Solving Linear Systems Using Matrix Algebra - Problem 1

Transcript

One of the really neat things about using the matrix method, for solving systems of equations, is suppose you have two systems with the exact same left hand sides. The same kinds of linear expression 4x plus 11y, 3x plus 8y we have both equations. We’ll use the same inverse matrix to solve both systems at the same time. So first of all recall that you have to turn the system of equations into a matrix equation.

So you start with matrix, I’ll call it a, the coefficient matrix; 4 11, 3 8. Then you got the variable matrix xy, then you’ve got the constant matrix 10, -5.

So you have to find the inverse matrix and then multiply. But luckily we are actually given the inverse matrix -8 11, 3 -4. So x is going to be equal to -8 11, 3 -4 times 10 -5. So we just multiply, we get -80 minus 55. And we get 30 plus 20. And that’s going to be -135, and 50. So that means x equals -135, and y equals 50. That’s our answer.

Well what gives you the exact same matrix to solve this system? Matrix equation for this system is 4 11, 3 8 that’s the same matrix a, and then x y equals -1, 3.

And so x y equals the inverse of a which Is again -8 11, 3 -4, -1 3. And remember you have to multiply this inverse matrix on the left of your constant matrix. So you are going to get 8 plus 33, you are going to get -3 minus 12. So 8 plus 33 that’s 41, -15. And that means that x is 41 and y equals -15.

Very powerful method you can use the same inverse matrix to solve several systems as long as they are the same left hand side.

Tags
matrices square matrices matrix equation coefficient matrix inverse matrix matrix multiplication