Solving Systems of Equations Using Matrices Section 12.4 Solving Systems of Equations Using Matrices

Objectives Define a matrix and determine its order Write the augmented matrix for a system Perform elementary row operations on matrices Use matrices to solve a system of two equations Use matrices to solve a system of three equations Use matrices to identify inconsistent systems and dependent equations

Objective 1: Define a Matrix and Determine Its Order A matrix is any rectangular array of numbers arranged in rows and columns, written within brackets. Some examples of matrices are: Each number in a matrix is called an element or an entry of the matrix. A matrix with m rows and n columns has order m  n, which is read as “m by n.”

Objective 2: Write the augmented matrix for a system To show how to use matrices to solve systems of linear equations, we consider the system which can be represented by the following matrix, called an augmented matrix: Each row of the augmented matrix represents one equation of the system. The first two columns of the augmented matrix are determined by the coefficients of x and y in the equations of the system. The last column is determined by the constants in the equations.

EXAMPLE 1 Represent each system using an augmented matrix: Strategy We will write the coefficients of the variables and the constants from each equation in rows to form a matrix. The coefficients are written to the left of a vertical dashed line and constants to the right. Why In an augmented matrix, each row represents one equation of the system.

EXAMPLE 1 Represent each system using an augmented matrix: Solution Since the equations of each system are written in standard form, we can easily write the corresponding augmented matrices.

Objective 3: Perform Elementary Row Operations on Matrices To solve a system of linear equations using matrices, we transform the entries to the left of the dashed line in the augmented matrix into an equivalent matrix that has 1’s down its main diagonal and 0’s in all the remaining entries directly above and below this diagonal. A matrix written in this form is said to be in reduced row-echelon form. We can easily determine the solution of the associated system of equations when an augmented matrix is written in this form. Reduced row-echelon form a, b, and c represent real numbers. Main diagonal Main diagonal

Objective 3: Perform Elementary Row Operations on Matrices To write an augmented matrix in row echelon form, we use three operations called elementary row operations. Type 1: Any two rows of a matrix can be interchanged. A type 1 row operation corresponds to interchanging two equations of the system. Type 2: Any row of a matrix can be multiplied by a nonzero constant. A type 2 row operation corresponds to multiplying both sides of an equation by a nonzero constant. Type 3: Any row of a matrix can be changed by adding a nonzero constant multiple of another row to it. A type 3 row operation corresponds to adding a nonzero multiple of one equation to another.

EXAMPLE 2 Perform the following elementary row operations. a. Type 1: Interchange rows 1 and 2 of matrix A. b. Type 2: Multiply row 3 of matrix C by . c. Type 3: To the numbers in row 2 of matrix B, add the results of multiplying each number in row 1 by –4. Strategy We will perform elementary row operations on each matrix as if we were performing those operations on the equations of a system. Why The rows of an augmented matrix correspond to the equations of a system.

EXAMPLE 2 Solution Perform the following elementary row operations. a. Type 1: Interchange rows 1 and 2 of matrix A. Solution

EXAMPLE 2 Solution Perform the following elementary row operations. b. Type 2: Multiply row 3 of matrix C by . Solution

EXAMPLE 2 Perform the following elementary row operations. c. Type 3: To the numbers in row 2 of matrix B, add the results of multiplying each number in row 1 by −4.

Objective 4: Use Matrices to Solve a System of Two Equations We can solve a system of two linear equations using a series of elementary row operations.

Objective 4: Use Matrices to Solve a System of Two Equations When a system of linear equations has one solution, we can use the following steps to solve it. Write an augmented matrix for the system. Use elementary row operations to transform the augmented matrix into a matrix in reduced row-echelon form with 1’s down its main diagonal and 0’s directly above and below the 1’s. When step 2 is complete, write the resulting equivalent system to find the solution. Check the proposed solution in the equations of the original system.

EXAMPLE 3 Use Gauss-Jordan elimination to solve the system: Strategy We will represent the system with an augmented matrix and use a series of elementary-row operations to produce an equivalent matrix in row-echelon form. Why The values of the variables x and y that solve the original system of equations will be in the last column of the reduced row-echelon form matrix.

EXAMPLE 3 Use Gauss-Jordan elimination to solve the system: Solution We can represent the system with the following augmented matrix: First, we want to get a 1 in the top row of the first column where the shaded 2 is. This can be done by applying a type 1 row operation and interchanging rows 1 and 2.

