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 1 0- 3 4 0- 1 3 - 5

Determinant exercise for college algebra, chapter 7, question 25 with a 3x3 matrix.

Accepted Answer

Hey everyone, welcome back in this problem. We are asked to evaluate the following determinant through the expansion by minors. When we want to expand by minors, we want to choose one row or one column to do the expansion by in this case we have zeros in the first column. And so let's choose to expand through the first column. Okay. And you'll see why in a minute choosing a column or a row with zeros makes it a little bit easier. You can choose any column or any row. All right. So I'm going to rewrite the matrix that we're given three times so I can show it what we're doing here. Okay? So we have 00 negative two in the first row, negative four negative 65 in the second row And 3 1 -7 in the 3rd row. Reading This it one more time. And you don't need to write this out every time that you do a problem like this. Okay. But it can help if you get a little bit confused or just when you're learning can help to write it out and just keep your work organized. All right, So, we're starting with the first column. We're expanding along the first column. So choose the first entry in the first column are minor. We're gonna subscript 11. Okay, Because we're in the first row in the first column. This is just going to be equal to the value that we're at, which is zero. Okay, then we're gonna get rid of everything else in the same row. Everything else in the same column and take the determinant of the two by two matrix that's left over. In this case it gets it's minus 65 in the first column and one negative seven in the second column. Alright well we're just multiplying whatever this determinant is by zero. So we don't even need to calculate this determinant, we're just going to end up with zero. All right now the second one, we're gonna just work down along the column network expanding. Okay, so we're gonna be here, we're going to be in row two column one. So we're going to have M 21. Okay and again the value of the entry which is zero. Okay. Get rid of everything in the same row, everything in the same column. And multiplied by the determinant of the two by two matrix that's left in the first row we get minus 43 in the second row we have five negative seven. Okay? And again we're multiplying by zero so it doesn't matter what the value of the determinant is. We're going to get zero. Alright, last step. Okay we've done the first two rows. Now we have this last value in the first column. Okay so we have M 31. The third row. The first column, our value is negative two. Okay. We eliminate everything in the same row, everything in the same column and we take the determinant of the remaining two by two matrix negative 43 negative 61. Taking the determinant. A two by two matrix we multiply. Okay. The corners and subtract. And so we're gonna get negative four times one minus negative six times three. This is going to give us negative two times negative four plus 18. For a final minor of - minor 3 1. Alright so now we have all of our miners and we just need to calculate the determinant that we were asked to find. Alright, so giving ourselves more room. The determinant we were asked to find is a determinant of row 10 -43. Okay. Row 20-6 1. Row 3 -25 -7. Well we're gonna get us we're gonna have a negative one. We get to the exponents I plus J. And I'll explain in a minute why we use I plus J Times our first major m. 1 1 plus negative one. I plus J. Of our second Major M. +21 plus negative one. I plus J. Times are third Major M. 31. Okay. And these I's and J's represent the column and row that we're at. Okay. And so they're going to correspond to the subscript we've used for our miners and so for the first one we have negative one. We're in column one. Row one. So we get to the exponent one plus one. Many times are minor which is zero. Okay. Second one. Same thing two plus one are minor zero. And in the third one we have three plus one with a minor of negative 28. Okay so these first two terms they just go to zero. We're left with negative one to the exponent three plus one. That's gonna be four. That's even. So we're gonna get a positive value times a negative. So we're gonna get negative 28. Okay so our determinant is negative 28. That's gonna correspond with answer. A. Okay. Thanks everyone for watching. See you in the next video.

In Exercises 23–30, use expansion by minors to evaluate each determinant. 3 1 0- 3 4 0- 1 3 - 5

Key Concepts

Determinants

Expansion by Minors

Cofactors

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