After looking in my book for a couple of hours, I'm still confused about what it means for a $(n\times n)$-matrix $A$ to have a determinant equal to zero, $\det(A)=0$.

I hope someone can explain this to me in plain English.

Rodrigo de Azevedo
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    If you are talking about the determinant of a square matrix $A$, a very useful characterization is $\det A=0$ if and only if $A$ is not invertible. – Julien Apr 09 '13 at 06:09
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    Your answer is already solved, but I would like to add a trick. If the rank of an nxn matrix is smaller than n, the determinant will be zero. – tashuhka Feb 19 '14 at 09:14
  • One has to presume that they wanted to know the significance rather than the definition, since the definition of $det(A)=0$ follows trivially from that of $det(A)$ Given the level of the question, I feel that the most illuminating equivalent consequence is that there are two different vectors which become the same when multiplied by $A$, i.e. an elementary formulation of what it means for $A$ not to be invertible. – PJTraill Jan 18 '18 at 14:55
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    The columns and rows of the matrix are linearly dependent,,, – DVD May 31 '18 at 02:39
  • There are infinite solutions to the matrix... – omkaartg Jan 11 '19 at 05:03

10 Answers10


For an $n\times n$ matrix, each of the following is equivalent to the condition of the matrix having determinant $0$:

  • The columns of the matrix are dependent vectors in $\mathbb R^n$

  • The rows of the matrix are dependent vectors in $\mathbb R^n$

  • The matrix is not invertible.

  • The volume of the parallelepiped determined by the column vectors of the matrix is $0$.

  • The volume of the parallelepiped determined by the row vectors of the matrix is $0$.

  • The system of homogenous linear equations represented by the matrix has a non-trivial solution.

  • The determinant of the linear transformation determined by the matrix is $0$.

  • The free coefficient in the characteristic polynomial of the matrix is $0$.

Depending on the definition of the determinant you saw, proving each equivalence can be more or less hard.

Ittay Weiss
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    "The system of homogenous linear equations represented by the matrix has a non-trivial solution." Does this mean it has an infinite number of solutions? – Mateen Ulhaq Jul 24 '15 at 05:51
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    @MateenUlhaq Yes: if the kernel is non-empty, as it's a vector space, there's an infinity of solutions ($Ax=0 \Rightarrow A(2x)=0$). – anderstood Oct 21 '15 at 20:38
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    @MateenUlhaq Well, that is true if your matrices are defined over an infinite field (like the rational, real or complex numbers), but false if they are defined over a finite field! – Jose Brox Nov 07 '17 at 21:45
  • @JoseBrox Interesting. I'd never thought about that. :-) – Mateen Ulhaq Nov 07 '17 at 23:31
  • Of course we cannot omit the fact that we "solve" a polynomial = 0 when we compute eigenvectors and of course det(A)=0 only when 0 is an eigenvalue. – David Dyer Feb 26 '18 at 19:40
  • @anderstood is the kernel ever empty (doesn't it contain at least one 0)? – ThatsRightJack Jun 19 '20 at 04:08
  • @ThatsRightJack You're right I meant "if the kernel is not reduced to $\{0\}$" – anderstood Jun 19 '20 at 06:02

For me, this is the most intuitive video on the web that explains determinants, and everyone who wants a deep and visual understanding of this topic should watch it:

The determinant by 3Blue1Brown

The whole playlist is available at this link:

Essence of linear algebra by 3Blue1Brown

The crucial part of the series is "Linear transformations and matrices". If you understand that well, everything else will be like a piece of cake. Literally: plain English + visual.

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    Do you think that is an answer to this three-year old question ? – Claude Leibovici Mar 21 '17 at 07:00
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    As I saw this question 3 years after it's asked, someone else may see it 3 or 30 years from now, struggling to understand why the determinant is equal to zero. I don't mind if my answer is not the "accepted" one. My point was to share a helpful resource. – ebu_sho Mar 21 '17 at 08:18
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    Thank you for these links! Definitely gets my vote for "should be accepted answer". Even though I'd used these videos many times many years ago — probably to answer a related question (it's a great series) — and (at least) at one time I'd already looked up the answer to this very question, the clarity & unassuming nature of these tutorials is exactly the refresher I needed. (I would just up-vote, but also wanted to add counterbalance to the uselessly snarky comment on the value of this answer.) – michael Nov 30 '18 at 03:28
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    As a time traveler from the year 2021, I must commend @ebu_sho for the video link. Much has changed since 2017, but I'm pleased to say matrix math has NOT changed. This is a great video for math students of all eras. (The answer by Ittay Weiss is great, too.) YouTube didn't exist when I was in school back in the late 20th century, and attempts to travel back that far in time and share the video have failed. I'm happy that the 2017 version of StackOverflow has survived to my time period. Much obliged! – user148097 Jan 12 '21 at 13:37
  • @ClaudeLeibovici The applicability of an answer to this question does not vary with time. – Kröw Jan 26 '22 at 03:06

