Question Video: Computing Area of Triangle Using Determinant of a Matrix | Nagwa Question Video: Computing Area of Triangle Using Determinant of a Matrix | Nagwa

Question Video: Computing Area of Triangle Using Determinant of a Matrix Mathematics

Use determinants to find the area of the triangle with vertices (0, −1), (0, 2), and (5, 0).

04:43

Video Transcript

Use determinants to find the area of the triangle with vertices zero, negative one; zero, two; and five, zero.

So in order to solve a problem like this, what we have is a formula. This formula tells us the area of a triangle. And it’s a way of finding it using determinants. So what we’ve got is that the area of a triangle is equal to positive or negative a half. Now this doesn’t mean there are two answers, positive and negative a half. What it means is that we use either positive or negative a half, whichever one gives us the positive outcome. And that’s because we’re dealing with an area. So we want a positive outcome. And that’s this multiplied by the determinant of the matrix 𝑥 one, 𝑦 one, one; 𝑥 two, 𝑦 two, one; 𝑥 three, 𝑦 three, one. Where 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two; and 𝑥 three, 𝑦 three are the coordinates of each of the vertices of our triangle.

So what I’ve done now is I’ve labeled each of the coordinates. So what we have is 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two; and 𝑥 three, 𝑦 three. So now if we substitute in our values, what we get is that our area is equal to positive or negative a half multiplied by the determinant of the matrix zero, negative one, one; zero, two, one; five, zero, one.

So now let’s just quickly remind ourselves how we find the determinant of a three-by-three matrix. So if we’ve got the matrix 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, then the determinant of this matrix is equal to 𝑎 multiplied by this submatrix determinant 𝑒, 𝑓, ℎ, 𝑖 minus 𝑏 multiplied by the determinant of the submatrix 𝑑, 𝑓, 𝑔, 𝑖 plus 𝑐 multiplied by the determinant of the submatrix 𝑑, 𝑒, 𝑔, ℎ.

So what we’re gonna have is the area is equal to positive or negative a half multiplied by, and then we’ve got zero. That’s because that’s our 𝑎 because it’s the first element in the first row and column. Then multiplied by the determinant of the submatrix two, one, zero, one. And we get this because what we do is we delete the row and column that our first element is in.

And when we do that, this is what’s left over. And then we’ve got minus. Then we’ve got negative one multiplied by the determinant of the submatrix zero, one, five, one. And again, we got that by deleting the row and column that our element was in.

It’s worth noting that the coefficients have a sign before them. And that’s determined by the column. First column, positive; second column, negative; third column, positive; et cetera. But because we’re subtracting a negative, what we do is we rewrite this is add one multiplied by the determinant of the submatrix zero, one, five, one. And then, finally, we’ve got add one multiplied by, and again we’ve got the determinant. This time, it’s the submatrix zero, two, five, zero.

Now in order to complete the next step, what we need to do is remind ourselves how we find the determinant of a two-by-two matrix. And the way we do this is if we look at the determinant of the matrix 𝑎, 𝑏, 𝑐, 𝑑. And this is equal to 𝑎𝑑 minus 𝑏𝑐. So what we do is we cross-multiply and then subtract. Okay, great.

So now let’s use this on our example. So then what we’re gonna have is area is equal to positive or negative a half multiplied by, then we’ve got a zero. That’s cause zero multiplied by anything is zero. Plus then we’ve got zero multiplied by one minus one multiplied by five. And we got that from our cross-multiplication. And because the coefficient or the value that’s multiplied by the determinant was just one, we didn’t have to do anything extra. And this is plus zero multiplied by zero minus two multiplied by five. And again, we use cross-multiplication here. And because the coefficient or the value that was multiplied by the determinant was one, we didn’t have to multiply three by anything else again.

So we get the area is equal to positive or negative a half multiplied by negative five minus 10. So then this means we can decide what our a half is going to be. And I’m gonna make it negative. And that’s because negative five minus 10 is negative 15. So therefore, we need a negative to be multiplied by this to give us a positive.

So therefore, the area is equal to negative a half multiplied by negative 15. So therefore, this is gonna give us the final area for the triangle with vertices zero, negative one; zero, two; and five, zero of 7.5.

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