Carl Horowitz

**University of Michigan**

Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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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.