I have a proof for the method to find the inverse by taking the adjugate of the matrix over the determinant. I believe that it is correct; however, I would like to have it checked over before I submit it for bonus marks. The proof goes as follows:

Let $A\in\mathbb{M}_{n\times{n}}(\mathbb{R})$ be an $n\times{n}$ invertible matrix. Let $A^{-1}=\begin{bmatrix}\vec{x_1}\cdots\vec{x_n}\end{bmatrix}$.


$\therefore A\vec{x_i}=\vec{e_i}^T\text{for } 1\leq{i}\leq{n}$

By Cramer's Rule, $x_{ji}=\frac{det(A_j)}{det(A)}\text{for }1\leq{i,j}\leq{n}$ where $A_j$ is the matrix formed by replacing the jth column of $A$ with the solution vector; in this case, $\vec{e_i}$.

By properties of determinants, we can add a multiple of a column to another column without changing the determinant.

$\therefore det(A_j)\\=det(\begin{bmatrix}\vec{a_1}\cdots\vec{a_{j-1}}\space\space\space\space\vec{e_i}\space\space\space\space\vec{a_{j+1}}\cdots\vec{a_n}\end{bmatrix})\\=det(\begin{bmatrix}\vec{a_1}-a_{1i}\vec{e_i}\cdots\vec{a_{j-1}}-a_{j-1,i}\vec{e_i}\space\space\space\space\vec{e_i}\space\space\space\space\vec{a_{j+1}}-a_{j+1,i}\vec{e_i}\cdots\vec{a_n}-a_{ni}\vec{e_i}\end{bmatrix})$

This matrix has all 0s in the ith row except for a 1 in the jth column. Thus, if we expand $det(A_j)$ along the ith row, we get:

$det(A_j)\\=0\space det(A_{i,1})\times(-1)^{i+1}+\cdots+1\space det(A_{i,j})\times(-1)^{i+j}+\cdots+0\space det(A_{i,n})\times(-1)^{i+n}\\=det(A_{i,j})\times(-1)^{i+j}$

where $A{i,j}$ is the matrix formed by removing the ith row and jth column of A.

$\therefore x_{ji}=\frac{det(A_j)}{det(A)}$

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


My questions are: - Can I use Cramer's rule in the proof? - Is this proof correct? - Are there any terminology issues or blatant inconsistencies/errors?


  • 6
    $\begingroup$ Usually the Cramer rule is proven by using this formula. $\endgroup$ – N. S. Jun 9 '17 at 18:34
  • $\begingroup$ @N.S. Oh well... rip. As long as it's provable otherwise though it should be fine. I hope. $\endgroup$ – HyperNeutrino Jun 9 '17 at 20:10
  • $\begingroup$ LOL. I gave @HyperNeutrino the idea to use crammers rule. Oh well, it still works. $\endgroup$ – Radek Martinez Jun 11 '17 at 15:01
  • $\begingroup$ @RadekMartinez Haha yes :P Too bad the teacher is indecisive though -.- $\endgroup$ – HyperNeutrino Jun 11 '17 at 18:19

The formula $A^{\operatorname{adj}}A = \det(A) I$ is Cramer's rule so whether or not you can use Cramer's rule depends on how you proved it.

Your proof looks correct but could be written better if you had a better understanding of the key ideas.

The canonical proof is to expand the left hand side of $A^{\operatorname{adj}}A = \det(A) I$ in terms of matrix multiplication. Let $A^{\operatorname{adj}} = (b_{i,j})$ where $b_{i,j} = (-1)^{i + j}\det(A_{j,i})$ then the $(i,j)$ entry of the left-hand side is

$$ \sum_{k = 1}^n b_{i,k}a_{k,j} = \sum_{k = 1}^n (-1)^{i + k}\det(A_{k,i})a_{k,j}$$

which is exactly the expansion by minors formula (along the $i$-th column) for the determinant of $A$ with the $i$-th column replaced by the $j$-th. So for $i = j$ this gives you $\det A$ and for $i \ne j$ this gives zero since in that case two columns are equal. Hence the formula $A^{\operatorname{adj}}A = \det(A) I$.

If you looked at $AA^{\operatorname{adj}}$ instead, you would get the expansion by minors formula for the rows instead.

This seems like from my cursory look, to be what you are doing. But it is worth pointing out what is actually going on.

  • $\begingroup$ Oh, that proof makes a lot more sense. Thanks! $\endgroup$ – HyperNeutrino Jun 9 '17 at 20:11

It is perfectly fine to use Cramer's rule for such a purpose. Here it is a self-contained approach.

Lemma 1. If $A$ is an invertible $n\times n$ matrix, $v=(x_1,x_2,\ldots,x_n)^T$ and $w=(y_1,y_2,\ldots,y_n)^T$, the solution of $Av=w$ is given by $x_k = \frac{\det(A_k)}{\det(A)} $, where $A_k$ is the original $A$ matrix with the $k$-th column being replaced by $w$.

Proof: Let $e_1,e_2,\ldots,e_n$ be the canonical base of $\mathbb{R}^n$ and $a_1,a_2,\ldots,a_n$ be the columns of $A$.
The linear map $f$ represented by $A$ sends $e_j$ into $a_j$ for any $j\in[1,n]$.
The linear map $f_k$ represented by $A_k$ does the same, with the exception of sending $e_k$ into $w$.
Let us consider what $f^{-1} f_k$ does: for any $j\neq k$, it sends $e_j$ into $e_j$.
$f^{-1}f_k$ sends $e_k$ into $v$, hence the matrix representing $f^{-1} f_k$ has the $k$-th column equals to $v$ and the remaining columns equal to the columns of the identity matrix. In particular $\det(f^{-1}f_k)=x_k$ and the claim follows from Binet's formula $\det(AB)=\det(A)\det(B)$.

Lemma 2. If $A$ is an invertible matrix, in order to find $A^{-1}$ it is enough to find $A^{-1}e_1,\ldots,A^{-1}e_n$. That can be done through Lemma 1 and a Laplace expansion along the proper column, leading to the adjugate formula.

  • $\begingroup$ Okay. Thanks for the help! $\endgroup$ – HyperNeutrino Jun 9 '17 at 20:11
  • $\begingroup$ @HyperNeutrino: you're welcome. $\endgroup$ – Jack D'Aurizio Jun 9 '17 at 20:13

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