Evaluate each determinant in Exercises 49–52.

- 2 - 3 3 5
1 - 4...

Question

Evaluate each determinant in Exercises 49–52. - 2 - 3 3 5 1 - 4 0 0 1 2 2 - 3 2 0 1 1

Matrix for Exercise 51 in college algebra, chapter on systems of equations.

Accepted Answer

welcome back. I am so glad you're here. We are given a four by four matrix. I'll copy that a little bit bigger. So the first column negative 4133. The second column negative six negative 240. The third column 6042. And the fourth column 70 negative 6, 10. And we are asked to solve for the determinant of this matrix. All right. We need to recall previous lessons on solving the determinant of a four by four matrix. What we want to do is choose any row or column. And then we are going to expand it and we are going to multiply each term in that row or column by its minor and it's a little bit easier when we choose a row or a column that has a lot of zeros in it because zero times anything is going to be zero. And then we don't have to do a lot of work on those. We look at this and we've got this second row that has two zeros in it already and we can do some column addition in order to get a third zero there. So looking at column one and column two, we can replace column two By multiplying column one By a negative or by a positive two. Excuse me by positive too. And then adding that to column two. So because we are replacing column two we can copy over columns 13 and four. So I'll do that negative 4133 is column one, column three is 6042, column four is 70 negative 6, 10. And now doing the work to replace column two. We're taking column one times two. So negative four times two is negative eight plus negative six is negative 14. One times two is two plus negative two is the zero that we want. Ye three times two is six plus four is and three times two is still six plus zero is six. Okay, so now we've got this beautiful row two and recalling those previous lessons. What we're going to do is we're going to take that number. We're going to take each number in that row and we'll start with one. We're going to take one and we're going to multiply that by negative one raised to the exponent of the number row and column of the position of the number we're looking at. So that one is in the first column. So one plus and it's in the second row, one plus two. Okay, so that's the first part. And now this is all getting multiplied by the minor matrix to find that we're going to cross off the rest of the numbers that are in the first column and second row and we're going to create a minor matrix using the numbers that aren't crossed off. So our first column will be starting in the upper left of any numbers that aren't crossed off yet. So negative 14. And then going down 10 6. The second column will be 64 to the third column will be seven negative 6, 10. And that's the minor. Now we would be doing this for each position in the row that we've chosen to expand. We would be doing this same thing with the zero we'd say okay, zero times negative one and then that's in the second row, second column and then we would create a minor for it. However, zero times anything is just going to be zero. So that's why we wanted to create as many zeros in this row as possible because we don't have to do the work for the rest of this row, it's going to be zero. So we now have this simplified, we can simplify what's in front of our minor matrix one times negative one raised to the one plus two. Well one plus two, that's three negative one raised to the third is still negative one, so one times negative one is negative one. So it will be a negative one times reading the columns negative 14 10, 66427 negative 6, 10 for our minor matrix. And now we are going to go through the same process with our three by three matrix. We're going to choose a row or a column. We're going to expand it and it's a lot easier if our row or column has mostly zeros in it and just one number that we have to work with. And looking at this third row here, the two in column two can go into both six and 10. So we can use column two to get the six and the 10 into zeros. So let's start off replacing column one. We're going to do that by multiplying column two times negative three and then adding that to column one. Then we'll have a zero in this position where the six is because we're replacing column one, we can copy over columns two and three. So I'll do that 642 and seven negative 6, 10. And now doing the work to replace Column one. So negative three times six is negative 18 plus negative 14 gives us negative 32 four times negative three is negative 12 plus 10 is negative two, two times negative three is negative six plus six is the zero that we want. All right now let's work on getting a zero in the position where this 10 is. We will replace column three in order to do that. We'll take column two times negative five and add that to column three. Because we're replacing column three, we can copy over columns one and two. So we have negative 30 to negative 20642. And now doing the work to replace column three six times negative five is negative 30 plus seven is negative four times negative five is negative 20 plus negative six is negative 26 and two times negative five is negative plus 10 is the zero that we want. Ye all right now we have this beautiful row three that when we expand it multiplying those zeros by our negative one raised to a power and our minor will just be zero. So we only have to deal with this too. So to expand that we're taking to multiplying it by negative one, raise to the power of the number row and column added together. So this is the second column and the third row. And now creating our minor using the numbers that aren't part of the second row and the third column. So we cross those off. So starting in the upper left hand corner of numbers that aren't crossed off. We have negative 32 going down negative two And then our second column will be -23 and -26. We can simplify what's in front and it's really important to recall that this whole thing is still being multiplied by that negative one in front but we can simplify what's here. So negative one raised to the two plus 32 plus three is five, negative one to the fifth. That's still negative one. So we have two times negative one. Well that's negative two. So we've got a negative one times a negative two times it's minor which is negative 30 to negative two and negative 23 negative 26. And next we've got a two by two matrix. Yeah, because with the two by two matrix we can much more quickly solve for its determinant and then just do the multiplication. So I'm going to rewrite just what's inside here, negative 30 to negative two and negative 23 negative 26. And doing the work to expand this we take negative 32 times negative 26 that is 832. Then we take negative two times negative 23 and that is a positive 46. Then we are going to subtract those two numbers. So 832 -46 gives us 786. Okay, that's what's inside here in the green. So that is 700 Oops. I wrote 768 and said 786. It's 786. Okay, so 786, times negative two Also Times -1. And doing what's inside -2 times is negative. 1572 And that's times the -1. And that's just going to give us a positive 1572. That's our answer. That is the determinant of that. Four by four. Matrix 1572. So we scroll back up here where you can see all the answer choices and that matches with answer choice. C Well done. We'll catch you on the next one

Evaluate each determinant in Exercises 49–52. - 2 - 3 3 5 1 - 4 0 0 1 2 2 - 3 2 0 1 1

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