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### Course: Algebra (all content) > Unit 20

Lesson 15: Determinants & inverses of large matrices- Determinant of a 3x3 matrix: standard method (1 of 2)
- Determinant of a 3x3 matrix: shortcut method (2 of 2)
- Determinant of a 3x3 matrix
- Inverting a 3x3 matrix using Gaussian elimination
- Inverting a 3x3 matrix using determinants Part 1: Matrix of minors and cofactor matrix
- Inverting a 3x3 matrix using determinants Part 2: Adjugate matrix
- Inverse of a 3x3 matrix

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# Inverting a 3x3 matrix using determinants Part 2: Adjugate matrix

Sal shows how to find the inverse of a 3x3 matrix using its determinant. In Part 2 we complete the process by finding the determinant of the matrix and its adjugate matrix. Created by Sal Khan.

## Want to join the conversation?

- At0:20, what is meaning of det(c)?(20 votes)
- det(c) is the abbreviation for "the determinant of c."(5 votes)

- What if you got a non regular matrix ( det(c)=0 )). Then you can use this example ?(12 votes)
- A matrix has no inverse if its determinant is zero.(55 votes)

- If matrices were defined by humans, why on earth would we design such a convoluted way to find the inverse of a matrix? What problems does this complicated process solve?(9 votes)
- If you could provide a less convoluted solution, then I think that you might have the Field's medal coming your way in the future :-)(19 votes)

- What if the determinant is 0? Wouldn't that make everything in the inverse matrix undefined?(6 votes)
- Yes. The inverse is undefined.(since we don't know how to divide by 0)(10 votes)

- Why is the adjugate (transpose) for a 2x2 matrix different from that of a 3x3? Wikipedia shows the 3x3 method shown here as applicable for a 2x2 as well.

http://en.wikipedia.org/wiki/Transpose

Look at how Sal writes the adjugate for a 2x2 in the link below.

https://www.khanacademy.org/math/algebra/algebra-matrices/inverting_matrices/v/inverse-of-a-2x2-matrix(3 votes)- An adjugate matrix is the transpose of the cofactor matrix, not of the original square matrix.(11 votes)

- Is there a general method of inverting n*n matrix?(5 votes)
- As far as I understood this methode can be used as a general method of inverting an n*n matrix. (But of course the technique sal used to get the determinant can't be used for n*n matrices, so you would have to find another technique there.)(6 votes)

- 4:06: Wouldn't it be easier to say TRANSPOSE instead of ADJUGATE?! At least would be better to say that T is something that is more used when it comes to studying matrices more? (Í know its the same thing but might confuce students like me who are having exam about this and never heard of adjugate and instead of TRANSPOSE T)(2 votes)
- Adjugate of matrix A is "Transpose of Cofactor matrix of A"(7 votes)

- What is the Gauss–Jordan elimination? (mentioned by Aron Radojčić). Is there a video of it in Khan Academy ? Thanks.(4 votes)
- Can this same strategy be applied to solve for the inverse of a 4x4 matrix? I imagine there will be more intermediate steps, but is the principle the same?(3 votes)
- Yes the process is the same. The method Sal uses for the determinant of a 3x3 does not have a pretty analog in the 4x4 case however, so you would need to calculate the determinant using the other method.(3 votes)

- Finally, I got the point. This lesson almost made me cry.(3 votes)

## Video transcript

We're nearing the home
stretch of our quest to find the inverse of this
three-by-three matrix here. And the next thing
that we can do is find the determinant
of it, which we already have a good bit
of practice doing. So the determinant
of C, of our matrix-- I'll do that same
color-- C, there are several ways
that you could do it. You could take this
top row of the matrix and take the value of
each of those terms times the cofactor-- times
the corresponding cofactor-- and take the sum there. That's one technique. Or you could do
the technique where you rewrite these
first two columns, and then you take the product
of the top to left diagonals, sum those up, and
then subtract out the top right to
the bottom left. I'll do the second
one just so that you can see that you
get the same result. So let's see. The determinant is going
to be equal to-- I'll rewrite all of these
things-- so negative 1, negative 2, 2,
2, 1, 1, 3, 4, 5. And let me now, just to make
it a little bit simpler, rewrite these first two columns. So negative 1,
negative 2, 2, 1, 3, 4. So the determinant
is going to be equal to-- so let
me write this down. So you have negative
1 times 1 times 5. Well that's just going
to be negative 5, taking that product. Then you have negative
2 times 1 times 3. Well that's negative 6. So we'll have negative 6. Or you could say plus
negative 6 there. And then you have
2 times 2 times 4. Well that's just 4 times
4, which is just 16. So we have plus 16. And then we do the top
right to the bottom left. So you have negative
2 times 2 times 5. Well that's negative 4 times 5. So that is negative 20. But we're going to
subtract negative 20. So that's negative 4
times 5, negative 20, but we're going to
subtract negative 20. Obviously that's going to
turn into adding positive 20. Then you have negative 1 times
1 times 4, which is negative 4. But we're going to
subtract these products. We're going to
subtract negative 4. And then you have 2 times
1 times 3, which is 6. But we have to subtract it. So we have subtracting 6. And so this simplifies
to negative 5 minus 6 is negative 11, plus 16
gets us to positive 5. So all of this
simplifies to positive 5. And then we have plus 20 plus 4. Actually, let me do
that green color, so we don't get confused. So we have plus
20 plus 4 minus 6. So what does this get us? 5 plus 20 is 25, plus 4 is
29, minus 6 gets us to 23. So our determinant right
over here is equal to 23. So now we are really
in the home stretch. The inverse of this
matrix is going to be 1 over our determinant
times the transpose of this cofactor matrix. And the transpose of
the cofactor matrix is called the adjugate. So let's do that. So let's write
the adjugate here. This is the drum roll. We're really in
the home stretch. C inverse is equal to
1 over the determinant, so it's equal to 1/23,
times the adjugate of C. And so this is going to be equal
to 1/23 times the transpose of our cofactor matrix. So we have our cofactor
matrix right over here. So each row now
becomes a column. So this row now
becomes a column. So it becomes 1, negative 7,
5 becomes the first column. The second row becomes
the second column-- 18, negative 11, negative 2. And then finally, the third
row becomes the third column. You have negative 4, 5, and 3. And now we just
have to multiply, or you could say divide, each of
these by 23, and we are there. So this is the inverse of
our original matrix C, home stretch. 1 divided by 23 is just 1/23. Then you have 18/23. Actually, let me give myself
a little bit more real estate to do this in. So there we go. So 1 divided by 23-- 1/23,
18/23, negative 4/23, negative 7/23, negative 11/23,
5/23, 5/23, negative 2/23. And then finally,
assuming we haven't made any careless mistakes,
which would shock me if we haven't, we get to 3/23. And we are done. We have successfully inverted
a three-by-three matrix. Once again, something I
strongly believe better done by a computer and
probably should not be part of a typical Algebra
2 curriculum, because it tends to be displayed in a,
non-contextual way.