Solve the following system of equations to find out where the intersection point is.
| x1 | + | 2x2 | = | 16 |
| x1 | + | x2 | = | 9 |
Take one of the equations and solve for one of the variables. Then substitute that into the other equation.
We'll take the second equation x1 + x2 = 9 and solve it for x2 = 9 - x1.
Substitute that into the first equation to get x1 + 2(9-x1) = 16.
Simplify both sides to get x1 + 18 - 2x1 = 16 or -x1 + 18 = 16.
Solve for x1 to get -x1 = -2 or x1 = 2.
Now that you know x1 = 2, we go back to either equation to find x2.
x1 + x2 = 9
2 + x2 = 9
x2 = 7
We have the solution x1 = 2, x2 = 7.
In the addition or elimination method, you multiply one or both equations by a constant and then add the results together to eliminate one of the variables.
To eliminate the x2, multiply the second equation by -2.
| x1 | + | 2x2 | = | 16 | stays | x1 | + | 2x2 | = | 16 | ||
| x1 | + | x2 | = | 9 | becomes | -2x1 | - | 2x2 | = | -18 |
Add the resulting equations together to get -x1 = -2 or x1 = 2.
To eliminate the x1, multiply the second equation by -1.
| x1 | + | 2x2 | = | 16 | stays | x1 | + | 2x2 | = | 16 | ||
| x1 | + | x2 | = | 9 | becomes | -x1 | - | x2 | = | -9 |
Add the resulting equations together to get x2 = 7.
We have the solution x1 = 2, x2 = 7.
Convert the system of linear equations into an augmented matrix.
| x1 | + | 2x2 | = | 16 |
| x1 | + | x2 | = | 9 |
Each equation becomes a row and each variable becomes a column. The resulting matrix is a 2×3 matrix.
| x1 | x2 | rhs | |||
|---|---|---|---|---|---|
| R1 | 1 | 2 | 16 | ||
| R2 | 1 | 1 | 9 |
Perform the row operation R2-R1 → R2 to clear out the 1 in the lower left.
| x1 | x2 | rhs | |||
|---|---|---|---|---|---|
| R1 | 1 | 2 | 16 | ||
| R2 | 0 | -1 | -7 |
Perform the row operation -R2 → R2 to change the -1 into a positive 1.
| x1 | x2 | rhs | |||
|---|---|---|---|---|---|
| R1 | 1 | 2 | 16 | ||
| R2 | 0 | 1 | 7 |
If we were to stop at this point, we would be performing Guassian elimination with back substitution. We might as well go ahead and completely reduce the matrix using Gauss-Jordan elimination and place the matrix into Reduced Row Echelon Form. There is no back substitution needed in this form.
Perform the row operation R1 - 2R2 → R1 to clear out the 2 in R1C2.
| x1 | x2 | rhs | |||
|---|---|---|---|---|---|
| R1 | 1 | 0 | 2 | ||
| R2 | 0 | 1 | 7 |
Convert the matrix back into a system of equations to get the solution x1 = 2, x2 = 7.
The TI83 and TI84 calculators have a RREF command built into them. RREF stands for Reduced Row Echelon Form and it does the entire Gauss-Jordan elimination in one step.
Watch the Flash video: How to row reduce a matrix using the calculator [564 KB].
When there are the same number of variables as equations, a system of linear equations can be written in matrix form as AX = B, where A is the coefficient matrix, B is the constant matrix (right hand side), and X is the variable matrix.
| A | = | 1 | 2 | X | = | x1 | B | = | 16 | ||||||||
| 1 | 1 | x2 | 9 |
The solution to the equation is X = A-1B.
A-1 can be found using the calculator.
| A-1 | = | -1 | 2 | ||
| 1 | -1 |
Then, multiplying A-1 by B we find the solution to the system.
| X | = | x1 | = | A-1B | = | -1 | 2 | 16 | = | 2 | |||||||||
| x2 | 1 | -1 | 9 | 7 |
We have the solution x1 = 2, x2 = 7.