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Anton Chapter 2, Section 2.1, Question 26Evaluate det(A) by cofactor expansion along row or column of your choice465]A =...

Question

Anton Chapter 2, Section 2.1, Question 26Evaluate det(A) by cofactor expansion along row or column of your choice465]A =

Anton Chapter 2, Section 2.1, Question 26 Evaluate det(A) by cofactor expansion along row or column of your choice 46 5] A =



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Evaluate det $(A)$ by a cofactor expansion along a row or column of your choice. $$4=\left[\begin{array}{rrr} -3 & 0 & 7 \\ 2 & 5 & 1 \\ -1 & 0 & 5 \end{array}\right]$$

Okay, This question wants us to find this determinant in two different ways using co factors. So the 1st 1 wants us to find the determinant using the second row. So remember to find the determinant. All you do is at the co factors from a certain row or column. And if we want wrote to, then we want the co factors from Row two column one plus row to count to plus row two column three So now we need a calculator. Co factors. So the co factor for specific country is he there plus or minus one, depending on our checkerboard pattern times that matrix entry times the determinant of the minor matrix that you get from deleting RO. IE COLUMN J So applying this process. Row two column one in our checkerboard pattern plus minus, plus minus plus minus plus plus row two Column one has a as a minus sign. So we get a minus one times. Well, our entry zero So we can stop there because if you have a zero anywhere that co factor and tree just become zero, then Row two column too has a plus sign. Our entry is 12 and our determinant is found by deleting, bro, too account, too. So we get five negative 313 and then we get a minus Sign from our checkerboard for Row two Column three So we get a minus for times the determinant of delete the row. Delete. The column could do that better. So we get 5016 All right, now let's do the algebra. So we have 12 times 15 plus three minus four times 30 minus zero. So this is 12 times 18 minus 1 20 which gives us an answer of 96 for a determinant. Now it wants us to do a different expansion and show that the answers are still the same. So this time it wants column, too. So this means that are determinant is equal to Row One column too. Plus row two column too Plus row three columns. Ooh, co factors. Then if we do this, we get well. Row one column, too has a minus sign in our entries again. Zero so we don't care about our determinant is that just vanishes then Row two column too. We already found the co factor for that which we found to be 12 times 18 so to 16. And then for 1/3 co factor, we get another minus sign. Our entry is six. And our determinant that we're evaluating is zero for 13 We're sorry. Sorry, we're in this room. So five negative 304 And this is equal to 16. Minus six times 20. And this gives us again are same familiar answer of 96. So our algebra is probably right because both of our determinants matched regardless of what Colin and roll we picked to expand over.

In the question. We have to find the determinant off the given metrics by the method off expansion expansion Bye, go factors, right? Yeah, on expanding using the indicated row or column for the metrics minus 321 456 two minor street one now moving towards the solution for the given. My tricks a part which WAAS expand according to the rogue one will be minus three 56 minus 31 minus two 46 to 1 last one 45 to minus three, which will give minus three five plus 18 minus 24 minus 12 plus minus 12 minus 10 which will be equal to minus 75 now moving towards part B which waas expand along. Call them toe. So it will be minus two 46 to 1 plus five minus 3121 plus three minus +31 46 So the answer to this will be minus 75 On this will be the final solution to the given question. Thank you

All right, So here is our given matrix A. So if we, um, find the determinant by expanding along the first row Well, then we would have our interest in the first row of three 31 so we would have three times wealth times the determinant of its, um, minor, which would be, um, zero negative four. Native, 35 cross out the road and the column that his entry is in at the remaining entries. The 35 Okay. And then we have minus, um, three times. It's minor, which would be one negative four if 15 and then you have plus just one time. It's matter. Be +101 native. Three. Okay, so what we get here, get three times. Well, the trend, Harriet. Zero times 50 minus negative. Four times. Negative. Three. So that's minus a positive 12. So we have three times a negative 12 and then we have minus, um, all three times five times one is five minus native, four times. Once. That's five minus minus four. So five plus four is nine. So we have made minus three times nine, and then we have, um, plus just one times It's minor here. So we have minus one times minus three. It's minus three minus one times zero. So minus three times zero is minus three. So we get three, um times native 12 minus, um three times nine plus one times negative three, which is negative. 64 plus 10 plus 15 to determine here. Evaluates to negative 66.

In the question. We have to find the determinant off the metrics by using the method off expansion off the co factors. Expansion off oh factors. And we have to expand using the indicator toe or column for the given my trucks, which is minus 342 631 four minus seven minus eight. And in the a part, we have to expand along Roto on DBI part along column three. Now moving towards this solution a part along Roto for the given matter except will be. That's pension will be minus six in tow. Four toe minus seven minus Sade. Last three, minus three to fourth minus said less. Oh, it will be minus 344 minus seven, which will be quarto 108 plus 48 minus five. That is 151 now moving towards Part B. A Long column three, which will be 2634 minus seven minus minus 344 minus seven minus said minus 3463 which will give 1 15 month and this filled with a solution to the given question. Thank you


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