Amy has a master's degree in secondary education and has taught math at a public charter high school.
Watch this video lesson to learn an easy way to solve a system of equations that involves manipulating a matrix. Learn the kinds of easy matrix manipulations that are needed to solve any system of equations.
In math, we come across equations by themselves with just one variable that we have to solve. And then we have linear systems, a collection of linear equations. Your linear equations are equations with variables that have no exponents. So 3x + 4x = 5 is an example of a linear equation, as is x + 3y - 4z = 3.
We need one equation for each variable in our system in order to solve the system. So if we have two variables, we need two equations. If we have three variables, then we need three equations, and so on. In this video lesson, we will learn about using Gaussian elimination, a method to solve a system of equations, to help us solve our linear system. This method requires us to know how to turn our linear system into matrix form and then use simple matrix manipulations. Let's look at solving this linear system using Gaussian elimination:
Remember that a matrix is just a rectangular array of values put into rows and columns. We first need to turn our linear system into matrix form by turning it into an augmented matrix. An augmented matrix is the combination of two matrices. In our case, we have a matrix for the coefficients of the left side of the equation and another for the right side of the equation.
Recall that turning a system of equations into matrix form involves isolating just the coefficients along with their appropriate signs after organizing them so that the x term is first followed by the y term followed by the z term, the equals sign, and then the constant. We can use a vertical line, or several dots in a vertical line, to represent our equals sign. Our linear system is already organized properly, so all we need to do is to isolate our coefficients. Our first row will have 1, 1, 1, | and then 5. Our second row has 2, 0, -1, | and 4. Our third row has 0, 3, 1, | and 2. Our matrix looks like this:
We can now use Gaussian elimination to help us solve this linear system. Gaussian elimination is about manipulating the augmented matrix until we have the matrix that represents the left side of the equations in upper triangular form. What this means is that we want all zeros below the main diagonal. This main diagonal starts at the top left and ends on the bottom right of the coefficients matrix. In other words, we want to manipulate the matrix so the 2 on the second row and the 0 and 3 on the third row are all 0s.
To change these numbers into 0s, we are going to use our matrix row operations. To turn our first 2 into a 0, we multiply our first row by a -2 and then we add it to the second row to create a new second row. We get a new second row of 0, -2, -3, | and -6. Now, to change the 3 in the third row into a 0, we will use this new second row combined with the third row. We will multiply the second row by 3 and add it to the third row multiplied by 2. We get a new third row of 0, 0, -7, | and -14.
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We didn't rewrite the first row because we don't need to change that equation. We only multiplied the first row by 3 just so we can subtract it from the second row. Remember that in algebra, whenever we multiply an equation by any constant, the equation doesn't change at all; the numbers just get bigger by a factor.
Now that we have zeros below the main diagonal, we are done using Gaussian elimination. We can now go on and solve our linear system.
Solving the System
Notice how easy it is to solve now. If we write out our linear equations, we get x + y + z = 5, -2y - 3z = -6, and -7z = -14. We can immediately solve the third equation for z to get z = -14/-7 = 2. We can then substitute this value for z into the second equation to solve for the next variable, y. We get -2y - 3(2) = -6. This turns into -2y - 6 = -6. To solve for y, we add 6 to both sides and we get -2y = 0. Dividing both sides by -2, we get y = 0. So now we have y = 0 and z = 2. To solve our last variable, x, we can use our very first equation. Plugging in these two values, we get x + 0 + 2 = 5. Solving this for x, we get x = 3. So our final answer is x = 3, y = 0, and z = 2. We can also write this in point form, like this: (3, 0, 2).
Let's review what we've learned. We learned that linear systems are collections of linear equations. A linear system has the same number of equations as there are variables, since we need one equation for each variable in order to solve such a system of equations. The method we talked about in this lesson uses Gaussian elimination, a method to solve a system of equations, that involves manipulating a matrix so that all entries below the main diagonal are zero. Upper triangular form is the term used to describe a matrix that has all zeros below the main diagonal. We then use algebra and substitution to finish solving our system of equations.
Completing this lesson can prepare you to:
Recite the definitions of linear systems, augmented matrix and upper triangular form
Demonstrate the use of the Gaussian elimination method to solve a system of equations
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