# Video: Relating Determinants of Planar Transformations to Area

Suppose the linear transformation 𝐿 sends (1, 0) to (−1, 5) and (1, 1) to (−6, 6). What is the absolute value of the determinant of the matrix representing 𝐿?

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### Video Transcript

Suppose the linear transformation 𝐿 sends one, zero to negative one, five and one, one to the negative six, six. What is the absolute value of the determinant of the matrix representing 𝐿.

In a general rule, we can say that a linear transformation that transforms the point 𝑃 which is 𝑥, 𝑦 into 𝑃 prime which is [𝑥 prime], 𝑦- 𝑦 prime, well, this can be written as 𝑥 prime over 𝑦 prime as a vector is equal to the matrix 𝑎, 𝑏, 𝑐, 𝑑 multiplied by 𝑥, 𝑦. And this is where the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑 is a transformation matrix. And it’s this transformation matrix that we want to find.

So let’s use the values we have to try and find it. So first of all, we can start with the first pair of points. So we have our matrix 𝑎, 𝑏, 𝑐, 𝑑 multiplied by one, zero is equal to negative one, five. Then what we do is we multiply the corresponding components. So we have 𝑎 multiplied by one and 𝑏 multiplied by zero which is gonna give us 𝑎 plus zero. And that’s because 𝑎 multiplied by one is just 𝑎 and 𝑏 multiplied by zero is just zero. And this is gonna be equal to negative one because that’s the corresponding value on the right-hand side.

And then I move on to the second row. But what I do first before I do that is I’m gonna label the first equation one. So then if I carry on the same process and when I have 𝑐 multiplied by one is just 𝑐, 𝑑 multiplied by zero is just zero, so I have 𝑐 plus zero is equal to five cause that’s our corresponding value. So then I can label this equation two. And from these equations, I can just summarise that 𝑎 is equal to negative one and 𝑐 is equal to five.

Okay, great, so we’ve found those two components. Now, what we want to do is find the 𝑑 and 𝑏 components of our transformation matrix. So now, to find this value, we’re gonna move on to our second coordinates. So we got one, one and negative six, six. So we can say that we have the matrix 𝑎, 𝑏, 𝑐, 𝑑 multiplied by one, one is equal to negative six, six.

So then using the same process as before, we’re gonna get 𝑎 plus 𝑏 is equal to negative six. And that’s cause we had 𝑎 multiplied by one. And that was the first one. And then you got 𝑏 multiplied by one which is the one at the bottom. So we get 𝑎 plus 𝑏. And that is equal to the corresponding value which is negative six. And then we have, finally, 𝑐 multiplied by one which is just 𝑐, 𝑑 multiplied by one which is just 𝑑. So we have 𝑐 plus 𝑑 is equal to six cause, again, this is the corresponding value.

So we now got equations three and four. We can use these to find 𝑏 and 𝑑. So now, what we’re gonna do is substitute in the values we have for 𝑎 and 𝑐. So first of all, we’re gonna substitute in 𝑎 is equal to negative one. And we’re gonna substitute it into our third equation. So when we do that, we get negative one plus 𝑏 is equal to negative six. So if we then add one to each side of the equation, we get 𝑏 is equal to negative five. So we’ve found 𝑏. So then finally, what we do is we substitute in our 𝑐-value. And we get five plus 𝑑 is equal to six, if we substituted it into equation four. So then if we subtract five from each side of the equation, we’re left with 𝑑 is equal to one. So we now have our 𝑎-, 𝑏-, 𝑐-, and 𝑑-values. So we’ve found our transformation matrix 𝐿.

So does that mean we’ve finished the question? Well, no, because what we want to do is find the absolute value of the determinant of the matrix. So what we want to do is find the absolute value of the determinant of the matrix negative one, negative five, [five], one. And what the absolute value means is we’re only interested in positive values because the absolute value, or another way of thinking about it, is the distance from a point. So, therefore, it doesn’t matter if it’s positive or negative. It’s just a distance to a point.

So we’re gonna use that at the end. And just to remind us how we find the determinant of a two-by-two matrix, if we have the matrix 𝑎, 𝑏, 𝑐, 𝑑, then the determinant of that matrix is 𝑎 multiplied by 𝑑, so top left multiplied by bottom right, minus 𝑏 multiplied by 𝑐, which is top right multiplied by bottom left. So, therefore, the determinant of the matrix negative one, negative five, five, one is gonna be negative one multiplied by one minus negative five multiplied by five. And we’ve got that via cross-multiplying as of shown on our example here.

So then we’re looking for the absolute value of negative one add 25. And that’s because you had negative one multiplied by one which is just negative one. Then we’ve got minus. And then we had negative 25. Well, if you subtract a negative, it turns to a positive. So, therefore, we can say that the answer is gonna be equal to the absolute value of negative 24, while the absolute value of negative 24 is just gonna be 24. So that means we can say that if the linear transformation 𝐿 sends one, zero to negative one, five and one, one to negative six, six, then the absolute value of the determinant of that matrix, which represents 𝐿, is 24.