### Video Transcript

Solve for 𝑥: the determinant of the matrix zero, five, negative five 𝑥, 𝑥, four, five, four, one, three is equal to 280.

So the first thing we need to do to solve this problem is work out the determinant of our matrix. And if we try to find the determinant of the three-by-three matrix, what we use is the first row. And we use the first row to be our coefficients or the values that we multiply our submatrices by, which are gonna be two-by-two submatrices. And I’m gonna show you how it’s gonna work. It’s worth noting also that these values are gonna be positive, negative, or positive, determined by the pattern that we have. Where the first column is positive, so it’ll be positive zero, the second one will be negative, the third one will be positive, et cetera. So what this means is where we have a negative above the column, it’s gonna change the sign of the value that we’ve got from our first row. If you’ve got a positive, then the sign is gonna stay the same.

So first of all, we’re gonna have zero multiplied by the determinant of the submatrix four, five, one, three. And we find that submatrix by deleting the row on column that the zero is in. And then, it’s what we’re left with afterwards. So 𝑥 would be left with four, five, one, three. Then next, what we have is negative five multiplied by the determinant of the submatrix 𝑥, five, four, three. Again, finding that in the same way, and it’s negative five because, as we said, the second column is negative. And what we mean by this is that the coefficient or our five is multiplied by negative one. So we get negative five.

Then, finally, we have minus five 𝑥 multiplied by the determinant of the submatrix 𝑥, four, four, one. And again, we’ve done this the same way. And this time, we’ve still got negative five 𝑥. And that’s because the third column is positive. So we multiply negative five 𝑥 by positive one, which gives us the same sign because it doesn’t change. And then, this is all equal to 280. So now, what we need to do is evaluate our submatrices, well, the determinant of our submatrices. And the way we do that is using this method.

So if we’ve got two-by-two submatrix, we can call this one 𝑎, 𝑏, 𝑐, 𝑑. Then, what we do is we multiply diagonally. So we have 𝑎 multiplied by 𝑑 and 𝑏 multiplied by 𝑐. And then, we take away 𝑏 multiplied by 𝑐 from 𝑎 multiplied by 𝑑. Well, for the first term, we don’t have to worry about the submatrix. And that’s because we’ve got zero multiplied by the determinant of the submatrix. So anything multiplied by zero is just zero. And then, we have minus five multiplied by. Now, we’ve got three multiplied by 𝑥 or 𝑥 multiplied by three, which gives us three 𝑥 minus. Then, we’ve got five multiplied by four, which is 20. And then, we have minus five 𝑥 multiplied by. Then, we’ve got 𝑥 and that’s because we had 𝑥 multiplied by one which is just 𝑥 minus 16 because we’ve four multiplied by four, which is 16. And this is all equal to 280.

So now, what we need to do is distribute across our parentheses. So first of all, we have zero minus. And then, we’ve got five multiplied by three 𝑥, which is 15𝑥 and then plus a 100. And that’s because we had negative five multiplied by negative 20. And a negative multiplied by a negative is a positive. And then, we have negative five 𝑥 squared plus 80 𝑥. That’s cause we’ve got negative five 𝑥 multiplied by 𝑥, which is negative five 𝑥 squared. And we’ve got negative five 𝑥 multiplied by negative 16. Negative multiplied by a negative is a positive. So it gives us positive 80𝑥. This is equal to 280. So now, if we simplify, we get negative five 𝑥 squared plus 65𝑥 plus 100 equals 280.

So now, what I wanna do is make this so that I have positive five 𝑥 squared or positive 𝑥 squared. So I’m going to add five 𝑥 squared to both sides of the equation, subtract 65𝑥 from both sides of the equation, and subtract 100 from both sides of the equation. So I wanna make our quadratic equal to zero. And when I do that, I get zero is equal to five 𝑥 squared minus 65𝑥 plus 180. So now, what we wanna do is solve this to find 𝑥. But the first thing we do to make it easier is divide through by five because five is a factor of each of our terms. So we’ve got zero is equal to 𝑥 squared minus 13𝑥 plus 36.

So now, we can solve this using factoring. And to factor, just to remind us what we need to do, we need to find two factors whose product is positive 36 and whose sum is negative 13. So we’re gonna have zero is equal to 𝑥 minus nine multiplied by 𝑥 minus four. And we get that because nine multiplied by four is 36. And we’ve got negative nine and negative four because we got positive 36. But I need to sum to negative 13. So we know they both need to be negative. And then, you’ve got negative nine and negative four gives us negative 13. So great, we factored.

So therefore, we can say that the solution for 𝑥 is gonna be equal to positive nine or positive four. And the way we got this is because we want to find the value where the quadratic is equal to zero. So that means that one of our parentheses needs to be equal to zero. So we can set them equal to zero. So we get 𝑥 minus nine equals zero and 𝑥 minus four equals zero. So therefore, if we add nine to each side of the equation, we get 𝑥 equals nine. If we add four to each side of the equation, we get 𝑥 equals four. So therefore, we’ve got a final answer 𝑥 equals nine or four.