[Math] When does system of equations have infinitely many solutions

Question

If for the system of equations,
$
a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\\a_3x+b_3y+c_3z=d_3$

the matix $A=\begin{pmatrix}a_1&b_1 &c_1\\ a_2 &b_2 &c_2\\a_3&b_3&c_3 \end{pmatrix}$ is singular and $(\text{adj}A)B=O$ where $B=\begin{pmatrix} d_1\\d_2\\d_3\end{pmatrix}$ and $X=\begin{pmatrix} x\\y\\z\end{pmatrix}.$

Then $AX=B$ $\Rightarrow$ $(\text{adj}A)AX=(\text{adj}A)B$ or $OX=O.$

Can we conclude here that the system of equations have infinitely many solution? If not, please give a counter example.

Answer 1

No. If the matrix is singular, i.e. if $\det A=0$, the system has $0$ solution or an infinity of solutions.

The general criterion is based on the augmented matrix $[A|B]$:

Let $A$ be an $m\times n$ matrix, $r$ its rank, $B$ an $m\times 1$ matrix. The system of equations $AX=B$ has a solution if and only if $\operatorname{rank}A=\operatorname{rank}[A|B]$. Furthermore the set of solutions is an affine subspace of $\mathbf R^n$ of dimension $n-r$.