Matrices Ax=b solutions

Question

For each of the following matrices $A$, deduce whether $A\vec {x} = \vec {b}$ has solutions for every $\vec {b}$ ∈ $\mathbb {R}^3$ or only for some $\vec {b}$:

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 3 & 3 \\ 3 & 4 & 5 & 6 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 3 \\ 3 & 3 & 3 & 4 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 \end{bmatrix}$

Here's my attempt at the second matrix. I really don't understand how to go about doing these!

Any tips or advice?

Answer 1

When you left-multiply a vector by a matrix, the result is a linear combination of the columns of the matrix. So, the equation $A\vec x=\vec b$ can be solved iff $\vec b$ is in the column space of $A$, that is, in the span of the columns of $A$. If this span is smaller than the codomain, then there will be some vectors $\vec b$ for which the equation has no solution. On the other hand, if the columns span the entire codomain, then there will be a solution for all $\vec b$.

This lets us answer the question for two of the matrices without much work. These matrices have three rows each, so their codomains are three-dimensional. All of the columns of the last matrix are identical, so its column space is one-dimensional. Three of the columns of the second matrix are identical, so its column space is at most two-dimensional.

The first matrix requires a bit more work. You’ll have to find the dimension of its column space (also known as the rank of the matrix), which you can do by row-reducing it. The rank of the matrix is the number of non-zero rows in the result. If there are any non-zero rows, then the matrix doesn’t have full rank and so there will be some vectors $\vec b$ for which $A\vec x=\vec b$ will not have a solution.

We could of course have row-reduced the other two matrices to find the answer for them, too, but if you develop an eye for dependencies among columns of a matrix, you can often save yourself a lot of work. It will also help you solidify the underlying concepts in your mind instead of blindly following a mechanical process.

Answer 2

For the last you have three equations which can all be put in the same form: $(6x_1 + 6x_2 + 6x_3 + 6x_4 = a_ib_i)$, so we only have freedom in one variable since we must have $b_3 = 3b_1$ and $b_2 = 2b_1$ (you can work this out by multiplying out the matrices and solving) so this gives a whole line of possibilities but restricts the number of elements of $\mathbb{R}^3$ we can use.

For the second we work analagously and get a dependency condition on one of the variables, but note that in the first all of the equations are independent so this is the only case for which we can choose any arbitrary vector in $\mathbb{R}^3$ and get solutions.

Answer 3

Note that $A$ spans $\mathbb{R}^3$ if and only if it has a pivot position in each row, or more generally, if $A$ is an $m \times n$ matrix, then $A$ spans $\mathbb{R}^m$ if and only if $A$ has a pivot in each row.

Anyhow, thats how you determine if it spans $\mathbb{R}^3$. Try to figure out what happens if $A$ has one or more rows of zeros.