I was studying inverse matrix. Suddenly I stumbled on the inverse of 3×3 K. And it involved a division by the determinant (Well, only with numbers it was). And it was also said, about involving a division by the determinant. Then I got the inverse of a 2×2 matrix,with variables as entries(That's the general form of the 2×2 real number matrix). And it also involved a division by ad-bc(the determinant). But why is that? Maybe it's the result of something in the matrix?

  • $\begingroup$ That is one way to get the inverse and it relies on properties of the adjugate (or also called classical adjoint) matrix if any square matrix. There is one other way to get the inverse by means of some simple row operations on it and on the unit matrix which doesn't require to know the matrix's determinant...though it still is there the fact that the determinant must be $\;\neq0\;$ (i.e., the matrix must be regular) $\endgroup$
    – DonAntonio
    Jun 7, 2021 at 10:49
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    $\begingroup$ Since $\det(A^{-1})=\frac{1}{\det(A)}$, it is not suprising that the determinant may appear for the formula of the inverse. $\endgroup$ Jun 7, 2021 at 10:50
  • $\begingroup$ That's what I'm asking- why det(A) appears? Kind of a question like why the natural numbers have the property that (n+(n+2))/2=n+1. $\endgroup$
    – user919525
    Jun 7, 2021 at 10:56
  • $\begingroup$ You could think about volumes of parallelepipeds. $\endgroup$ Jun 7, 2021 at 11:20
  • $\begingroup$ I haven't yet studied about determinants being volume. So this case isn't gonna work... $\endgroup$
    – user919525
    Jun 7, 2021 at 11:28

4 Answers 4


Question: "Why the inverse of a matrix involves division by the determinant?"

Answer: We use the adjugate matrix and the determinant to prove existence of an inverse of a matrix as follows:

The "adjugate matrix" $ad(A)$ has the property that $ad(A)A=Aad(A)=det(A)I$ where $det(-): Mat(n,k) \rightarrow k$ is a map with $det(AB)=det(A)det(B)$. Here $Mat(n,k)$ is the set of $n\times n$-matrices with coefficients in $k$. $det(A)$ is the "determinant" of the matrix $A$ as defined in your linear algebra course.

Lemma: A square matrix $A$ has an inverse iff $det(A)\neq 0$.

Proof: If $det(A)\neq 0$ it follows $A^{-1}:=\frac{1}{det(A)}ad(A)$ is an inverse. Conversely assume there is a matrix $B$ with $AB=BA=I$. It follows $det(AB)=det(A)det(B)=1$ and hence $det(A) \neq 0$.

Hence the adjugate matrix and the determinant map implies the existence of an inverse of $A$: The matrix $A$ has a unique inverse $A^{-1}$ iff $det(A)\neq 0$.

Example: Let \begin{align*} A= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*}

and define the adjunct matrix $ad(A)$ by

\begin{align*} ad(A)= \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \end{align*}

It follows

\begin{align*} ad(A)A=Aad(A)= \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} =\end{align*}

\begin{align*} det(A)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align*}

There is in general for any $n\times n$-matrix $A$ a unique matrix $ad(A)$ with $ad(A)A=Aad(A)=det(A)I$. This result is proved in any serious linear algebra course. Hence the above proves the Lemma explicitly for any $2\times 2$-matrix.


  • $\begingroup$ I didn't get the $det(A): Mat(n,k) \rightarrow k$ part. I mean I still don't understand this sort of writing. $\endgroup$
    – user919525
    Jun 7, 2021 at 11:19
  • $\begingroup$ @10209 all that notation means is that $\det$ is a function from the space of $n\times n$ matricies with coefficients in the field $K$, into the field $K$ (you can just replace $K$ with $\mathbb{R}$ if that makes you more comfortable, for now). So $\det$ takes in a square matrix $A$ and returns $\det(A)$. $\endgroup$
    – C Squared
    Jun 7, 2021 at 11:31
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    $\begingroup$ @C Squared Well, I need quite a bit of Wiki now. But thanks. $\endgroup$
    – user919525
    Jun 7, 2021 at 11:33

Here is an intuitive explanation: Remember that a matrix A regarded as a linear transformation scales up the area for 2x2 matrices, and volume for 3x3 matrices (and higher dimensional Lebesgue measure in those cases) by multiplicative factor of determinant of A.

Identity matrix preserves the area (volume, etc). So to undo the effect of A, its inverse has to involve 1/det(A).


The matrix $\begin{bmatrix} a & c\\ b& d \end{bmatrix}$ scales everything by ad-bc; The adjoint matrix brings the basis vectors back to the axes but also scales it again by ad-bc. Now the area is $(ad-bc)^2$ but also both have the same length of ad-bc. So, dividing both unit vectors by (ad-bc) gets us back to the original unit matrix.

