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Solving Linear Systems Using Matrix Algebra - Problem 2
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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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