For a $n \times n$ matrix $A$:$$\det (A) = \sum^{n}_{i=1}a_{1i}C_{1i}$$

where $C$ is the cofactor of $a_{1i}$. If the determinant is $0$, the matrix is not invertible.

Could someone an intuitive explanation of why a zero determinant means non-invertibility? I'm not looking for a proof, the book gives me one. I'm looking for intuition.

For example, consider the following properties of zero:

Property 1: If a factor of a polynomial is $0$ at $x$, then the polynomial is zero at $x$.

Intuition: Anything times zero, so if the remaining polynomial is being multiplied by zero, it has to be zero.

Property: $x + 0 = x$

Intuition: If you have something and don't change anything about it, it remains the same.

So why does having a zero determinant imply non-invertibility? Could someone give me a similar intuition for determinants?


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  • See also: [What's an intuitive way to think about the determinant?](http://math.stackexchange.com/q/668/3157) – Zaz Aug 18 '16 at 07:22

5 Answers5


There are several different ways to think about this intuitively, and which one you prefer may differ depending on your algebraic/geometric inclination.

The first interpretation is the one that tylerc0816 mentions. Each matrix induces a corresponding linear mapping. For such a map, the standard basis vectors (i.e. the unit cube) gets mapped to a parallelotope formed by the columns of your matrix. In order for this map to be invertible, the parallelotope must be non-degenerate. What is meant by degeneracy here is that the parallelotope must be "full"-dimensional, i.e. you cannot have a 3D parallelepiped collapse into a 2D parallelogram.

If such a collapse does happen, then intuitively you have multiple points stacked on top of each other. If this happens, then the map cannot be invertible. The condition for this collapse is equivalent to the parallelotope having $0$ volume, i.e. for the matrix to have zero determinant.

Secondly, you may view the determinant in terms of how it changes under elementary row operations:

  1. If you add a row to another row, then the determinant is unchanged.

  2. Multiplying a row by a non-zero scalar multiplies the determinant by the same non-zero scalar.

  3. Switching two rows changes the sign of the determinant.

Notice that all of these operations preserve the "zero-ness" of the determinant, i.e. the determinant will be zero after an elementary row operation if and only if it was zero before. With this view in mind, we see that the determinant of a matrix will be zero if and only if the determinant of its Reduced Row Echelon Form is zero. The RREF is either identity for invertible matrices, in which case the determinant is $1$, or it has an all zero row, which necessarily forces the determinant zero.

Alternatively, you may wish to look at the determinant in terms of the characteristic polynomial. The determinant is precisely the value of the characteristic polynomial evaluated at $0$ (perhaps up to some factors of $-1$ depending on how you define the characteristic polynomial). If the characteristic polynomial is $0$ when evaluated at $0$, then the matrix must have a zero eigenvalue which means a non-trivial nullspace. Again this forces the matrix to be non-invertible.

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The absolute value of the determinant is the volume of the mapping of the unit cube under $A$. If the determinant is $0$, then there had to be some 'collapsing' under $A$ (think dimension). Thus, $A$ is not injective, and thus not invertible.

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  • What do you mean by "collapsing"? – asdf Nov 13 '13 at 15:35
  • Specifically, I mean $A$ does not have full rank. For a concrete example, think about the unit square in $\mathbb{R}^2$. The only way for the area of the square to be $0$ is for one of the side lengths to be $0$. So, in a sense, $A$ collapses one of the sides to $0$. I was being slightly vague, but that is because you were just asking for intuition. – tylerc0816 Nov 13 '13 at 15:41

There is a simple property for determinants:

$Det(A)Det(b)=Det(AB)$, so if you take:

$Det(A A^{-1})= Det(A)Det(A^{-1})= Det(I) = 1$, from this follows than $det(A)$ must be different from $0$ to be invertible.

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Jorge Ramos
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I want to use a geometric approach: intuitive, informal and not so techie.

The determinant, usually algebraically defined, will be defined here as Gram Zeppi said, as representing an n-dimensional volume, from the abstraction of seeing the columns as n-dimensional vectors, forming the edges of a hyper parallelepiped expressed by determinant.