If the determinant of a square matrix $n\times n$ $A$ is zero, then $A$ is not invertible. This is a crucial test that helps determine whether a square matrix is invertible, i.e., if the matrix has an inverse. When it does have an inverse, it allows us to find a unique solution, e.g., to the equation $Ax = b$ given some vector $b$.

When the determinant of a matrix is zero, the system of equations associated with it is linearly dependent; that is, if the determinant of a matrix is zero, at least one row of such a matrix is a scalar multiple of another.

[When the determinant of a matrix is nonzero, the linear system it represents is linearly independent.]

When the determinant of a matrix is zero, its rows are linearly dependent vectors, and its columns are linearly dependent vectors.

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  • and what about a liniar system? does it mean it has a solution? – user2171775 Apr 09 '13 at 06:15
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    you needn't have that one row is a multiple of another, just that they are linearly independent. – Alex Youcis Apr 09 '13 at 06:16
  • @user2171775 yes...if the determinant is NOT zero. $Ax = b$ represents a linear system. – amWhy Apr 09 '13 at 06:21
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    "When the determinant of a matrix is zero, the system of equations associated with it is linearly dependent; that is, if the determinant of a matrix is zero, at least one row of such a matrix is a scalar multiple of another." If the determinant is zero, one of the rows doesn't need to be a scalar multiple of the others. For example $$\left(\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 7 & 9\end{array}\right)$$ has determinant zero but none of the rows are scalar multiples of another. – Cameron Williams Jan 22 '14 at 23:35
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    I will rephrase the sentence as: "at least one row/column of such a matrix is a linear combination of one or several rows/columns." The 3rd row is the sum of 1st and 2nd rows. – tashuhka Feb 19 '14 at 09:05

The determinant of a matrix is the oriented volume of the image of the unit cube. If it is zero, the unit cube gets mapped inside of a plane and has volume zero.

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Let $A$ be a matrix.

Then, $$A^{-1} = \dfrac{1}{detA} adjA$$

Hence, if determinant is zero, Inverse doesn't exist for that matrix.

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    If the OP doesn't know what a determinant is, what chance is there they know that the adjugate matrix is? – anon Apr 09 '13 at 06:17
  • @anon He did not say that he doesn't know what a determinant is, he said he is confused by the term 'determinant equal to zero'. – lsp Apr 09 '13 at 06:20
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    Indeed. At any rate, I don't see how this answer could be useful to someone who doesn't know what the phrase "determinant equal to zero" means, whether s/he meant they don't know what a determinant is or just that they don't know how to intuitively think about the situation where a determinant is equal to zero. This answer could be useful for, say, some sort of computational query, I think, which I don't think this is. – anon Apr 09 '13 at 06:26
  • Not a terribly convincing argument, as it might conceivably happen that $\det(A)$ *cancels out* with something in $\operatorname{adj}(A)$. Better note that if $A$ has an inverse $B$, so that $AB = BA = 1$, where $1$ is an appropriate identity matrix, then $\det(A) \det(B) = \det(1) = 1$, so $\det(A) \ne 0$. – Andreas Caranti Apr 09 '13 at 06:28
  • I think someone who is asking a question about determinant should be old enough to know that $\dfrac{1}{0}$ never exists and hence the inverse doesn't exist. Well my assumption might be wrong and yours might be right. But I leave it to @user2171775. – lsp Apr 09 '13 at 06:32
  • @AndreasCaranti Note that $det(A)$ cancels out only if it is not equal to $0$. – lsp Apr 09 '13 at 06:34
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    Sorry, perhaps I did not make myself clear. One might say correctly that the expression $f(x)/x$ makes no sense when $x = 0$. But then you might discover that $f(x) = 3 x$, so after cancellation $f(x) / x = 3$ makes sense. Or you might even think of a function like $\sin(x)/x$ which has a *removable singularity* at $x = 0$. – Andreas Caranti Apr 09 '13 at 06:40
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    If we want to show zero determinant implies nonexistence of inverses, it would be more elementary in my opinion to use multiplicativity of the determinant and look at $1=\det(A)\det(A^{-1})$, hence my confusion as to why the adjugate is being invoked here - it puts an unnecessary onus on the OP and similar readers to learn about adjugates and the relation between the adjugate and the determinant in addition to learning about just the determinant. The adjugate may be of independent interest, however, and perhaps you want to advertise that fact, which I would find understandable. – anon Apr 09 '13 at 07:30
  • This does not prove anything and you forgot to mention that this holds if $A$ is invertible. – anderstood Oct 21 '15 at 20:42

