Row Reduction Method Lesson 6.4

In Lesson 6.3, we used an inverse matrix to solve systems represented by matrices. This is a powerful strategy to use when the number of variables is equal to the number of equations. In many applications, however, the number of variables is not equal to the number of equations. The row reduction method is such a method. It requires only one matrix, and it extends the method of elimination in a systematic way.

In section 6.3 we replaced the system of equations with an equivalent matrix equation and then used the inverse matrix to solve the equation. In this lesson we’ll write an augmented matrix, which is a single matrix that contains columns for the coefficients of each variable and a final column for the constant terms.

The row reduction method transforms an augmented matrix into a solution matrix. Instead of combining equations and multiples of equations until you are left with an equation in one variable, you add multiples of rows to other rows until you obtain the solution matrix. The solution matrix for a consistent and independent system contains the solutions in the last column. The rest of the matrix consists of 1’s along the main diagonal and 0’s above and below it.

Example A Solve this system of equations. Because the equations are in standard form, you can copy the coefficients and constants from each equation into corresponding rows of the augmented matrix.

Using only the elementary row operations, you can transform this matrix into the solution matrix. You need both m21 and m12 to be 0, and you need both m11 and m22 to be 1. .

Add -2.5 times row 1 to row 2 to get 0 for m21 . Multiply equation 1 by 2.5 and add to equation 2 to eliminate x. Multiply row 2 by 2 to change m22 to 1. Multiply the equation by 2 to find y.

Multiply this new equation by -1, and add the result to equation 1 to eliminate y. Add 1 times row 2 to row 1 to get 0 for m12 . Multiply row 1 by 0.5. Multiply the equation by 0.5 to find x. The last column of the solution matrix indicates that the solution to the system is (2, 1).

You have learned how to interpret a typical solution matrix. For this matrix the solution is x = 12, y = 30, and z = 8. This matrix represents the equations x =12 and y + 3z = 54. These equations indicate that there was not enough information to find a single solution. Therefore, the system has infinitely many solutions. Can you find several ordered triples that satisfy the equations? possible answers (12, 0, 18), (12, 27, 9)

that the system of equations has no solution. Or the final matrix may look like The last row claims that 0 equals 5, which is not true. Therefore, this matrix shows that the system of equations has no solution.