Solving Systems Using Augmented Matrices
Date: 11/05/96 at 19:10:46 From: Dan Emmons Subject: Solving Systems Using Augmented Matrices Hi! I have been having trouble trying to figure out how to do this. Please help. Here are the directions that were given to me: Write the augmented matrix for the linear system. Use "elementary row operations" to write the system in triangular form. Then use substitution to solve for each variable. The elementary row operations of matrix terminology consist of these three things : 1. Interchanging 2 equations. 2. Multiplying an equation by a nonzero constant. 3. Adding a multiple of an equation to another equation. Here are the three equations to use: 2x - 3y - 5z = 1 1x - 2y = 12 -4x + 6y + 7z = -23 Thank you for your help.
Date: 11/06/96 at 11:36:06 From: Doctor Pete Subject: Re: Solving Systems Using Augmented Matrices I will solve a similar problem using the method you describe: Find x, y, z such that 2x - 3y + z = -1 y - z = 5 x + y + 4z = 0 . Step 1. Express the system of equations as an augmented matrix. To do this, just write down the coefficients, along with their signs, in the same position they appear (though it is important that you arrange the equations so that the variables are in the same order). We get: [ 2 -3 1 | -1 ] (row 1) [ 0 1 -1 | 5 ] (row 2) [ 1 1 4 | 0 ] (row 3) Step 2. Perform elementary row operations to obtain an upper triangular form. This means we want to use the row operations given to get a matrix that looks like this: [ ? ? ? | ? ] [ 0 ? ? | ? ] [ 0 0 ? | ? ] where the ?'s are numbers, which may or may not be zero. This solves your system because the last row tells you the value of z, which can be plugged into the second row, which gives you the value of y, and then the first row tells you the value of x. We'll see how this works in a moment. Let's perform some operations on our matrix. (a) I don't like the 2 in the first row, so I'll subtract two times row 3 from row 1 [abbreviated R1 --> R1 - 2R3]: [ 0 -5 -7 | -1 ] [ 0 1 -1 | 5 ] [ 1 1 4 | 0 ] (b) Now, I want the 1 in the top row, so I'll switch rows 1 and 3 [abbreviated R1 <--> R3]: [ 1 1 4 | 0 ] [ 0 1 -1 | 5 ] [ 0 -5 -7 | -1 ] (c) We're almost there; we don't want that -5 in the third row, so we get rid of it by adding five times row 2 to row 3 [R3 --> 5R2 + R3]: [ 1 1 4 | 0 ] [ 0 1 -1 | 5 ] [ 0 0 -12 | 24 ] Thus we have converted our matrix into upper triangular form. Step 3. Solve for the variables by substitution. Augmented matrices make things cleaner by serving as convenient bookeeping devices that make it unnecessary to write down variables. But this means that once we are done, we have to remember what the matrix *means* and think with variables again. The third row means 0x + 0y -12z = 24. Hence z = -2. Similarly, the second row means y - z = 5, and plugging in the value of z we obtained, we get y + 2 = 5 or y = 3. Finally, the first row gives x + y + 4z = 0, and substituting y and z, we obtain x + 3 - 8 = 0, or x = 5. Thus our solution set is (5, 3, -2). Notice that by making our matrix upper triangular, we are actually manipulating our equations so that we can readily obtain the solution. However, it is entirely possible that this process may fail; sometimes the last row turns to 0's, like [ 1 3 4 | 0 ] [ 0 1 -4 | 25 ] . [ 0 0 0 | 9 ] This is clearly wrong, because the last row means 0 = 9, which is impossible. Such a matrix corresponds to a system of equations which has no solution. Another case is when an entire row, including the augmented column, is 0. Then the last row gives 0 = 0, which tells us nothing about the variables invovled. In this case, the solution is not unique, and you can only solve for two variables in terms of the third. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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