3R3+R2 6R2+R3 Your Turn Problem #3 1. Get a 1 in the upper left hand column. 2. Get a 0’s in the first column beneath the 1. 3. Get a 1 in the second row, second column. 4. Get 0’s above and below the 1 in the second column. 5. Get a 1 in the third row, third column. 6. Get 0’s above the 1 in the third column. Then we have the solution x=3, y=-4, and z=2, (3,-4,2). Be sure to check answer in the system.

8. The matrix approach on inconsistent systems Instead of going through another example, suppose we have the following augmented matrix after row operations. The last row is 0x +0y +0z =11. Since this is impossible, the system is inconsistent, . Again, suppose we have the following augmented matrix after row operations. The bottom row is equivalent to 0x+0y+0z =0 which is true for all x, y and z. Get rid of the 2 by multiplying row 2 by -2 and adding it to row 1. The top two rows are equivalent to x-10z=21 and y +7z=-8 which can be rewritten as x=10z+21 and y=-7z-8. Therefore, if we let z = k, where k is any real number, the solution set of infinitely many ordered triples can be represented by {(10k+21, -7k-8, k)k is a real number} If we wanted to find specific solutions, let k be any number, for example, let k=2 and simplify: The End. B.R. 1-18-07 Then (41,-22,2) is a member of the solution set.