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Augmented Matrices  Problem 3
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Solving a system of linear equations using augmented matrices. So here we have a system in two variables and we want to solve it using a matrix. So the first thing we need to do is put it into augmented matrix form which basically means taking each element and putting it each coefficient and putting it into its own spot in the a matrix, 4 8 and 32, separating the variables from my answer.
Now the go is to use row operations to get 0’s for some of our variables. Typically we always start with the bottom left so I’m going to try get rid if this 4. Without introducing fractions it’s going to be really hard for me to do it. So what I’m going to end up doing is seeing that 4 and 3 have a common multiple of 12, I’m going to multiply this bottom row by 3. Okay so remember we can do scaling multiples and it won’t change. So we multiply that bottom row by 3, the top row doesn’t change. 3, 6, 24 and this then becomes 12, 24 and 32 times 3 is 96 and our dotted line is still there.
Okay I want to get rid of this 12 so if I multiply this top row by 4 I get 12 as well so I want to then subtract that to get 0. So I subtract 4 times row 1 and what happens there is our row 1 hasn’t changed I haven’t touched that at all, so that still remains the same. 3, 6, 24 and then our bottom row, 12 minus 4 times 3 so 12 minus 12 times 0. A little bit bigger number so let’s just write this out. So this turns into if we wrote 2 this is the second column. 24 minus 4 times 6 4 24 plus 24 that zeros out as well.
And lastly 96 minus 4 times 24, 4 times 24 is 96 as well so we end up with 96 minus 96 this goes to 0. Okay, we could have done from here to here in one step if you want. You could just say okay I want to multiply, we could say 3 row 2 minus 4 row 1 in other way you could do it all once but if you have trouble seeing that you can always break it up into two steps as well.
So let’s talk about our answer. We have zeros in this entire row. If we took that out of our augmented matrix into equation form this will tell us that we have 0x plus 0y is equal to 0, x and y don’t matter no matter what x is or y is this equation will hold. So actually what we end up having here is two equations out of the same exact line. So your answer is all reals on that line how you write that is going to depend on your teacher you could choose to write either of one of these equations so you could write 3x minus 6 is equal to 24, 4x minus 8 is 32, or if your teacher likes here settle in notation to be your x and y such that and then put in either equation because in essence they really are the same thing. So using augmented matrices we took our two equations put it in augmented matrix form and did a bunch of row operations to solve our equation.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Problem 2 3,370 viewsSolve for x and y.
3x − 5y = 76x + 10y = 12 
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Problem 3 3,212 viewsSolve for x and y.
3x − 6y = 244x − 8y = 32 
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Problem 4 234 views 
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Problem 5 245 views
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