# Intuition between determinants and matrix inverses

I have witnessed that the determinant of a matrix can be used to find the inverse of that matrix, however, I am unable to prove that this method of finding inverses works for any square matrix of dimension $m$. For, for me, it seems that although determinants can be used to find inverses their use in inverses are purely arbitrary and coincidental. So, I was wondering if there is any intuition for understanding the relation between determinants and inverses such as maybe a geometrical interpertation or something else.

• I'm not sure what answer you are looking for. At sounds like you are familiar with Cramer's rule and you're looking for the intuition behind this result? – Stefan Mesken Jun 21 '15 at 8:41
• No not really, I was just looking for an intuition as to why determinants can be used to find the inverses of a matrix because it seem s rather coincidental to me. – Reinhild Van Rosenú Jun 21 '15 at 9:30

The determinant can be interpreted as the signed volume of the epiped spanned by the images of the standard base vectors. Then "the volume is zero" is equivalent to "the image vectors live in a lower-dimensional subspace" is equivalent to "the linear map is not onto" is equivalent to "there is no inverse"

• Is this conversely true for when "there is an inverse", and could you possibly provide an example to show your explanation please? – Reinhild Van Rosenú Jun 21 '15 at 9:32
• Also, my question was asking as to why the determinant can be used to find the inverse of a matrix, not why determinants can show whether a matrix has an inverse or not. It would be very helpful for me if you could edit your question. – Reinhild Van Rosenú Jun 21 '15 at 16:41
• To your first Comment, yes, this is an if-and-only-if fact: A square matrix has an inverse if and only if its determinant is nonzero. More about this here. – hardmath Jun 22 '15 at 20:00
• There is an explicit formula for the inverse of $A$, any square matrix, in terms of signed cofactors divided by $\det A$. You may find this previous Question, Cramer's rule: Geometric interpretation, relevant. There $Adj(A)$ denotes the matrix of signed cofactors corresponding to the entries of $A$. – hardmath Jun 22 '15 at 20:06
• It might help to think of the inverse of $A$ as solving $AB = I$, so for each column of the identity matrix, we get the corresponding columns of $B = A^{-1}$. – hardmath Jun 22 '15 at 20:16

Consider a square matrix $A$ as a list of column vectors from a vector space $V$:

$$A = \begin{pmatrix} \vdots & \vdots & \ldots & \vdots \\ \mathbf{a}_{1} & \mathbf{a}_{2} & \ldots & \mathbf{a}_{n} \\ \vdots & \vdots & \ldots & \vdots \end{pmatrix}$$

The determinant $\det(A)$ is the coefficient of the wedge product of every column of $A$:

$$\mathbf{a}_{1} \wedge \mathbf{a}_{2} \wedge \ldots \wedge \mathbf{a}_{n} = \det(A) \; \mathbf{x}_1 \wedge \mathbf{x}_2 \ldots \wedge \mathbf{x}_n$$

And is said to live in the (1-dimensional) space $\wedge^n V$.

The wedge product is a generalization of the cross product that's good in any dimension. It is antisymmetric, meaning that $\mathbf{u} \wedge \mathbf{v} = - \mathbf{v} \wedge \mathbf{u}$, and $\mathbf{u} \wedge \mathbf{u} = 0$. Therefore if any factor appears in it twice it is 0: $\mathbf{v} \wedge \mathbf{a} \wedge \mathbf{b} \wedge \ldots \mathbf{v} = 0$ (via using the antisymmetry property to move one vector 'down the line' until it annihilates with its pair), hence the determinant property that if any column of $A$ is a linear combination of the others then $\det(A) = 0$.

If we consider the column vectors in $A$ as the sides of an $n$-dimension parallelepiped, gives the volume of that parallelepiped. See exterior algebra or this question for details.

Now consider wedge products of $n-1$ columns of $A$ at a time. For example, every column except the first, with alternating negative signs such that $\mathbf{a}_i \wedge \mathbf{c}_i = \det(A)$:

$$\mathbf{c}_1 = \mathbf{a}_{2} \wedge \ldots \wedge \mathbf{a}_{n} \\ \mathbf{c}_2 = - \mathbf{a}_1 \wedge \mathbf{a}_3 \wedge \ldots \mathbf{a}_n$$

If we wedge it with any other column, it will give 0, since that column will appear twice in the product. Therefore, we have:

$$\mathbf{a}_{i} \wedge \mathbf{c}_{j} = \det(A) \; \mathbf{1}_{i = j}$$

$\mathbf{c}_i$ is an element of $\wedge^{n-1} V$, but we can treat them as elements $\in V$, which makes the wedge product just a vector dot product, giving:

$$\mathbf{a}_i \cdot \mathbf{c}_j = \det(A) \; \mathbf{1}_{i=j}$$

$$C = (\mathbf{c}_1, \mathbf{c}_2, \ldots \mathbf{c}_{N})$$

This (considered as an element of $V$) is the cofactor matrix, and its transpose $C^T$ is called the adjugate matrix $\text{adj}(A)$.

Finally, we can write:

$$A \cdot C^T = A \cdot \text{adj}(A) = (\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n) \cdot \begin{pmatrix} \mathbf{c}_1 \\ \mathbf{c}_2 \\ \vdots \\ \mathbf{c}_n \end{pmatrix} = \det(A) I$$

Which shows that (assuming $\det(A) \neq 0$, which is the condition for $A$ being invertible anyway) the inverse of a matrix is its transposed cofactor matrix divided by its determinant:

$$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$

In summary, the fact that determinants can be used to find inverses is anything but arbitrary and coincidental. The two are fundamentally related -- the only thing that can annihilate every column of an invertible matrix except for one is a multiple of the wedge product of those columns.

(I am not entirely sure what is the best way to handle the minus signs in the cofactor matrix. We know that $A \cdot A^{-1} = A^{-1} \cdot A = I$, so the result must be the same, but it's not clear to me how to make that obvious in the construction.)