EXAMPLE 3 Use Gauss-Jordan elimination to solve the system: Solution To get a 0 in the first column where the shaded 2 is, we use a type 3 row operation and multiply each entry in row 1 by –2 to get

EXAMPLE 3 Use Gauss-Jordan elimination to solve the system: Solution To get a 1 in the bottom row of the second column where the shaded 3 is, we use a type 2 row operation and multiply row 2 by To get a 0 in the first row where the shaded –1 is, we use a type 3 row operation and add to those numbers the entries in row 2 to get This is R1 + R2.

EXAMPLE 3 Use Gauss-Jordan elimination to solve the system: Solution After simplifying the top row, we obtain an equivalent matrix in the desired reduced row-echelon form. This augmented matrix represents the system of equations Writing the equations without the coefficients, and dropping the 0y and 0x terms, we have

Objective 5: Use Matrices to Solve a System of Three Equations To show how to use matrices to solve systems of three linear equations containing three variables, we consider the following system that can be represented by the augmented matrix to its right. To solve the 3 x 3 system of equations, we transform the augmented matrix into a matrix with 1’s down its main diagonal and 0’s below its main diagonal.

EXAMPLE 4 Use Gauss-Jordan elimination to solve the system: Strategy We will represent the system with an augmented matrix and use a series of elementary row operations to produce an equivalent matrix in row-echelon form. Why The values of the variables x, y, and z that solve the original system of equations will be in the last column of the reduced row-echelon form matrix.

EXAMPLE 4 Use Gauss-Jordan elimination to solve the system: Solution This system can be represented by the augmented matrix To get a 1 in the first column where the shaded 3 is, we perform a type 1 row operation by interchanging rows 1 and 3. To get a 0 where the shaded 2 is, we perform a type 3 row operation by multiplying each entry in row 1 by –2 to get:

EXAMPLE 4 Solution and add these numbers to the entries in row 2. Use Gauss-Jordan elimination to solve the system: Solution and add these numbers to the entries in row 2. To get a 0 where the shaded 3 is, we perform another type 3 row operation by multiplying the entries in row 1 by –3 and adding the results to row 3.

EXAMPLE 4 Use Gauss-Jordan elimination to solve the system: Solution To get a 1 where the shaded –1 is, we perform a type 2 row operation by multiplying row 2 by –1. To get a 0 where the shaded 2 is, we perform a type 3 row operation by multiplying the entries in row 2 by –2 and adding the results to row 1. And to get a 0 where the shaded –5 is, we multiply the entries in row 2 by 5 and add the results to row 3.

EXAMPLE 4 Use Gauss-Jordan elimination to solve the system: Solution To get a 1 where the shaded 24 is, we perform a type 2 row operation by multiplying row 3 by . To get a 0 where the shaded 5 is, we multiply the entries in row 3 by –5 and add the results to row 2. And to get a 0 where the shaded –8 is, we multiply the entries in row 3 by 8 and add the results to row 1.

EXAMPLE 4 Use Gauss-Jordan elimination to solve the system: Solution The final augmented matrix represents a simpler system of equations that is equivalent to the original system:

Objective 6: Use Matrices to Identify Inconsistent Systems and Dependent Equations In the next example, we will see how to recognize inconsistent systems and systems of dependent equations when matrices are used to solve them.

EXAMPLE 5 Use Gauss-Jordan elimination to solve the system: Strategy We will represent the system with an augmented matrix and attempt to use Gauss-Jordan elimination to produce an equivalent matrix in row-echelon form. Why It is easy to recognize an inconsistent system or dependent equations during that process.

EXAMPLE 5 Use Gauss-Jordan elimination to solve the system: Solution a. The system can be represented by the augmented matrix Since the matrix has a 1 in the top row of the first column, we proceed to get a 0 where the shaded –3 is by multiplying row 1 by 3 and adding the results to row 2. The equation in the second row of this system simplifies to 0 = –8. This false statement indicates that the system is inconsistent and has no solution. The solution set is Ø.

EXAMPLE 5 Use Gauss-Jordan elimination to solve the system: Solution b. The system can be represented by the augmented matrix To get a 1 where the shaded 2 is, we perform a type 2 row operation by multiplying row 1 by . To get a 0 where the shaded –6 is, we perform a type 3 row operation by multiplying the entries in row 1 by 6 and adding the results to the entries in row 2.

EXAMPLE 5 Use Gauss-Jordan elimination to solve the system: Solution The second row of this augmented matrix represents the equation 0 = 0. This true statement indicates that the equations are dependent and that the system has infinitely many solutions. The solution set is {(x, y) | 2x – y = 4}.