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Suppose $A$ is $5$ by $4$ with rank $4.$ Show that $\mathbf{Ax = b}$ has no solution when the $5$ by $5$ matrix $\begin{bmatrix} \mathbf{A} & \mathbf{b} \\ \end{bmatrix}$ is invertible. Show that $\mathbf{Ax = b}$ is solvable when $\begin{bmatrix} \mathbf{A} & \mathbf{b} \\ \end{bmatrix}$ is singular.

Answer : If the $5$ by $5$ matrix $\begin{bmatrix} \mathbf{A} & \mathbf{b} \\ \end{bmatrix}$ is invertible, $\mathbf{b}$ is not a combination of $\mathbf{A}$'s columns.
If $\begin{bmatrix} \mathbf{A} & \mathbf{b} \\ \end{bmatrix}$ is singular, and the $4$ columns of $A$ are independent, then $\mathbf{b}$ is a combination of the columns of $\mathbf{A}.$ In this case, $\mathbf{Ax = b}$ has a solution. $\qquad \square$

P27 : A singular matrix has dependent columns.
P144: The rank of a matrix is its number of pivots.

$A$ is given as $5$ by $4$ and has rank $4$, so it has $4$ pivots which means that one of the rows must be linearly dependent. So REF(A) contains only $1$ zero row.
Then $rank(A) = 4 < \text{ # of rows in A } = 5 \implies A$ has $0$ or $1$ solutions.

How would I continue from here? I also don't see how the answer relates to my work hitherto?

This question precedes dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please omit them from answers.

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You need to concentrate your attention on the number of linearly independent columns/ columns of $A$. Thinking of rows is less helpful. Also $Ax=x_1c_1+\dots x_4 c_4$, where $x_i$'s are components of clmn vector $x$ and $c_i$'s are columns of $A$. –  Any Nov 26 '13 at 7:30

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