Algebra (all content)
- 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
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.
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- At0:20, what is meaning of det(c)?(20 votes)
- What if you got a non regular matrix ( det(c)=0 )). Then you can use this example ?(12 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.
Look at how Sal writes the adjugate for a 2x2 in the link below.
- 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)
- 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)
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.