CHAPTER 9: Systems of Equations and Matrices

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1 MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 9: Systems of Equations and Matrices 9.1 Systems of Equations in Two Variables 9.2 Systems of Equations in Three Variables 9.3 Matrices and Systems of Equations The TI Calculator tutorials on the Important Links webpage include instructions for Matrix Operations with the TI calculator. This tutorial is available at 9.3 Matrices and Systems of Equations Solve systems of equations using matrices. Matrices The system can be expressed as where we have omitted the variables and replaced the equals signs with a vertical line. Matrices A rectangular array of numbers such as is called a matrix (plural, matrices). The matrix is an augmented matrix because it contains not only the coefficients but also the constant terms. The matrix is called the coefficient matrix. Matrices continued The rows of a matrix are horizontal. The columns of a matrix are vertical. The matrix shown has 2 rows and 3 columns. A matrix with m rows and n columns is said to be of order m by n. When m = n the matrix is said to be square. 1

2 Gaussian Elimination with Matrices Row Equivalent Operations 1. Interchange any two rows. 2. Multiply each entry in a row by the same nonzero constant. 3. Add a nonzero multiple of one row to another row. We can use the operations above on an augmented matrix to solve the system. Example Solve the following system: First, we write the augmented matrix, writing 0 for the missing y term in the last equation. Our goal is to find a row equivalent matrix of the form. Example continued New row 1 = row 2 New row 2 = row 1 Example continued We multiply the second row by 1/5 to get a 1 in the second row, second column. We multiply the first row by 2 and add it to the second row. We also multiply the first row by 4 and add it to the third row. We multiply the second row by 12 and add it to the third row. Now, we can write the system of equations that corresponds to the last matrix above: 2

3 Example continued We back substitute 3 for z in equation (2) and solve for y. Next, we back substitute 1 for y and 3 for z in equation (1) and solve for x. Row Echelon Form 1. If a row does not consist entirely of 0 s, then the first nonzero element in the row is a 1 (called a leading 1). 2. For any two successive nonzero rows, the leading 1 in the lower row is farther to the right than the leading 1 in the higher row. 3. All the rows consisting entirely of 0 s are at the bottom of the matrix. If a fourth property is also satisfied, a matrix is said to be in reduced row echelon form: The triple (2, 1, 3) checks in the original system of equations, so it is the solution. 4. Each column that contains a leading 1 has 0 s everywhere else. Example Which of the following matrices are in row echelon form? a) b) c) d) Gauss Jordan Elimination We perform row equivalent operations on a matrix to obtain a row equivalent matrix in row echelon form. We continue to apply these operations until we have a matrix in reduced row echelon form. Example: Use Gauss Jordan elimination to solve the system of equations from the previous example; we had obtained the matrix Matrices (a) and (d) satisfy the row echelon criteria. In (b) the first nonzero element is not 1. In (c), the row consisting entirely of 0 s is not at the bottom of the matrix. 3

4 Gauss Jordan Elimination continued Gauss Jordan Elimination continued We continue to perform row equivalent operations until we have a matrix in reduced row echelon form. Next, we multiply the second row by 3 and add it to the first row. Writing the system of equations that corresponds to this matrix, we have We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row echelon matrix. Special Systems When a row consists entirely of 0 s, the equations are dependent. For example, in the matrix Special Systems When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix the system is equivalent to the last row corresponds to the false equation 0 = 9, so we know the original system has no solution. 4

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