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## Precalculus

### Unit 7: Lesson 14

Finding inverses of 2x2 matrices# Finding inverses of 2x2 matrices

Sal gives an example of how to find the inverse of a given 2x2 matrix. Created by Sal Khan.

## Want to join the conversation?

- What if the determinant is 0? Is the answer undefined?(67 votes)
- When the determinant for a square matrix is equal to zero, the inverse for that matrix does not exist.

So yes, there is no inverse if the determinant is 0.(86 votes)

- Shouldn't A and A' be separeted from each other when writing the equations? Cause A couldn't be equal to its inverse...?(15 votes)
- You are right. After he wrote A=, he should have started a new line and wrote A'=....and then continued with everything he does. However, it's clear what Sal is doing if you listen to what he is saying, but it would get confusing if you were just looking at the equations.

Applied mathematicians (physicists, computer scientists, engineers etc) are often sloppy with mathematical notation. The emphasis is often on getting the result rather than on the rigorous proof of each step. If you search elsewhere online, you will find that there is a lot of criticism for Khan Academy for its inconsistant use of proper mathematical notation. (But imperfect as it may be, it still is making math accessible to lots of people)(52 votes)

- Inverse of a 3x3 matrix(12 votes)
- What Sal introduced here in this video, is a method that was 'woven' specially for finding inverse of a 2x2 matrix but it comes from a more general formula for determining inverse of any nxn matrix
**A**which is:**A**⁻¹ = 1/det(**A**) * adj(**A**)

where adj(**A**) - adjugate of**A**- is just the transpose of cofactor matrix**C**ᵀ.

Cofactor matrix**C**of matrix**A**is also nxn matrix whose each entry (Cᵢ,ⱼ for example) is the determinant of the submatrix formed by deleting the i-th row and j-th column from our original matrix**A**multiplied by (-1)^(i+j).

Saying all of that, let's try it on 3x3 matrix. Suppose we have a matrix**B**

⌈a b c⌉

| d e f |

⌊g h i ⌋

the first thing is to find determinant of**B**:

det(**B**) = a(ei - fh) - b(di - fg) + c(dh - eg)

second thing is to find the cofactor matrix**C**of**B**(which takes a lot of effort):

⌈C₁,₁ = (-1)² * (ei - fh) C₁,₂ = (-1)³ * (di - fg) C₁,₃ = (-1)⁴ * (dh - eg) ⌉

|C₂,₁ = (-1)³ * (bi - ch) C₂,₂ = (-1)⁴ * (ai - cg) C₂,₃ = (-1)⁵ * (ah - bg)|

⌊C₃,₁ = (-1)⁴ * (bf - ec) C₃,₂ = (-1)⁵ * (af - cd) C₃,₃ = (-1)⁶ * (ea - bd)⌋

finally, we transpose our last matrix**C**to get the adjugate of**B**:

⌈(ei - fh) (-bi + ch) (bf - ec) ⌉

|(-di + fg) (ai - cg) (-af + cd)|

⌊(dh - eg) (-ah + bg) (ea - bd)⌋

so**B**⁻¹ = 1/det(**B**) * adj(**B**)

usually, you let the computer calculate the inverses for you.(16 votes)

- how does inversing a matrix help?(9 votes)
- You can add, subtract, and multiply matrices. Division, however, is not defined. Thus to undo matrix multiplication, you need to multiply by the inverse matrix. It is thus a pretty fundamental operation.

One early application for inverse matrices is to solve systems of linear equations. You can express the system as a matrix equation AX=B, then solve it by multiplying by the inverse of the coefficient matrix to get X = A^(-1)*B(16 votes)

- What are some of the practical applications for this?(3 votes)
- Do you like Pixar films?

Matrix math is used intensively in making 2D and 3D graphics - they have developed special chips just to do this type of math as fast as possible. Back when the original Toy Story came out, it took**24 hours**just to do the math (using matrices) to render**one**frame of animation.

Matrices are also a great way to solve simultaneous systems of linear equations. These are used in as many different fields as you can think of, from medicine and all its applications, finances, forecasting, trend analysis, system optimization and on and on.

Since there is no direct way to do matrix division, we take the inverse of a matrix and then multiply the matrix we want to divide by the inverse, just like 3÷2 = 3×½.(17 votes)

- How can we verify the answer?(8 votes)
- You can check your work by multiplying the inverse you calculated by the original matrix. If the result IS NOT an identity matrix, then your inverse is incorrect.

If A is the matrix you want to find the inverse, and B is the the inverse you calculated from A, then B is the inverse of A if and only if AB = BA = I(7 votes)

- Is there a video on Khan Academy where Sal proves the inverse matrix equation? If there is, I'd like to know.(9 votes)
- why do we have to take the adjugate (sp?) of A to find the inverse? How does this come about?

Is there a video where this is explained someone could direct me to?(6 votes)- Hi Mel. Go here: https://www.khanacademy.org/math/algebra/algebra-matrices/inverting_matrices/v/inverse-matrix--part-1

This video goes into the whole idea of inverting matrices. :)(2 votes)

- Can calculators work out matrices?(5 votes)
- Yes, if you are using a TI-84 you can click 2nd inverse and it will take you to the matrix menu. Here you can use the edit menu to create different matrices and you can use the main menu to use the matrices in any way that you want.(1 vote)

- i'm applying this solution in the unit tests and none of the solutions worked because the other solutions are solved using the gaussian approach; i had to calculate the inverse of [[10 0][-8 1]] and i got a wrong answer: i had 1/18 [[1 0] [8 10]].(5 votes)

## Video transcript

Let's attempt to take the
inverse of this 2 by 2 matrix. And you'll see the
2 by 2 matrices are about the only
size of matrices that it's somewhat pleasant
to take the inverse of. Anything larger than that,
it becomes very unpleasant. So the inverse of
a 2 by 2 matrix is going to be equal to 1 over
the determinant of the matrix times the adjugate
of the matrix, which sounds like a very fancy word. But we'll see for by a 2 by 2
matrix, it's not too involved. So first let's think about what
the determinant of this matrix is. Well, we've seen this before. We just look along
the two diagonals. It's 3 times 2 minus
negative 7 times 5. So this is going to be equal to
1 over 3 times 2 minus negative 7 times 5. And then the adjugate
of A-- and here I'm really just teaching
you the mechanics of it. And it's a little unfortunate
that in a typical algebra II class you kind of just go
into the mechanics of it. But at least this will get
us to where we need to go. So the adjugate of
A, you literally just need to swap the two
elements on this diagonal. So put the 2 where the 3 is
and the 3 where the 2 is. So this element right
over here, this 3 will go right over there. This 2 will go right over here. And then these two elements, you
just take the negative of them. So the negative-- let
me do a new color. Actually, I'm running
out of colors. The negative of
that is negative 5. The negative of
that is positive 7. So we are left
with-- this is going to be equal to 1
over-- 3 times 2 is 6. Negative 7 times
5 is negative 35. But then we have this
positive over here. So this whole thing
becomes plus 35. So 6 plus 35 is 41. So the determinant
of our matrix is 41. We're going to take 1
over the determinant and multiply it times our
adjugate, times 2, negative 5, 7, and 3. So we get -- so this is the
drum roll part -- 2 over 41, negative 5 over 41. I'm just multiplying each of
these elements times 1 over 41. 7 over 41, and 3 over 41. And we are done.