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. 0.5 7 50.5 3 90.5 1 3

Accepted Answer

hey everyone in this problem we are asked to evaluate the following determinant. Okay. Using the expansion by minors. When we think of miners essentially thinking of taking the determinant of a smaller matrix made by removing a particular row and column of the matrix of interest. Okay. Alright. So we see this um let's choose to expand about the second column. Okay. Just because of that 0.25, if we have to multiply that with every determinant it's a little bit finicky but you can choose any row in any column. So let's choose the second column. As I said. Alright, so we're gonna get three values. Okay? And we have we can think of the matrix here. We have 0.25 0. 0.25 175. And I'm just gonna repeat this three times because we have a three by three matrix we're gonna have three minors. 2640.2 phone, Whoops. And we almost forgot the last column here. 175. There we go. And one last time to 640.25. And you don't need to write this out every time. Once you understand the process and you can do it more efficiently, you don't need to write this out every time. But it does help When you're doing this when you're starting to learn. Okay, so we're expanding along the second column. So we're gonna do is we're going to take the first value in the second column. Okay, so that's gonna be 0.25. We're gonna get rid of the rest of this row and column. Alright so this is gonna be our first major and we're going to label it M. 12. Okay it's in the first row in the second column. So M. 12. And the value of the major is just going to be the value Of the entry that we're at. So 0.25 times the determinant of the four by four matrix that's left over. The determinant of what's left over which is six four 75. Okay. So now instead of having to calculate determinant of this three by three matrix we just have to do a series of two by two determinant calculations which we know how to do and are just simpler. Okay so 0.25 times we're gonna get six times five. And when we do the determinant calculations six times five and then we're gonna subtract four times seven. Okay so six times five Subtract four times 7. You just multiply on the diagnose attracting the second. So we get 0.25 times two Which is gonna give us 0.5. Yes we found our first major. Now we can do the same for the second. So now we just moved down in the column we've chosen so we've chosen to do column two. So we've done the first entry. Now we do the second entry again we get rid of the row and the column that we're in. In this case we're in the second row and second column. So we have M22. That's going to be equal to the value there. 0.25 times the determinant of what's left over which is 2415. This is going to be 0.25 times two times five -4 times one. Which is going to give us 0. times six, Which is 1.5. And finally for our last major M in this case we're in the third row, second column. So we have em 32, Okay, we're going to be 0.25 times the determinant of what's left over when we eliminate the column and the row that we're in. Okay. And what's left over is 2617. And multiplying we get 0.25. The determinant two times 7 - times one. This is going to give us 0. times eight Which is equal to two. Okay, so we found our three majors, like going along the second column and now we just have to write the result. Ok, so the determinant of this three by three, matrix 2640. point 250. is equal to And we're going to have negative one to the exponents. I plus J. Okay, times our first major plus negative one times I plus J Times our second major plus negative one. I plus J. Times our third major. And let me explain what these eyes and J. R. Okay these are going to correspond to the row and column of that value we took. Okay so they're gonna be corresponding to these subscription. Rm. Okay so in the first major we have row one column two. So we're gonna have one plus two eight times m. 1 2 plus negative one. In the second one we have two plus two Cheechoo plus negative one. Three plus two and 32. Okay. And this is gonna tell us which of those values is going to be positive and which of those values is gonna be negative in our determinant calculation. So in the first case we have an exponent that's odd. So we're gonna end up with negative so we're gonna have negative 0.5. Okay. Two plus two. That's four. That's even. So we're going to end up with positive. Okay 1.5 and for the last 13 plus two that's five. That's odd. So we're gonna end up with negative two. This is going to give us a value of negative one. Okay and so the determinant of this three by three matrix is negative one. That's answer. C. That's it for this one. I hope this video helped see you in the next one

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In Exercises 23–30, use expansion by minors to evaluate each determinant. 0.5 7 50.5 3 90.5 1 3

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