here's a system of linear equations which is slightly different from the ones we've seen so far in class it has in it as coefficient what we might call a parameter namely it's a unknown real number as the coefficient so question is for which values of rule number a either exists as usual no Solutions a unique solution or infinitely many solutions now we consider here a is being a fixed real number it's not a variable x y and Zed here are the variables and we just want to know fix an A and for any given fixed a which of these three cases is true so we have to break it down into different cases depending on what a is so firstly we write it into into an augmented Matrix so we have over here have coefficient X is 1 coefficient Y is z coefficient of Z is a and - 2 and 0 1 0 a 0 0 a^ 2 - A and 2 1 - A so this is the augmented Matrix this is the augmented Matrix for our system of linear equations now we must importantly we must remember that we can't ever divide 3 by0 so for instance if a^ 2 - A was Zero we couldn't divide through to get a one a one and a one this is important to remember so when we break down cases we need to consider this so the most obvious raow operation we might want to do is to make this a one and then cancel off the a so we'll just forget about this a in top corner here just for the time being but as I said if a 2 - a = 0 we can't divide the last row by a^ 2 - A and this is the case if a is = 1 or0 so we could Factor this as a * 1 - a = 0 and so either a is z or a is 1 so if a is 1 so let's say if a = 1 then this is zero as we saw this is also zero so the last row is all zeros 0 0 0 Z so we're in a case where we have a free variable namely this rows or zeros Zed is now a free variable we have one pivot we have a second pivot and we have a free variable so we have infinitely many solutions and we can write them down if we set Z equal to t a real number then we find that y = a you could have seen that before or rather in fact we've set a = to 1 so we don't have y = a have y = 1 and then the top row is then remember a is 1 we have x + z = - 2 so x = - 2 - t and then this is our solution where there is infinitely many solutions we can write this out as a vector as we want second case a = 0 then here is zero on this side we have 0 0 0 but on this side we don't have a zero on this side we have we have two and so we're trying to solve essentially 0 = 2 which we cannot do so if AAL 0 there are no Solutions so now if a is not equal to Z not equal to 1 we can divide through by a 2us a because this term is not zero and so we can you put it into a row Echelon form and solve from there so I'm not going to bother canceling out this a here but I'm just going to solve it from row Echelon form not reduced ration form just make a bit more space so now if a is not equal to Z or 1 we can do some raow operations namely send Row three goes to 1 on a^ 2 - A Row 3 therefore we get 1 0 a - 2 0 1 0 a 0 0 1 we find to 1 - A over a * a - 1 and we just cancel out these at the expense of a minus sign - 2 on a and we can write out the equations that these correspond to so we have a unique solution we have a single pivot in each column so we can write out X plus a z = - 2 we have y = a we have Z = - 2 on a you can see here why when a is zero we don't have a solution because that would imply that Z here was infinite and we can rearrange this one so we get if we stick zal -2 on a in here you get x = - 2 - 2 is here cancel in fact we get x = x = to0 given that Z isus 2 a and there you have it just a quick example supplement what of did lectures and I hope to do more of these soon thank you