Volume and Determinant

In this case there are basic properties, which serve as a starting point to deduce the Laplace Expansion, usually taught in the definition of Determinants in schools and universities.

1) The volume of a transposed matrix (when one changes rows to columns) is ​​equal to the original matrix volume. In fact, it is expected by symmetry. Thus, the properties valid for columns, described below, are also valid for rows.

2) The vectors in the columns must be independent linearly, so that the volume is different from $0$. For example, In $2D$, two collinear vectors (multiples) don’t form a parallelogram. In the $3D$ case, three coplanar vectors (the situation that happens when one vector can be obtained by the linear combination of the other 2) or 2 collinear vectors generates a null volume.

In general case ($N$ dimensions) any subset of columns from n-dimension determinant where one column can be obtained by linear operators (sequence of sums and scalar multiplications) using the rest of colums subset, make the volume flat in $N$ dimensions, because, in that case, at least 1 dimension will be missing because at least 1 vector is redundant and don’t points to a different dimension.

$\bigstar \bigstar \bigstar $

Before proceeding with more properties, one must learn to calculate volume with $N$ dimensions using $N-1$ dimensions volumes.

First of all, we don’t use here cross product, because it only exist for 3 dimensions (and for 7 … (https://en.wikipedia.org/wiki/Seven-dimensional_cross_product)). Also, we don’t use more determinants, because it would be a recursive reasoning.

For clearer understanding, let’s imagine a matrix 3x3 with a left vector $\mathbf {\vec{a} = [a_1,a_2,a_3]}$and to the right two additional vectors: $\mathbf {\vec{b}[b_1,b_2,b_3]}$ and $\mathbf {\vec{c} = [c_1,c_2,c_3]}$ in 3D space. Let’s forget the left vector $\mathbf {\vec{a}}$ for a while.

Parallelepiped volume with vectors

If the vector $\mathbf {\vec{b}}$ and $\mathbf {\vec{c}}$ are independent linearly, they are contained in a plane inside 3D space. Now we will find the area spanned by $\mathbf {\vec{b}}$ and $\mathbf {\vec{c}}$. It wouldn’t be necessary, because we could use 2D space using vector module, but it’s interesting to see how it’s possible to avoid cross product usage at all.

First, we will find a normal $\mathbf {\vec{n} = [n_1,n_2,n_3]} $that is perpendicular to $\mathbf {\vec{b}}$ and $\mathbf {\vec{c}}$ simultaneously. So $\mathbf {\vec{n} \bullet \vec{b} = 0}$ and $\mathbf {\vec{n} \bullet \vec{c}= 0}$.


$\mathbf {b_1 n_1 + b_2 n_2 + b_3 n_3 = 0\quad}$ and$ \mathbf {\quad c_1 n_1 + c_2 n_2 + c_3 n_3 = 0} $

As it only matters the normal direction, one can consider $\mathbf {n_1 = 1}$ and solve the above equations.

Now, we need find an orthogonal $\mathbf {\vec{o}}$ coplanar to $\mathbf {\vec{b}}$ (could be $\mathbf {\vec{c}}$ , whatever) and orthogonal to $\mathbf {\vec{n}} $

To find $\mathbf {\vec{o}}$, we use the same kind of equations above with $\mathbf {\vec{n}}$ and $\mathbf {\vec{b}}$ coefficients. After we find $\mathbf {\vec{o}}$, we normalize to $\mathbf {\|b\|}$ magnitude. So

$ \mathbf {Area_{bc} = \vec{o} \bullet \vec{c}}$

Why? Because dot product projects $\mathbf {\vec{c}}$ in $\mathbf {\vec{o}}$ (normal to $\mathbf {\vec{b}}$) multiplying by $\mathbf {cos \theta}$, the angle between $\mathbf {o}$ and $\mathbf {c}$, resulting in the parallelogram height that will be multiplied by $\mathbf {\|\vec{o}\|}$, that is $\mathbf {\vec{b}}$ basis.

