Solving a linear system of equations using an augmented matrix. So in this case we have a linear equation two variables behind me and we want to solve it using an augmented matrix. The first thing we have to do is take this equation, we just take two equations and put them into matrix form. So each element gets its own spot in the matrix we end up with 3 -5 7, -6 10 and -12.
Okay, now using our row operations what we want to end up doing is trying to eliminate an x or y, typically the x is done first just so to figure what this all means. So I’m looking at this and I say okay I see -6 and 3, if I add 2 times the first row to the second row, -6 plus 2 times 3 is 6 as well those will disappear, so if I add 2 times row 1 to row 2 that first x will disappear which is what I want okay.
So we are doing this, my first row doesn’t change and then my second row I’m adding 2 times row 1. So I have -6 plus 2 times 3, so -6 plus 6 that disappears to 0, 10 plus 2 times -5, 2 times -5 is -10, so 10 minus 10 is 0 and then leaving us here with -12 plus 2 times 7 -12 plus 14 is 2.
So what does this mean we have and we are trying to just get rid of our x but on the process we got rid of both r x and our y. So what this equation actually tells us is we took it back from a matrix line into equation line if this is 0x plus 0y is equal to 2.
Let’s think about that for a second, we have no x’ s we have no y’ s is it possible to get 2? No. So what this tells us is that actually no solution to this equation, there is no x that we could have or y that we have if we have none of them we end up with 2. So in every CA matrix you go through it and you have a entire row that 0 short of its answer and its answer is another number this tells us that we have no solution to this equation.
So I take in the equation bring into augmented matrix form using our row operations we can solve that equation even if it has no solution.