There is a geometrical interpretation for the determinant. In addition to the interpretation in the another answer, yet another appealing one related to the determinant is its interpretation as the volume of a $N$ dimensional parallelopipped. This is more expressed in 3-Dimensions. If you take $3$ vectors in 3-D, they may or may not form the corners of a parallelopipped, if you take the determinant of a matrix with this 3 vectors as the columns (or rows), if the determinant is zero, it means, they don't form a parallelopipped together, if it is non-zero, it means, they indeed form the 3 edges of a parallelopipped with the volume given by the determinant. The sign of the value of determinant gives a kind of information on the orientation of this body.

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When you think a matrix as transformation, the determinant is the area(or volume in higher dimension) made by basis after transformation. This type of thinking will give you visual aid. For example for $n=3$, determinant of a matrix equals zero means, the map by that matrix transforms a plane into a line(the area made by the basis in transformed plane is zero). The idea can be generalized for higher dimension.

Note that, it is not possible to map back a line to a plane. So, inverse of a matrix, which has determinant zero, doesn't exists.

tarit goswami
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Take a 2 x 2 matrix, call it A, plot that in a coordinate system.

A= [[2,1],[4,2]] . --> Numpy notation of a matrix

Following two vectors are written from A



If you plot that, you can see that they are in the same span. That means x and y vectors do not form an area. Hence, the det(A) is zero. Det refers to the area formed by the vectors.

Proton Boss
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Another way of putting it:

If you take $2$ vectors in $2D$ space, you can show that the area of the parallelogram formed is simply the determinant of the matrix formed by those two vectors. This is a general result for $n$-dimensions - the determinant of a matrix is the volume of the $n$-parallelogram formed by the rows of the matrix.

If the determinant is zero, this means the volume is zero. This can only happen when one of the vectors "overlaps" one of the others or more formally, when two of the vectors or linearly dependent.

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    not necessarily two of the vectors need to be linearly dependent. Rather, if the determinant is $0$, then the volume of the parallelepiped is $0$, which means that at least one vector lies in the hyperplane spanned by the others, thus the set of vectors is dependent. It's still possible that every two vectors are independent (in $R^3$ and above). – Ittay Weiss Apr 09 '13 at 06:22

If a matrix A has an inverse, it means that the matrix multiplication AB = I, the identity matrix. In plain English then, if a matrix is invertible then it may have a solution.

If a matrix's determinant is nonzero, the matrix may have a solution. If the determinant is zero, then the matrix is not invertible and thus does not have a solution because one of the rows can be eliminated by matrix substitution of another row in the matrix.

Common reasons for matrix invertibility are that one or more rows in the matrix is a scalar of the other.

EXAMPLE for a 3x3 matrix: Row 1: x + y + z = P versus Row 2: Qx + Qy + Qz = QP. Row two can be rewritten as Q*(x + y + z = P) or Q*(Row 1)]

OR one of the rows can be created by addition of two or more of the other rows.

EXAMPLE for a 3x3 matrix: Row 1: x + 2y + 3z = P, Row 2: 4x + 5y + 6z = Q, Row 3: 7x + 8y + 9z = P+Q. You see that Row 3 can be duplicated by adding Row 1 and Row 2.

In short, if the determinant of a matrix is zero, the matrix does not have a solution because the matrix cannot be inverted.

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