To get the volume (our initial goal) the normal $\mathbf {\vec{n}}$, has to be normalized with magnitude equal to area spanned by $\mathbf {\vec{b}}$ and $\mathbf {\vec{c}}$ producing $\mathbf {\vec{n_{bc}}}$, so

$ \mathbf {Volume =\vec{n_{bc}} \bullet \vec{a}}$

For what reason? if $\mathbf {\vec{a}}$ in column $1$ was a vector with magnitude $1$ normal to plane that contains $\mathbf {\vec{b}}$ and $\mathbf \vec{c}}$ , the volume generated would be equal (in value) to $\mathbf {Area{_bc}}$ in plane. It’s a $3D$ solid with height $1$ at right angle so the volume is the basis area $\times $height.

Generally the vector $\mathbf {\vec{a}}$ is oblique to the basis $\mathbf {\vec{b} \vec{c}}$, so if the angle between $\mathbf {\vec{a}}$ and basis $\mathbf {\vec{b} \vec{c}}$ is $\mathbf{\phi}$, the volume is $\mathbf {Area_{bc} \, * \|\vec{a}\|cos\phi}$

We know to calculate $3D$ volumes. By induction, we want to show that if we know calculate hyperplane volume in $N-1$ dimensions, we will know to calculate in $N$ dimensions.

The argument below is a kind of repetition, but within an abstract dimension, as any dimension above $3$.

We get a determinant with $N$ dimensions. In the same way, we separate the first column vector $\mathbf {\vec{a}}$ in the left (the same reasoning is also true for lines for symmetry ), Analogously the other vectors to the right forming the edges of a hyperplane $V$ with $N-1$ dimensions, the we know how to calculate the volume!

So, in the same way, we find the normal vector $\mathbf {\vec{n}}$ with a similar procedure seen above (just more equations) and we normalize magnitude vector $\mathbf {\vec{n}}$ to $N-1$ dimensions volume obtaining $\mathbf {\vec{n_v}}$.

The same above procedure can be done. The volume for $n$ dimensions with additional vector $\mathbf {\vec{a}}$ is simply:

$ \mathbf {Volume =\vec{n_v} \bullet \vec{a}}$

At short, knowing how to calculate volume and normal in $n-1$ dimensions, allow easily calculate volume in $n$ dimensions.

So far, we have said that it is possible to calculate the volume from a determinant , but we don’t yet link this to the determinant calculation, which we will do later.

$\bigstar \bigstar \bigstar$

All that reasoning helps to justify the properties below:

3) Out of any signal convention (discussed later), simple column switching (base volume vectors that defines a volume) does not change the volume in module, because different vector picking orders don’t change the geometric form.

4) If a column is multiplied by $\mathbf {\alpha}$, the determinant is also multiplied by $\mathbf {\alpha}$. According to property 3, one can change the column multiplied by $\mathbf {\alpha}$ by the first column and use the linearity of the scalar product:

$\mathbf {\vec{N_V} \bullet \alpha \,\vec{a} = \alpha (\vec{V_{N-1}} \bullet \vec{a})}$

5) If 3 matrices $\mathbf {A}$, $\mathbf {B}$ and $\mathbf {C}$ differ only by column $\mathbf {k}$ where

$\mathbf {C [., t_1 + t_2, .] = A [., t_1, .] + B [., t_2, .]}$ then $\mathbf {Det (C) = Det (A) + Det (B)}$

( . indicates 0 or more columns, $\mathbf {t_1+t_2}$, $\mathbf {t_1}$ and $\mathbf {t_2}$ are in column $\mathbf {k}$)

According to property 3, change the column $\mathbf {k}$ to the first one, then use the linearity of the scalar product:

$ \vec{N_V} \bullet (\vec{a_1} + \vec{a_2}) = \vec{N_V} \bullet \vec{a}$

6) The inversion of 2 columns (or lines) reverses the value of determinant $\mathbf {A}$.

This shows that some signal convention needs to be adopted in order to respect this property.

Let’s prove it.

