I know how to calculate a determinant, but I wanted to know what the meaning of a determinant is? So how could I explain to a child, what a determinant actually is.

Could I think of it as a measure of independence of the columns or rows? Or is there another interpretation that is even simpler?

Edit: The explanation should also explain, the role of the determinant for the solvability of linear equations.

  • 127
  • 1
  • 6
  • 15,530
  • 3
  • 18
  • 47

4 Answers4


The determinant is the volume of parallelepiped formed by the vectors.

Wikipedia is more precise:

Determinants express the signed $n$-dimensional volumes of $n$-dimensional parallelepipeds.

The determinant is zero iff the vectors are linearly dependent. In this sense, it is a coarse measure of the independence of vectors.

The finer measure of independence of the columns or rows is the rank.

A system of linear equations $Ax=b$ is solvable iff $rank(A)=rank(A\mid b)$, where $A\mid b$ is the augmented matrix.

Determinants appear in the actual solution of systems of linear equations via Cramer's rule (but this is mainly an important theoretical result, because it is not pratical, except for very small systems).

  • 208,399
  • 15
  • 224
  • 525
  • I like this, but it seems hard for me to connect this to solving linear equastions. Why should the volume have something to do with the solvability of a system. Still a nice interpretation. – MrYouMath Apr 12 '16 at 14:38

Adding to lhf's answer, you can think of the 'geometric' meaning of this volume interpretation: think of how $A$ acts on the vector space $V$.

The volume of the parallelepiped is connected to this interpretation in that the parallelepiped is the image under $A$ of the unit cube. In other words, the parallelepiped is a small-scale model of how $A$ acts on the whole of $V$ (because of linearity!).

When the volume is $0$, it means one or more dimensions were lost in the transfomation (the 'area' of a line or the 'three dimensional volume' of a plane are both zero). This means that, for $Ax = b$, some vectors $b$ – those that have a component in the 'lost dimensions' – won't be reachable at all, while any others will admit multiple solutions (because adding to $x$ some $u$ that lies in those dimensions that are discarded by $A$ does not change the result).

  • 22,641
  • 2
  • 23
  • 42

I think @lhf 's answer is the best intuitive explanation.

In response to your edit: a system of $n$ linear equations in $n$ unknowns will always have a unique solution if and only if the determinant is not $0$, because that is the condition that says the rank is $n$ so the corresponding linear map is one to one and onto. But in practice you don't use determinants to solve the equations - although students find it cool, Cramer's rule is really only of theoretical interest. Gaussian elimination (and more advanced numerical methods) do the job better, and give you more information - even when the determinant is $0$ the equation will sometimes have solutions, and when it does they won't be unique.

Ethan Bolker
  • 80,490
  • 6
  • 95
  • 173

The intuition I have stuck with once I discovered it is the following. Let $f(x) = Ax$ be a linear transform where $A$ is an $n \times n$ matrix. Then the determinant $|A|$ is the volume scaling factor of the transform $f$. That is, if $S$ is a set in $n$-space that has volume $V$, then $f(S)$ is a set in $n$-space that has volume $V \cdot |A|$.

Timothy Shields
  • 744
  • 4
  • 7