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

Question

In Exercises 23–30, use expansion by minors to evaluate each determinant. $\begin{vmatrix}1 & 1 & 1 \2 & 2 & 2 \-3 & 4 & -5\end{vmatrix}$

Accepted Answer

Hey everyone. Welcome back in this problem. We are asked to evaluate the determinant K through the expansion by minors. All right, matrix were given as three by three. Okay. So in the first row we have 333. The second row. We have 444 and the third row we have two negative six and eight. All right. Now, when we're looking at this matrix and we're trying to determine which row or column. We want to do our expansion by minors from notice that the first two rows are the same here. Okay, so let's expand along the third row. Okay. And you'll see why we choose this in a minute. Alright, so again, we're doing our expansion by minors. What we're doing is we're essentially eliminating a roman column and taking the determinant of what's left over. Okay, so we have 34234 minus 6348. Okay. And I'm just gonna write this three times so that we can use it for or three miners that we will need to find. Okay, we have a three by three matrix. So we have three minors. Alright.  aN:aN:000NaN minus 68. Alright, beautiful. So for our first minor. Okay. We're expanding along the 3rd row. So we're gonna start here with the first value. Okay. We eliminate everything in that column in that row we take The value to we multiplied by the determinant of what's left in this case. It's three 434. Now we're gonna label this based on the row and column. So we're in the third row, first column. So this is going to be M31 and you'll see that we get two times well the determinant of a two by two. We multiply the diagonals. Okay? And subtract. So we get three times four minus three times four. Okay. Well that's just gonna be zero. Mhm. All right, so that 10. Let's move on to the second one. Okay, So we're going to be still in the third row but this time we're gonna be in the second column. Okay? So we're just working our way along that bottom row. We eliminate everything from that row and column. We multiply the value Times the determinant of what's left over. And in this case we get 334. Alright? And you'll notice that we have the same determinant as last time. Okay. We're gonna get three times 4 -3 times before we're going to get zero. And this is why we're expanding along that third row because those top two rows are equal. We're always gonna have this zero determinant. So it makes it easier if you didn't notice that right off the bat, that's okay, you can expand along any row or any column. You're going to get the same final answer. You just might have to do a little bit more arithmetic in between. Alright and last but not least. We're gonna be in the third row still and in the third column. Okay with this eight we're gonna get rid of everything in the row, Everything in the column. We're going to take the value and then multiply by the determinant of what's left over. Okay. And again here we're gonna get a determinant of zero. Alright, So when we're looking at the determinant of this total three by three, matrix three, 33 in the first row, 444 in the second row, and two negative 68 in the third row. Well this is going to be equal to negative one to the I. Plus J. Of our first major plus negative one to the I plus J. Of our second major plus negative one to the I. Plus J. Of our third major. Okay. All of our majors are zero. And so this determinant is just going to be equal to zero. Okay. And so the answer here is D the determinant is zero. That's it for this one. Thanks everyone for watching. See you in the next video.

In Exercises 23–30, use expansion by minors to evaluate each determinant. $\begin{vmatrix}1 & 1 & 1 \\2 & 2 & 2 \\-3 & 4 & -5\end{vmatrix}$

Key Concepts

Determinant of a Matrix

Expansion by Minors

Properties of Determinants

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