# Show that Ax = b is solvable when [A b] is singular.

I have the following problem:

Review: Suppose A is 5 by 4 with rank 4. Show that Ax = b has no solution when, the 5 by 5 matrix [A b] is invertible. Show that Ax = b is solvable when [A b] is singular.

I don't know how to do that. Any hint or full solution is welcomed.

Thank you

• Since $A$ is rank $4$, what does that imply about the linear dependence or independence of the columns of $A$? If $[A~b]$ is invertible, what does that imply about the linear dependence or independence of its columns? What does that imply about $b$ being in the image of the matrix $A$? If $[A~b]$ is singular, what does that imply about the rank of $[A~b]$? Since $A$ is rank $4$, what does that imply about $b$? – JMoravitz Mar 9 '16 at 22:42

Let $a_1, \dots a_4$ be the column vectors of $A$ with entries from some field $K$.
1. Suppose $C := [A\ b]$ is invertible: Then the column vectors $a_1, \dots a_4, b$ of $C$ are linearly independent, thus there are no $x_1, \dots, x_4 \in K$ with $b = \sum_{i=1}^{4}{x_ia_i}$. But this is equivalent to that $Ax = b$ has no solution.
2. Suppose $C := [A\ b]$ is singular: The rank of $A$ is 4, so $a_1, \dots , a_4$ are linearly independent. But $C$ is singular, so $b$ is a linear combination of $a_1, \dots, a_4$, i.e. $\exists x_1, \dots, x_4 \in K$ with $b = \sum_{i=1}^{4}{x_ia_i}$. Which means that $Ax = b$ has a solution.
$$\exists x=(x_1, \dots, x_4): Ax=b \iff \exists x_1, \dots, x_4 \in K: b = \sum_{i=1}^{4}{x_ia_i} \iff b\ \text{is linear combination of }a_1, \dots, a_4 \iff \text{(because A has rank 4)}\ [A\ b] \text{ is singular}$$