# When does system of equations have infinitely many solutions?

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.

• do you mean by O the ) matrix? we have two O's!. Also )X=O is a triviality! Commented Nov 6, 2015 at 11:38
• @Adelafif $O$ is zero matrix. Commented Nov 6, 2015 at 11:48
• $\operatorname{adj}(A)$ denotes the adjugate matrix of $A$? Commented Nov 6, 2015 at 12:37
• @Bernard Yes it denotes adjugate matrix of A. Commented Nov 6, 2015 at 12:39

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 $$n-r$$.

• You're welcome. Always glad to help! Commented Nov 6, 2015 at 12:59
• I haven't heard the term "bordered matrix" yet. Is this the same as the augmented matrix $[A | b]$? Commented Nov 6, 2015 at 13:12
• That's what I meant. Maybe bordered’ is a slightly different notion used in another context. I'll rephrase it. Commented Nov 6, 2015 at 13:17
• @SiamHabib: Thank you for the edit. I retained the term augmented matrix and its notation. However I had to reject it for the end of the edit, which might be confusional, as the rank $r$ of an $m\times n$ matrix can't be $>n$ (nor $m$). Commented Mar 11, 2019 at 18:02

The technical definition is given by Bernard. Notice that this also entails that all non-square matrices have either 0 or infinitely many solutions.

But notice that in general you actually don't have to invert a matrix to find its solutions. Row reduction will bring the matrix in a form where it is easy to read off a) a particular solution (if any such exists) and b) a basis of the null space.

The complete solution of a matrix equation is $\{\lambda + v | v \in N(A)\}$ for any particular solution $\lambda$.

• In practice, calculating the rank or the inverse of a matrix often come down to row-reduction. Commented Nov 6, 2015 at 13:21
• True, but you have to keep track of what you do to the identity matrix, which is slightly more cumbersome if you don't care about the inverse. Of course, this only matters when you do it by hand, on a computer, the algorithmic complexity is basically the same. Commented Nov 6, 2015 at 13:23
• Right. What I wanted to point is there's no fundamental difference between completing the matrix with a column vector or with $n$ column vectors. Commented Nov 6, 2015 at 13:27