00:00 Intro

00:30 Solutions for systems of linear equations

01:11 Visualization of range and kernel

02:46 Existence needs range of A

03:01 Uniqueness needs kernel of A

03:13 The set of solutions is an affine subspace

04:18 Concrete description of the set of solutions

05:45 Proof of S = v + Ker(A)

08:52 Row operations don’t change the set of solutions

09:51 Recipe for finding the set of solutions

11:00 Credits

Q1: Which of the following solution sets is never possible for a system of linear equations $A \mathbf{x} = \mathbf{b}$?

A1: There are exactly two solutions.

A2: There is no solution.

A3: There are infinitely many solutions.

A4: There is exactly one solution.

A5: The solution set is a one-dimensional subspace.

Q2: Let $\mathbf{v}_0$ be a solution of $A \mathbf{x} = \mathbf{b}$ and $\widetilde{\mathbf{x}}$ be an element of $\mathrm{Ker}(A)$. Which claim is always correct?

A1: $\mathbf{v}_0 + \widetilde{\mathbf{x}}$ is also a solution of $A \mathbf{x} = \mathbf{b}$.

A2: $\mathbf{v}_0 + \widetilde{\mathbf{x}}$ is also an element of the kernel of $A$.

A3: $\mathbf{v}_0 + \widetilde{\mathbf{x}}$ lies in the range of $A$.

A4: $\mathbf{v}_0$ also lies in the kernel of $A$.

Q3: What is the solution set of the system $$ \left( \begin{array}{cccc|c}1 & 0 & 3 & -1 & 0 \ 0 & 1 & 1 & -1 & 0 \ 0 & 0 & 0 & 0 & 1 \ \end{array} \right) $$

A1: $\mathrm{S} = \emptyset$

A2: $\mathrm{S} = { 0 }$

A3: $\mathrm{S} = \left{ \begin{pmatrix} 2 \ 2 \ 1 \end{pmatrix} \right}$

A4: $\mathrm{S} = \left{ \begin{pmatrix} 2 \ 2 \ 1 \ 0 \end{pmatrix} \right}$

A5: $\mathrm{S} = \mathbb{R}^4 $