 In this video, we're going to present the solution to question 11 from the practice midterm exam number one for math 2270 and so we were given a transformation which goes from z7 to To z7 3 and it's given by the following rule T of x1 x2 is given as x1 plus 4x2 2x1 plus 5x2 and 3x1 plus 6x2 Now our function is something that's going to be working mod 7 here So we go from mod 7 with two entries to mod 7 with three entries But the fact that we're working mod 7 really isn't going to have much bearing on this calculation whatsoever So for this consideration, let say x be the vector x1 x2 and let y be the vector Y1 y2 and so these are vectors that belong to z7 squared And so what we're going to do is we're going to take a look at t of x plus y to show to prove that this is a linear transformation We have to show that the transformation preserves vector addition, which is what we're going to be doing right now So notice that t tx by give above is just x1 x2 and y is just y1 y2 and So adding these together we end up with the vector x1 plus y1 and then x2 plus y2 and So now we're going to apply the definition of the Function itself. So this function will create a vector in z7 3 so there's three components where the first component you're going to take the first entry plus four times the second entry So we're going to get x1 plus y1 plus four times x2 plus y2 So for the second entry, we're going to take two times the first input plus five times the second input So we get two times x1 plus y1 plus five times x2 plus y2 and Then for the last one we're going to take three times the first input plus six times the second input So we get three times x1 plus y1 plus six times x2 plus y2 so that's what it means to evaluate the Transformation at the input vector x1 plus y1 and x2 plus y2 So what I'm going to next do is distribute these coefficients. So we're going to end up with an x1 plus y1 plus 4x2 plus 4y2 Next we're going to get 2x1 plus 2y1 we're going to get 5x2 plus 5y2 and then lastly we're going to get 3x1 plus 3y1 plus 6x2 Plus 6y2 like so. So we just distribute these coefficients. I'm then going to add together The common x's and y's right so notice we can break this this thing up in the following way We're going to have an x1 plus 4x2 We're going to have a 2x1 Plus 5x2 and we're going to have a 3x1 Plus 6x2 so I just wrote all of the x's in a single vector And then I'm going to do the same thing for y you have the y1 plus 4y2 We have a 2y1 Plus 5y2 and then lastly we have a 3y1 Plus a 6y2 Which the first vector listed that's just t of x x1 x2 and then the second one is just t of y1 y2 which We know whoops, which we know that's just going to be t of x Plus t of y so we've now demonstrated that this transformation preserves vector addition We wanted the same thing for scalar multiplication, so we're going to take t of c times x well x remember is Just the vector x1 x2, so if you times that by a scalar you end up with cx1 and Cx2 and So let's apply the definition of the function here So our function is going to take the first entry Cx1 and add to it four times the second input Cx2 so for the for the second output we're going to take two times Cx1 plus five C of x2 and then for the third output we get three times Cx1 plus six times Cx2 Now notice that everything in entry one of this vector is divisible C, so let's factor that out So you get C times x1 plus 4x2 Everything in the second entry is divisible by C, so we can factor that out C times 2x1 plus 5x2 and then everything in the third entry is also divisible by C, so we factor that out leaving 3x1 plus 6 6x2 now since everything in all three entries is divisible by C. We can take it out So that's just scalar multiplication right there. So you get x1 plus 4x2 2x1 plus 5x2 And then 3x1 plus 6x2 and you can see that this vector right here is just T of x1 x2 and what do you know that is just C of T of x and so then what we have now done is that this shows That T is a linear transformation. So therefore T is a linear transformation. We just have to show that it preserves addition and scalar multiplication and that's what we've now done transformation