Suppose that the column $\mathbf {J}$ = column $\mathbf {K}$ (where $\mathbf {J < K)}$. In this case, by property 2, the determinant is $0$

Assume two values ​​$\mathbf {t}$ and $\mathbf {u}$ that add up to give the value of the repeated column. (In the below formulas, dot ($\mathbf {.}$) indicates 0 or more hidden columns and the displayed columns are respectively the columns $\mathbf {J}$ and $\mathbf {K}$).

$\mathbf {\mid.,\,t + u,\,.,\,t + u,\,.\mid\quad(A )\quad = }$ $\mathbf {\qquad \mid.,\, t,\, .,\, t,\,.\mid\quad(A_1)\quad+}$ $\mathbf {\qquad \mid.,\, t,\, .,\, u,\,.\mid\quad(A_2)\quad+}$ $\mathbf {\qquad \mid.,\, u,\, .,\, t,\, .\mid\quad(A_3)\quad+}$ $\mathbf {\qquad \mid.,\, t,\, .,\, t,\, .\mid\quad(A_4)}$

$ \mathbf {A_1}$ and$ \mathbf {A_4}$ are null because they have equal columns. Thus $\mathbf {A_2 = -A_3}$, so that the determinant cancels out. The only difference between $\mathbf {A_2}$ and $\mathbf {A_3}$ is the swap of 2 columns.

$\bigstar \bigstar \bigstar$

We define above the fundamental determinant properties, based on the geometric paradigm. Now we are ready to prove the Laplace Expansion, again using the finite induction.

Imagine a matrix $M$ with two dimensions $\mathbf {[\vec{v}, \vec{w}]}$, Its determinant defines a parallelogram. The vector $\mathbf {\vec{v} = [v_1 v_2]}$ has a normal given by $\mathbf {[-v_2, v_1]}$, which is the basis of the parallelogram formed by$ \mathbf {\vec{v}}$ and $\mathbf {\vec{w}}$ in the direction of the parelogram.

Thus the scalar product $\mathbf {[-v_2, v_1]\bullet [w_1 w_2] = v_1w_2 - v_2w_1}$, which corresponds to the projection of the lateral of the parelogram in the height multiplied by the base of the parallelogram, gives the parallelogram area.

$\mathbf {\begin{bmatrix} v_1&w_1\\v_2&w_2 \end{bmatrix} = \begin{bmatrix} v_1&w_1\\0&w_2 \end{bmatrix}+\begin{bmatrix} 0&w_1\\v_2&w_2 \end{bmatrix}}$

It is clear that the same can be found by the Laplace Expansion with $\mathbf {v_1 w_2}$and $\mathbf {v_2 w_1}$, except that in the second case there was one line change (2 with 1), according to Theorem 6, by inverting the signal of the determinant, because there are odd permutations, so determinant is $\mathbf {v1.w2 - v2.w1}$

Let's insert a new dimension from the left and the base in the dimension $N-1$, which is supposed to apply the Laplace Expansion correctly: From this it must be proved for case $N$, illustrated here by a matrix $M$ with $N = 3$.

$\mathbf {\begin{bmatrix} u_1&\mid&v_1&w_1\\u_2&\mid&v_2&w_2 \\u_3&\mid&v_3&w_3\end{bmatrix}}$

Using the sum theorem, we separate the matrices

$\qquad \qquad M_1 \qquad\qquad\qquad\qquad M_2 \qquad\qquad\qquad\qquad M_3 \\ \begin{bmatrix} u_1&\mid&\mathbf{v_1}&\mathbf{w_1}\\0&\mid&v_2&w_2 \\0&\mid&v_3&w_3\end{bmatrix}+\begin{bmatrix} 0&\mid&v_1&w_1\\u_2&\mid&\mathbf{v_2}&\mathbf{w_2}\\0&\mid&v_3&w_3\end{bmatrix}+\begin{bmatrix} 0&\mid&v_1&w_1\\0&\mid&v_2&w_2\\ u_3&\mid&\mathbf{v_3}&\mathbf{w_3}\end{bmatrix}$

Looking at the matrices above $\mathbf {M_1}$, $\mathbf {M_2}$ and $\mathbf {M_3}$, it is clear that the bold components of $ \mathbf {[v_1, v_2, v_3]}$ and $\mathbf {[w_1, w_2, w_3]}$ in the same line of non-zero value of column 1 don’t contribute at all for volume formation, by not bringing the point out of the pure axis expressed in column 1.

