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# Exploring the solution set of Ax = b

Exploring the solution set of Ax=b (non homogeneous equations). Created by Sal Khan.

## Want to join the conversation?

• What is a homogeneous equation vs. a non-homogeneous equation? • Is that always the case Sal, for a non-homogenous linear equation? or are there exceptions? • When Sal tries to define the actual vectors in the domain that are mapped to the co-domain, using the b1 as a variable, isn't it corresponding to the inverse of the transformation? And we know that this transformation is NOT invertible! • I think the distinction is that to be the inverse of a transformation you need to map to a unique x in the domain. Instead, here we are mapping to an infinite number of x's in the domain that fall on the (5, 0) + x2(3, 1) line for that particular b (5,-5).

Also in this case if you took the following A^(-1)Ax you should get x back if A in invertible, but in this case you would not get x back (if this were even possible) because you would instead get a line.
(1 vote)
• So I came across 2 questions that said,

2x+3z=4
-x+4y-2z=1
x-3y+5z=2

write in the form of ax=b
does that basically mean:

2 0 3 x = 4
-1 4 -2 y = 1
1 -3 5 z = 2

and then it asks what is the condition that allows matrix A to be inverted?

What does that mean?

Thank you.
(1 vote) • `⎡ 2 0 3⎤⎡x⎤ ⎡4⎤`
`⎢-1 4 -2⎥⎢y⎥ = ⎢1⎥`
`⎣ 1 -3 5⎦⎣z⎦ ⎣2⎦`
This is the Ax = b form. The brackets are important, indicating which set is A, x, and b respectively. A is the 3x3 matrix containing the 9 numbers.

By the definition of invertibility, A is considered invertible if there exists a matrix `B`, such that `AB = BA = I` where `I` is the identity matrix (in this case the 3x3 identity). All this means for our problem is that If we can solve for x, y, and z (all three, not just some of them), then the matrix is invertible. The easiest way to solve for x, y, and z is to augment the matrix and put it into some row echelon form (Sal prefers reduced row echelon form):
`⎡ 2 0 3 | 4⎤`
`⎢-1 4 -2 | 1⎥`(this is the augmented matrix)
`⎣ 1 -3 5 | 2⎦`
(skipped some legwork with the magic of code; put into rref)
`⎡1 0 0 | 0.92⎤`
`⎢0 1 0 | 0.84⎥`
`⎣0 0 1 | 0.72⎦`
You'll notice that the A portion of our augmented matrix now looks like the identity matrix with 1's down the diagonal and 0's everywhere else. This alone proves that there is a matrix `B` such that `AB = BA = I`, so the matrix is invertible. However, putting the matrix in this form has the added bonus of solving for x, y, and z for us (which is the right column from top to bottom).
• so is this example one-to-one? I would think it is since it's null space is zero but from the geometric interpretation it does not look like it's one-to-one. I'm confused
(1 vote) • In this example, the null space isn't the zero vector and it's not one-to-one.
Given the equation:
`T(x) = A x = b`
All possible values of b (given all values of x and a specific matrix for A) is your image (image is what we're finding in this video). If b is an Rm vector, then the image will always be a subspace of Rm. If we change the equation to:
`T(x) = A x = 0`
Then all possible values of x (we have to solve for x) is known as the nullspace of A, or as the kernel of T. If x is an Rn vector, then the nullspace of A will always be a subspace of Rn.
• At about the vector x1,x2 is set equal to the vector b1,0 plus x2 times the vector 3,1. Does this imply the x1,x2 is equal to the augmented matrix specified earlier in the lecture?
(1 vote) • At and I'm really confused on what exactly is the matrix and what is the vector and by not knowing that it gets me confused throughout the whole video.Can someone help me understand.
(1 vote) • What is the solution set of X+2=2+X
(1 vote) • If a non-homogeneous linear system Ax=b has the solution set x_p+s(u_1)+t(u_2), where x_p is a particular solution vector of Ax=b, and s(u_1)+t(u_2) is an arbitrary vector in the solution space of Ax=0, then the dimension of the null space will be 2, but will the solution set be a vector space as well and have a dimension?
(1 vote) • Around Sal talks about how to find the solutions when a row of your matrix is 0. But in the last video, "Determining whether a transformation is onto," Sal said that case would imply that there are no solutions. The case I mean is when you have a zeroed out row in your matrix but there's some expression of your b variables in the augmented part. What am I not getting?
(1 vote) 