If the determinant $\det(A)$ of the matrix $A$ of a non-homogeneous system of equations is $0$, then how do we know if it has no solutions or infinitely many solutions?

And while we are at it, kindly answer the following "sub-questions" arising from it. I shall be really grateful to you as it will be crucial to my understanding of the whole thing:

a) Since the determinant being zero means that a situation of "Division by zero" arises (using Cramer's Rule), the "no solution" option is understandable as division by zero is not defined. But it confuses me how then, in any circumstance, the system can have infinitely many solutions. I mean, won't we encounter division by zero in all cases when determinant is zero? So please give me an intuitive and insightful explanation to it.

b) Will I be wrong to assume that, in a case when determinant is equal to zero, there are infinitely many solutions if and only if it's a homogeneous system of equations? Please explain why or why not.

And kindly don't forget the main question--"for determinant $=0$, how to know if there are no or infinitely many solutions?"

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    $\begingroup$ A matrix $\,A\,$ cannot have solutions, finite or infinite. You have to make clear what you mean: do you mean a homogeneous system $\,A\vec x=\vec 0\,$ , or a non-homogeneous system $\,A\vec x=\vec b\,\,,\,\,\vec b\neq\vec 0\,$ ? $\endgroup$ – DonAntonio Dec 3 '12 at 12:24
  • $\begingroup$ @DONANTONIA Sorry,I mean whether the system of equation represented by A has solutions,finite or infinite.Thanks for pointing it out.It's my first question on this forum.And yes, I have mentioned in the first line that I mean non-homogeneous system. $\endgroup$ – Ivy Mike Dec 3 '12 at 12:26
  • $\begingroup$ You might want to test your question on simple $2\times2$ systems. $\endgroup$ – Did Dec 3 '12 at 12:27
  • $\begingroup$ I mean, like in a homogeneous system of equations,if det(A)=0,then the system has infinite number of solutions else if det(A) is not zero then it has one a unique,but trivial solution.I want to know what happens for the case of non-homogeneous equations.Thanks. $\endgroup$ – Ivy Mike Dec 3 '12 at 12:30

For any square linear system $\,A\vec x=\vec b\,$ over some field, there exists a unique solution iff $\,\det A\neq 0\,$ , as then we can use the inverse matrix:

$$A\vec x=\vec b\Longleftrightarrow A^{-1}A\vec x=A^{-1}\vec b\Longleftrightarrow A^{-1}\vec b=\vec x $$

As for (a) and your "main question": if $\,\det A=0\,$ one still may have to check whether there are no solutions or infinite solutions (assuming we're working on an infinite field). For example, if the system is homogeneous (over an infinite field) it must have infinite solutions, whereas if the system is non-homogeneous it may have no solutions or several:

$$\begin{cases}x+y=1\\x+y=1\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{1}\longrightarrow\,\,\text{infinite solutions}$$

$$\begin{cases}x+y=1\\x+y=0\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{0}\longrightarrow\,\,\text{no solutions at all}$$

and, of course, in both cases above we have $\,\det A=0\,$

  • $\begingroup$ Thanks Boss!!And please don't mind my clumsy ways.I am totally new to this forum and I'll learn the etiquette soon. $\endgroup$ – Ivy Mike Dec 3 '12 at 12:48
  • $\begingroup$ Any time, @IvyMike. We all had to learn. Remember to upvote all the questions you find useful (not necessarily only one), and then eventually choose one as your "accepted answer" $\endgroup$ – DonAntonio Dec 3 '12 at 12:59
  • $\begingroup$ Can I say for the non-homogeneous linear system that if $\det A=\det A_x=\det A_y=0$, then the system has infinite solutions?($A_x$ => replace the $x$ coefficients with the constant terms in the determinant $A$) $\endgroup$ – user42912 Aug 10 '17 at 5:43

For the case of a linear system of non-homogeneous equations, you need to consider the augmented matrix and compare its rank to the rank of the coefficient matrix of the system.

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    $\begingroup$ Thanks for your answer Mhenni.I was really depressed that something simple is holding me down.On the bright side this led me to sign up with this fine maths community..... $\endgroup$ – Ivy Mike Dec 3 '12 at 12:43
  • $\begingroup$ @IvyMike: You are welcome. $\endgroup$ – Mhenni Benghorbal Dec 3 '12 at 12:45

Let be the system of equations: $$ \left\{ \begin{array}{l} ax+by=e \\ cx+dy=f \end{array} \right. $$ If we multiply the determinant by $x$ and use some properties of determinants we get: $$x\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} ax & b \\ cx & d \\ \end{array} \Rightarrow$$

$$\Rightarrow x\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} e-by & b \\ f-dy & d \\ \end{array} \Rightarrow$$ $$\Rightarrow x\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} e & b \\ f & d \\ \end{array}+ \begin{array}{|cc|} -by & b \\ -dy & d \\ \end{array} \Rightarrow$$ $$ \Rightarrow x\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} e & b \\ f & d \\ \end{array}+ 0 \Rightarrow$$ $$\Rightarrow x\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} e & b \\ f & d \\ \end{array} $$ If we have a system where $\begin{array}{|cc|} a & b \\ c & d \\ \end{array}=0$ and $\begin{array}{|cc|} e & b \\ f & d \\ \end{array}=0$ then $x$ can be any real number.

We can do the same with $y$, but in that case we get: $$\Rightarrow y\begin{array}{|cc|} a & b \\ c & d \\ \end{array}= \begin{array}{|cc|} a & e \\ c & f \\ \end{array} $$ Again if $\begin{array}{|cc|} a & b \\ c & d \\ \end{array}=0$ and $\begin{array}{|cc|} a & e \\ c & f \\ \end{array}=0$, $y$ can assume many values depending on $x$ values or vice versa.


There are two cases actually: If the vector b is not in the column space of the matrix A, it will have no solutions. If the b is in the column space of A, and since det(A)=0, then it will have infinitely many solutions. Hoping this can be a good starting point for you.


If you're after a high school level answer:

There are an infinite number of solutions when the equations that go into the matrix would make exactly the same graph.


x + y = 1

2x + 2y = 2

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    $\begingroup$ This is nonsense. $\endgroup$ – DonAntonio Jan 2 '17 at 17:43

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