Why discard some determinant values?

The figure above helps the reader understand because an pure dimensional vector like $\mathbf {\vec{A}}$ (parallel to a canonical axis, $z$ in this case) don’t need to handle with same dimension components from other vectors. The only effect is shearing through this dimension, what don’t create or destroy any volume.

Then we have

$\mathbf{M_1}:\quad\begin{bmatrix} v_2&w_2\\v_3&w_3 \end{bmatrix}$ is a parallelogram of area $\mathbf {v_2w_3 - v_3w_2}$ in $\mathbf {plane_{23}}$ in a coplanar vector of this magnitude $\bullet$ (scalar product) a vector $\mathbf {\bot}$ to $\mathbf {plane_{23}}$ with magnitude $\mathbf {u_1}$ that results in $\mathbf {u_1 (v_2w_3 - v_2w_3)}$ with order: $\mathbf {123}$ (no row swap)

$\mathbf{M_2}:\quad \begin{bmatrix} v_1&w_3\\v_1&w_3 \end{bmatrix}$

is a parallelogram of area $\mathbf {v_1w_3 - v_3w_1}$ in $\mathbf {plane_{13}}$ in a coplanar vector of this magnitude $\bullet$ (scalar product) vector $\mathbf {\bot}$ to $\mathbf {plane_{13}}$ with magnitude $\mathbf {u_2}$ that results in $\mathbf {u_2 (v_1w_3 - v_1w_3)} $with order: $\mathbf {213}$ (1 row swap)

$\mathbf{M_3}:\quad \begin{bmatrix} v_2&w_3\\v_2&w_3 \end{bmatrix}$

is a parallelogram of area $\mathbf {v_1w_2 - v_2w_1}$ in $\mathbf {plane_{12}}$ in a coplanar vector of this magnitude $\bullet$ (scalar product) vector $\mathbf {\bot}$ to $\mathbf {plane_{12}}$ with magnitude $\mathbf {u_3}$ that results in $\mathbf {u_3 (v_1w_2 - v_1w_2)} $with order: $\mathbf {321}$ (2 rows swaps)

In $\mathbf {M_2}$ there is a single row swap that reverses the signal, thus

$\mathbf {Det(M) = u_1 (v_2 w_3 - v_2 w_3) - u_2 (v_1w_3 - v_1w_3) + u_3 (v_1w_2 - v_2w_1)}$

This corresponds to the Laplace Expansion for $\mathbf {N = 3}$ because $ \mathbf {Det (M) = u_1 Minor_{11} - u_2 Minor_{21} + u_3 Minor_{31})}$

Where the first subscript in Minor is the line reference and the second subscript is column reference.

In the general n-dimension matrix case, first separate a matrix $M$ into a sum of M matrices, each one with just one value of column 1.

$ \mathbf{M}:\quad \begin{bmatrix}M_{11}&...&M_{1N}\\...&...&...\\M_{LI}&...&M_{LN}\\...&...&...\\ M_{N1}&...&M_{NN}\end{bmatrix} \quad=\quad \Sigma_{I=I,N} \begin{bmatrix}0 \; or\; M_{1,1}(I=1) &...&M_{1N} \\...&...&... \\0 \; or\; M_{L,1}(I=L)&...&M_{LN}\\...&...&... \\0 \; or\; M_{N,1}(I=N)&...&M_{NN}\end{bmatrix}$

The volume of the total determinant, will be the sum of several volumes obtained for each matrix, multiply the column value (only non-zero component of first column in each matrix) by the hypervolume (determinant) associated in n-1 dimensions submatrix, with the right convention of signal.

As above, this submatrix from column 2 until column N is a square matrix, because it discard the line containing values of the same line to the non-zero value in the first column, as it does not contribute to moving away from the pure axis expressed in column 1.

