# Is this a well known determinant identity? Are there any generalizations?

Let $A$ be a $3\times3$ matrix and for any $i,j\subseteq\{1,2,3\}$, let $A^{i,j}$ denote the $2\times2$ matrix resulting from removing row $i$ and column $j$ from $A$. Then:

$\det\left(\begin{array}{ccc}\det(A^{1,1})&\det(A^{1,2})&\det(A^{1,3})\\ \det(A^{2,1})&\det(A^{2,2})&\det(A^{2,3})\\ \det(A^{3,1})&\det(A^{3,2})&\det(A^{3,3})\end{array}\right)=\det(A)^2$

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This is actually exactly the same question as this one (which was asked yesterday!), although it is not obvious, because of different notation. – Hans Lundmark Oct 27 '11 at 6:12

Let $A$ be an $n \times n$ matrix. Let $C_{ij}$ be the $ij^{th}$ cofactor of $A$, defined by $C_{ij} = (-1)^{i+j} \det A^{ij}$ and let $$C = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & \cdots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{bmatrix}$$ be the cofactor matrix. Also let $M$ be the matrix of minors; that is, the matrix that appears in the left hand side of the equation in the question. We will show that

1. $\det C = (\det A)^{n-1}$, and
2. $\det M = \det C$,

which together establish the claim.

Claim 1. From this wikipedia page, we know that $$C^T A = A C^T = (\det A) I_n.$$

If $A$ is invertible, then taking determinants on both sides, we get: $$\det (C^T) \cdot \det A = (\det A)^n \cdot 1.$$ Since $\det (C^T) = \det C$ and $\det A \neq 0$, we get $\det C = (\det A)^{n-1}$, and we are done.

On the other hand, if $A$ is noninvertible, then $\det A = 0$ and $C^T A = AC^T = 0$. In this case, (WLOG assuming $n \geq 1$) it suffices to show that $\det C = 0$, which is equivalent to showing that $C$ is noninvertible as well. Here, we have two subcases:

• If $A$ is the zero matrix, then $C$ is also the zero matrix, and hence non-invertible.

• Suppose $A$ is nonzero. Then if $C$ is invertible, then $C^T A = 0$ implies that $$A = \Big((C^{-1})^T C^T \Big) A = (C^{-1})^T \cdot 0 = 0 ,$$ which is a contradiction. Hence $C$ is noninvertible, and we are done.

Claim 2. We now prove that $\det M = \det C$. We have the relation $C_{ij} = (-1)^{i+j} M_{ij} = (-1)^{i+j} \det A^{ij}$. So, if we take the matrix $M$, and multiply the $i^{\rm th}$ row by $(-1)^i$ and $j^{\rm th}$ column by $(-1)^j$, then we end up with the matrix $C$. In this process, the determinant gets multiplied by $$\left(\prod_{i=1}^n (-1)^i \right) \left(\prod_{j=1}^n (-1)^j \right) = \left(\prod_{i=1}^n (-1)^i \right)^2 = \prod_{i=1}^n (-1)^{2i} = 1.$$ In other words, $\det C = \det M$.

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What if $\det(A)=0$? The formula still works, but the proof it doesn't... Anyhow, $AC=\det(A)I$ proves it amediatelly in this case too. – N. S. Oct 26 '11 at 22:24
I think you're forgetting a transpose. – Grasshopper Oct 26 '11 at 22:25
Magic trick when $\det A = 0$: Suppose entries of the matrix are $a_1, a_2, \dots, a_n$. Consider matrix from $R[x_1, x_2, \dots, x_n]$ where $a_i$ is replaced by new variable $x_i$ from polynomial ring. Now $\det A$ is a nonzero polynomial. Proceed with proof and then use evaluation homomorphism in final equality. Doesn't need continuity. – sdcvvc Oct 26 '11 at 22:30
@Srivatsan Narayanan: If $\det(A)=0$ then $AC=0_n$. Then $C$ cannot be invertible cause then $A=0$ which would make $C=0$ ;) – N. S. Oct 26 '11 at 22:30
@user9176 Thanks for the comment. I rewrote the answer completely incorporating it. – Srivatsan Oct 27 '11 at 0:28