In Exercises 23–30, use expansion by minors to evaluate each dete...

Question

In Exercises 23–30, use expansion by minors to evaluate each determinant. 3 0 02 1 - 52 5 - 1

Determinant matrix for exercise 23 in college algebra, chapter 7 on systems of equations.

Accepted Answer

Hey everyone, Welcome back in this problem. We are asked to evaluate the following determinant through expansion by minors. Okay. We're asked to do expansion by minors. We want to choose one of the rows or columns to do the expansion through. Okay. We see that row two has these zeros. So let's go ahead and choose row two. Okay. And you'll see why um as we work through the problem, choosing a row or column with Zeros. Makes it a little bit easier. Okay, you can choose any row or any column, you'll get the same result, but this just will make it a little bit easier on yourself. Okay? So choose second row. All right. Now we're going to rewrite this matrix three times. Okay. One for each of our miners and it's just going to help us um work through the calculation for our miners. Okay. And we have three because this is a three by three matrix. Okay, so three by three Matrix. Three minors. Alright. And one last time writing out this matrix. Okay. Row one is one negative three negative two, row two is 500, and row three is 34 negative five. Okay, So for our first major we're starting on the left. Okay. We have the major and we're choosing to do the second row. So we're going to start with the first entry in the second row. And the subscript we're gonna use is going to correspond to the row and the column. Okay, so we're in the second row. So we're gonna have em too and we're in the first column. So M- 21. Alright. What we're gonna do is we're gonna take the value at that entry which is five. Okay then we're gonna get rid of everything in the same row and everything in the same column. We're gonna multiply by the determinant of the two by two matrix that's left over. So in this case the matrix that's left over is negative three negative two in the first row, four negative five in the second row. Okay determinative of two by two. Matrix we know how to do that. Okay recall that you just multiply the diagonals and subtract. So we get five times negative three times negative five minus four times negative two. This is going to give us five times minus negative. Eight minus minus is going to be positive. We get five times 23 which is 115. Okay so that's the first minor we found em 21. Alright, moving on to our next minor. Now we're just gonna move along that road that we chose. Okay so we're gonna move across to the next term in this case we're second row. Second column. So we have M. 22, we take the value at that entry and we multiply the determinant of the two by two matrix left over by eliminating the row and column that we're at and so we get one negative 23 negative five. Okay we're multiplying by zero. So it doesn't matter what this determinant is our minor will be zero. And then for the third minor in this case, again, working along that second row, eliminating everything in the column and row that we're in. Okay, We're in the 2nd row. 3rd column, our value is zero and the determine of what's left over. One negative 334. Okay, we're multiplying by zero. So again, it doesn't matter what this determinant is. We get a value of zero. Okay. And this is why it's helpful to choose a row or column that has zero in it because you'll end up with a lot of these terms where you get zero times some determinant. You don't even need to do the determinant calculations. Okay, so this will make that easier. Okay, so we have our M. 2 to 0 and rm 23 is zero. Alright, let's give yourself some more room. And we're going to do the final calculation. So we found all of our miners. Now we need to find the determinant of that three by three matrix that we were looking for. Ok, so the determinant of 153 in the first column, negative 304. In the second column, negative 20, negative five in the third column. This is going to be equal to negative one to the I plus J. Times our first minor. Okay. And I will explain what this I and J is in just a minute. Okay, I get negative one to the I. Plus J. Of our second minor and negative one to the I. Plus J. Of our final minor. Okay now the 2nd and 3rd minor there is zero. So these terms are going to go to zero. Okay The first one this I plus J. This corresponds to the row and column index. Okay so they're going to correspond with the subscript of our minor. Okay so we're at row two column one. So this is going to be negative I. Two. The two plus one. Okay. Row index plus column index of M. +21 negative one to the exponent three. Well that's gonna give us a negative. Okay so we have negative one times 115. Which is gonna give us a determinant of negative 115. Okay so that's the determinant of the three by three matrix we were given that's going to correspond with answer. D. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 23–30, use expansion by minors to evaluate each determinant. 3 0 02 1 - 52 5 - 1

Key Concepts

Determinants

Expansion by Minors

Cofactors

Watch next