From top to bottom, the number of swaps is incremented by 1, alternating the positive (initial) signal with the negative signal in the corresponding part.

In the end, the term is positive with an even number of row swaps (permutations) and negative the term with an odd number of row swaps or permutations.

So we have the Laplace Expansion:

$\mathbf {Det (M) = \Sigma_{L = 1, N} (-1)^{L + C} M[L, 1] Minor_{L1}}$

With transpose and column or row swap properties, it’s easy to proof that this formula can by applied to any row or column basis.


We will now prove one of the most important properties form the determinants that states:

A determinant of a matrix product is the product of the determinants of the matrices.

Linear Transformations are one the most important fields in Linear Algebra and all applied sciences. The geometric approach for determinants and this theorem are essentials to Linear Transformations because help to understand and value the transformations and its compositions.

Below I display a geometric proof what can be formally formulated with integrals.

Suppose an matrix $\mathbf {A}$ with $\mathbf {N}$ dimensions.

a) If $\mathbf {B}$ is a matrix representing a cube with all dimensions = 1, this corresponds to the identity matrix $\mathbf {I}$ with volume 1. It’s easy to notice this, because this represents the standard axis of all dimensions.

When we multiply a matrix $\mathbf {A}$ by $\mathbf {I}$ we get the matrix $A$ itself.

Since the determinant represents the volume of matrix $\mathbf {A}$, then in this case

$ \mathbf {Det (A * I) = Det (A) * Det (I) = Det (A)}$

b) If the cube can be represented by a form with all dimensions $\mathbf {=\beta}$, that is, a matrix $\mathbf {B}$ with all its main diagonal $= \mathbf {\beta}$, then by linearity, and considering that all $\mathbf {N}$ lines are multiplied by $\mathbf {\beta}$, we have :

$\mathbf {Det (A * B) = Det (A * \beta I) = \beta^N Det (A * I) = \beta^ N Det (A)}$ $\mathbf {Det (A) * Det (B) = Det (A)\beta^N Det (I) = \beta^N Det (A)}$

**c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\mathbf {s}$ of small cube matrices $\mathbf {B_k}$ ($\mathbf {k}$ is an index from 1 to s) of dimension $\mathbf {\beta}$ (diagonal matrix with only $\mathbf {\beta}$value), where $\mathbf {\beta}$ is arbitrarily small and therefore $\mathbf {s}$ arbitrarily large.

Look at the volume below the colorful surface, above the plane $\mathbf {z=0}$ (vertical axis), below the plane $\mathbf {y=0}$ (negative y) to down and left, right to the place $\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\mathbf {s}$) of tiny cubes.

product of matrices vs determinants

$\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \beta^N \qquad}$ (c1) $\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)} $

By (b) above

$\mathbf {= \beta^N Det(A) + ... + \beta^N Det(A)\qquad}$ ($s$ times) $\mathbf {= s Det (A) \beta^N \qquad}$(c2)

So by (c1) and (c2)

$ \mathbf {Det (A * B) = Det (A) * Det (B)}$

$\bigstar \bigstar \bigstar$

Notice that based on a few properties, which make very logical in the context of the geometric interpretation of determinants, we went able to deduce the Laplace Expansion and the important property above that relates Linear Transformations with Determinants.

On the other hand, when a determinant is defined by the Laplace Expansion, that seems convolute for the layman, all properties are derived from it, including the geometric interpretation above. (and it is not an easy proof because it depends on many matrix transformations).

However this is an inverted approach because Laplace Expansion is a consequence from geometric properties of determinants that is highly useful and desirable to Linear Algebra, not the opposite (Laplace Expansion would be an exoteric truth and, thanks God, geometric interpretation emerges as a derivation.

The conventional way it’s a very demotivating approach for those who study Math.


The determinant is the product of the eigenvalues of the matrix. If the matrix is invertible, the eigenvalues of that inverse matrix are the reciprocals of the eigenvalues of the original matrix -- this fails if one of them is zero. Hence,

($A$ is non-invertible) $\leftrightarrow$ ($A$ has a zero eigenvalue) $\leftrightarrow$ (det $A=0$).

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