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Solving Linear Systems Using Matrix Algebra - Problem 2

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

After you’ve seen an example where we use inverse matrices to solve a two by two system, you might say oh well so what. Those are actually pretty easy to solve anyway. But the same method works equally well on a system of 3 equation with 3 unknowns and that’s why it's so powerful this method. Because it works on a system no matter how big it is.

Let’s first convert this system of equations into a matrix equation. So it will become, first I deal with the coefficient matrix. I have -1 3 0, 1 2 -2, and I have 0 -2,1. That’s a coefficient matrix. The variable matrix would just be x y z, and that’s going to equal. And then I have a constant matrix 5 1 -2. So this is my matrix a, this is matrix x, and this is b my coefficient matrix. A times x equals b. And so to solve this x equals a inverse times b. So I just need a-inverse. I’m not actually given that. So let’s use that now.

X is just x y z equals a-inverse 2 3 6, 1 1 2, 2 2 5. And then b is 5 1 -2 .So we just have to multiply these two matrices to get the answer. We get 10 plus 3 minus 12. 10 plus 3 is 13, minus 12, 1. And I get 5 plus 1, 6 minus 4, 2. Then I get 10 plus 2, 12, minus 10, 2. 1 2 2, so the solution is x equals 1, y equals 2. And z equals 2.

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