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# Solving Linear Systems Using Matrix Algebra - Concept

###### 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.

One of the most commonly used applications of square matrices is solving systems of linear equations. The methods of solving systems of linear equations using **matrix algebra** are much more efficient than hand calculating the systems using substitution. This is especially true when dealing with systems of 3 or more variables. Two methods of matrix algebra include row reducing and finding the inverse.

It's possible to use matrix Algebras to solve linear systems of equations, we'll show you how in a moment but first of all let me take a look at 3 matrices a, x and b. And let's notice what the equation, the matrix equation ax=b stands for by multiplying this out, now ax is 5, -3, 7, -4 right x is xy it's the 2 by 1. And if I multiply this I get 5x-3y and I get 7x-4y right I get this 2 by 1 matrix and that matrix equals b 5,6. Now you know the two matrices are only equal if they have the same dimensions they're both 2 by 1 and they both have the same exact equal entries so 5x-3y has to equal 5 and 7x-4y has to equal 6 so let me just write that down 5x-3y=5, 7x-4y=6 now this system is a system of linear equations this linear system is completely equivalent to the matrix equation ax=b and this suggests a way that we can use matrix equations to solve linear systems so let's focus on solving this linear system.

Now first when you're turning a linear system into a matrix equation let's make the observation that on the left hand side you want your x and y terms and you want in that order so these are this these numbers here are called coefficients and what you want to do is you want to come up with the coefficient matrix for the system and that could be 5, -3, 7, -4 that's matrix a and then the variable matrix xy that's this and then the constant matrix 5,6 is is matrix b.

Now how do we solve a matrix equation ax=b? Well it turns out that this is basically like a linear equation with real numbers the only difference is that we can't divide by matrices there's no way to divide by matrices but you can multiply by its inverse so if if matrix a has an inverse you can multiply by the inverse and remember when you multiply both sides by a matrix you've got to multiply either both sides on the left or both sides on the right because matrix multiplication is not commutative well what are you going to get here a inverse times a is the identity matrix times x and the identity matrix times x is going to be x so this is x equals a inverse b that's actually your solution you'll have xy equals and then some numbers so all you have to do is find out what the inverse matrix of a is. Now let's say that we know that a inverse is -4, 3 -7, 5 okay say say that we've solved for the inverse matrix and we found this, then x is going to equal a inverse b which is this -4, 3, -7, 5 times b which is 5,6. Let me just do that multiplication it's -20+18 -2, -35+30 -5, right and this is x,y so that's our that's gives us a solution x=-2, y=-5 that's it.

So when you're solving a linear system using matrices the first step is to write a matrix equation like this right you'll have ax=b. And the second step is just to find the inverse of matrix a the coefficient matrix. Then the third step is to multiply that inverse matrix by the constant matrix b this guy right? Multiply on the left and you'll get your solutions. It's as easy as that so it's very important to know how to invert matrices to do this prob- to use this process but you'll find it's a very quick process if you good at inverting matrices.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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