Inverse visulaization

Let me explain:

First you need to understand what a matrix is and what a determinant is.

A matrix represents a Linear transformation. We can think about a Cartesian coordinate system and a linear transformation applied to it which lands $\hat{i}$ to (a,b) and $\hat{j}$ to (c,d). This transformation can be represented by a matrix $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$ .

A vector which was originally at (x,y) lands to new postion $$x \begin{bmatrix} a \\ b \end{bmatrix} + y \begin{bmatrix} c \\ d \end{bmatrix} $$ $$= \begin{bmatrix} ax+by \\ cx+dy \end{bmatrix}$$ So the new coordinates of the vector is $(ax+by,cx+dy)$. This is explained neatly by 3b1b in this video


When we change the basis vectors and hence the coordinate system, any square or circle or any kind of area in the coordinate system also gets scaled by some amount. This is called determinant.

If we multiply the initial area to the determinant, we get the final area.


Finding the inverse

The inverse of a matrix A is the matrix when multiplied to A gives the identity matrix.

Or from our understanding of matrix as transformations, The inverse of a matrix A (a transformation) is the transformation when applied to A brings back all the vectors to the initial position.

If i and j was initially at $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$ and later transformed to$\begin{bmatrix} a & c\\ b& d \end{bmatrix}$ then we can say that the inverse of A is the transformation which brings the i and j back to their initial coordinates.

So currently the basis vectors is like this:

drawn diagram

Now, we should bring both the vectors to the axes. For that we should change the y-coordinate of x and x-coordinate of y. But how much? Remember that x and y are the new basis vectors and all the transformation is done according to these.

Initially the tip of X vector is has a y-coordinate of $b \hat{j}$ where $\hat{j}$ is the unit vector in the y direction.

But now Y is the new basis vector. So, $b Y$ is the current y coordinate of X. So to bring it back, you must scale it by $-b Y$. Similarly, You must multiply $-c X$ to the X vector to bring it to the X axes. So, the corresponding matrix is:

$\begin{bmatrix} ? & -c\\ -b& ? \end{bmatrix}$

Now we have two vectors in the x and y axes but they are of unequal length. First lets find their length:

Image here

So, vector X has length $a - \frac{bc}{d}$ and

Vector Y has length $d - \frac{bc}{a}$

To make both equal multiply X with d and Y with a.

Now, we have two vectors with length $ad-bc$.

So, the matrix to bring the X and Y back to the orthogonal axes and to make it equal is :

$\begin{bmatrix} d & -c\\ -b& a \end{bmatrix}$

This is also called the adjoint of the matrix. But we are not done yet: the unit basis vectors has a length of 1.

To make it the original unit matrix, we have to divide both by ad-bc. since this is just the determinant of the matrix,

$$inv(A) = \frac{adj(A)}{det(A)}$$


To understand it intuitively, you can consider the real number $x$. Its inverse is $\frac1x$. i.e. we are dividing by $x$. Similiary, to find the inverse of matrix $A$, we divide by the representation of matrix $A$, i.e. its determinant. The difference here being, instead of dividing the identity ($I$), we are dividing the adjoint matrix.

  • $\begingroup$ >>representation of matrix A, i.e. its determinant.>>The determinant does not identify a matrix.Determinant is the area in case of a 2 by 2 matrix .The question of OP is why the we divide by the determinant of A instead of some other quantity $\endgroup$
    – Sophile
    Sep 14, 2021 at 12:13
  • $\begingroup$ @AbhinavPB area of what? $\endgroup$
    – aarbee
    Sep 14, 2021 at 12:17
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    $\begingroup$ Determinant of a matrix [a c //bd] is the area of a parallelogram with the adjacent vectors having coordinates (a,d) and (c,d). I suggest you to watch this video:youtu.be/Ip3X9LOh2dk $\endgroup$
    – Sophile
    Sep 14, 2021 at 12:20
  • $\begingroup$ In case of 3 by 3 matrix, determinant is the volume of the paralleopiped formed by three vectors having coordinates as the columns of the matrix. $\endgroup$
    – Sophile
    Sep 14, 2021 at 12:26
  • $\begingroup$ That is why I said determinant doesn't uniquely identify the matrix.because two parallelograms can have the same area.that is, two different matrices can have the same determinant. For example consider [1 0 //1 1] and [1 0//0 1]. Both have a determinant of 1 $\endgroup$
    – Sophile
    Sep 14, 2021 at 12:28

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