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Current time:0:00Total duration:9:13

I have this 4 by 4
matrix, A, here. And let's see if we can figure
out its determinant, the determinant of A. And before just doing it the way
we've done it in the past, where you go down one of the
rows or one of the columns-- and you notice, there's no 0's
here, so there's no easy row or easy column to take
the determinant by. We could have gone down this
row and do all the submatrices, but this becomes
pretty involved. You end up doing four 3 by 3
determinants, and then each of those are composed of three
2 by 2 determinants. It becomes a pretty
hairy process. Let's see if we can use some
of the realizations we've discovered in the last
two videos to simplify this process. Well, one of the realizations is
that row operations, or if you subtract-- let me write it
this way-- if you replace row j with, let's say, row j minus
some scale or multiple, times row i, it does not change
the determinant of A. We saw that, I think it
was two videos ago. So this is a pretty
big realization. We can do these type of row
operations and it won't change the determinant. The other realization we had
was that these upper triangular matrices, you can
figure out their determinant. So what does upper triangular
look like? Let me just review it. The upper triangular--
everything below the diagonal. So let's say the diagonal has--
let me just draw its terms like that. These are some non-zero terms.
Oh, they don't have to be. Then upper triangular,
everything below the diagonal is a 0, and everything above the
diagonal probably isn't a 0, but you never know. But they're non-zero terms, so
all the red stuff here is non-zero, all this stuff
in green is 0. And I didn't touch on it in that
video, but there is also such a thing as a lower
triangular, that you might have guessed how it looks. Everything above the main
diagonal is 0, so this is the main diagonal right here, all
the way down like that. All of these guys are going
to be non-zero. All of that's going to be
non-zero, and then the 0's are going to be above the
diagonal, like that. We saw in the last video that
the determinant of this guy is just equal to the product of the
diagonal entries, which is a very simple way of finding
a determinant. And you could use the same
argument we made in the last video to say that the same is
true of the lower triangular matrix, that its determinant
is also just the product of those entries. I won't prove it here, but you
can use the exact same argument you used in the video
that I just did on the upper triangular. So given this, that the
determinant of this is just the product of those guys, and
that I can perform row operations on this guy and not
change the determinant, maybe a simpler way to calculate this
determinant is to get this guy into an upper
triangular form, and then just multiply the entries
down the diagonal. So let's do that. So we want to find the
determinant of A. Let me rewrite A right here. It's 1, 2, 2, 1, 1, 1, 2, 4,
2, 2, 7, 5, 2, minus 1, 4, minus 6, 3. Now let's try to get this into
upper triangular form. So let's replace the second
row with the-- so I'm just going to keep the first
row the same. 1, 2, 2, 1. And let's replace the second row
with the second row minus the first row. The second row minus the first
row is going to be equal to 1 minus 1 is 0. So in this case the constant
is just 1. So 1 minus 1 is 0. 2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row
with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. Let me get a good color here. I'll do pink. Let's replace the last row
with the last row, essentially, plus
the first row. You could say minus minus 1
times the first row is the same thing as the last row
plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6. Minus 6 plus 2 is minus 4. And then 3 plus 1 is 4. So there we have it like that. And this guy has two 0's
here, so maybe I want to swap some rows. So let me swap some rows. So if we swap rows,
what happens? I'm going to swap the middle
two rows just for fun. Well, not just for fun. Because I want a pivot
entry right here. I shouldn't say a pivot entry. I want to do it in upper
triangular form. So I want a non-zero
entry here. This is a 0, so I'm going
to move this guy down. So I'm going to keep the
top row the same. 1, 2, 2, 1. I'm going to keep the
bottom row the same. 0, 0, 6, minus 4, 4. And I'm going to swap these
guys right here. So this is going to
be 0, 3, 1, 0. And then 0, 0, 2, 1. Now, can I just swap
entries like that? Well, I can, but you have to
remember that when you swap entries, your resulting
determinant is going to be the negative of your original
determinant. So if we swap these two guys,
the determinant of this is going to be the negative
of this determinant. When you swap two rows, you
just flip the sign of the determinant. We saw that. That was one of the first videos
we did on these, kind of messing with the
determinants. Now, what do we want
to do here? To get this guy into upper
triangular form, it'd be nice to get this to be a 0. So to get that to be
a 0, let me keep everything else the same. So I have a 1, 2, 2, 1. I have a 0, 3, 1, 0. The third row is 0, 0, 2, 1. And now this last row, let me
replace it with the last row minus 3 times this row. So let me write it like this. I have to carry that negative
sign as well. So I'm going to replace this
last row with the last row minus 2 times the second row. want to zero it out. So 0 minus 2 times 0 is 0. 6 minus 2 times 3 is 0. Minus 4 minus 2 times
1 is minus 6. And then 4 minus 2 times
0 is just 4. We're almost there. Now we want to zero this
guy out, so let's replace this one. So I'm going to keep my top
three rows the same again. And let me see if I can write
it a little bit neater. So my first row is 1, 2, 2, 1. My second row is 0, 3, 1, 0. Fourth row is 0, 0, 2, 1. And I'm going to take
the matrix. I haven't written the
fourth row yet. And, of course, the negative
of this is going to be determinant of our original
matrix, because we had swapped those rows. But let's replace this last row
with the last row, plus 3 times the third row. So we get 0 plus
3 times 0 is 0. 0 plus 3 times 0 is 0. Minus 6 plus 3 times 2 is 0. 4 plus 3 times 1 is 7. And just like that, we have a
determinant of a matrix in upper triangular form. So this is going to be equal to
the product of these guys. We can't forget our
negative sign. Let's throw our negative sign
out there and put a parentheses just like that. This is going to be
the product of that diagonal entry. 1 times 3, times 3, times 2,
times 7, which is 6 times 7, which is 42. So the determinant of this
matrix is minus 42, which was pretty fast. This was a
pretty fast shortcut. And it actually turns out it
tends to be computationally more efficient to use these
takeaways to put things into upper triangular form first. And
then, you know, if you do swaps, you have to remember to
make the determinant negative. And then just multiply
down the diagonal. And we did that there, and
we got the determinant as being minus 42.