# To prove an identity involving determinants

Suppose, I have a $$3 \times 6$$ matrix of the form,

$$A=\left( \begin{array}{ccccc} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & a_{1,6} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & a_{2,6} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & a_{3,5} & a_{3,6}\\ \end{array} \right)=\left( a_1 \, a_2 \, a_3 \, a_4 \, a_5 \, a_6\right)$$

where $$a_i$$ denotes a column matrix, and A is of rank 3.

Now, let me denote $$A_{i,j,k}=( a_i \, a_j \, a_k)$$. Then, how can I prove the following identity?

$$\left| \begin{array}{ccccc} |A_{4,2,3}| & |A_{5,2,3}| & |A_{6,2,3}| \\ |A_{1,4,3}| & |A_{1,5,3}| & |A_{1,6,3}| \\ |A_{1,2,4}| & |A_{1,2,5}| & |A_{1,2,6}| \\ \end{array} \right|=\left| A_{1,2,3} \right|^2 \left| A_{4,5,6} \right|$$

I want an algebraic way to prove the above identity. This question is a generalization of the results in (How to prove the following Matrix Identity?) Any relevant references will be beneficial.

## 2 Answers

Note that the matrix $$\begin{pmatrix} \det A_{4,2,3} & \det A_{5,2,3} & \det A_{6,2,3}\\ \det A_{4,3,1} & \det A_{5,3,1} & \det A_{6,3,1}\\ \det A_{4,1,2} & \det A_{5,1,2} & \det A_{6,1,2}\\ \end{pmatrix}$$ is $$\begin{pmatrix} a_2\times a_3 & a_3\times a_1 & a_1\times a_2 \end{pmatrix}^T A_{4,5,6}$$ So take determinant and recall the identity $$[\mathbf{b}\times\mathbf{c},\mathbf{c}\times\mathbf{a},\mathbf{a}\times\mathbf{b}]=[\mathbf{a},\mathbf{b},\mathbf{c}]^2$$ to get the answer.

• Thanks! But I can't figure out how to derive the second equation from the matrix form. Sep 13, 2021 at 6:52
• Can me please show me that or any relevant reference? Sep 13, 2021 at 6:53
• @Epsilon Which second equation? The one with scalar triple product? Sep 13, 2021 at 6:55
• Going along the top row $(a_2\times a_3)^T$ of the $\begin{pmatrix}a_2\times a_3&\dots\end{pmatrix}^T$ and first column $a_4$ of $A_{4,5,6}$, for example, you have $a_4\cdot(a_2\times a_3)=\det A_{4,2,3}$. Similarly the other entries. Sep 13, 2021 at 7:02
• Note rank(A) can't be more than 3. If you want to take $A_{1,2,3}$ the identity properly, consider multiplying $A$ by $(A_{1,2,3})^{-1}$ on the left (and you need to deal with the case $A_{1,2,3}$ not invertible, there are various ways around this just pick one you are comfortable with -- density argument, explicit calculation for this case, etc.) and what effect it has on $\det A_{i,j,k}$ and what you need to prove. Sep 13, 2021 at 16:50

Here is an alternative proof:

With no loss of generality let me choose label $$a_i$$'s such that $$|A_{1,2,3}|\neq 0$$.

Then, I can write,

$$A= (a_1 \, a_2 \, a_3 \, a_4 \, a_5 \, a_6)= (a_1 \, a_2 \, a_3) A'= A_{1,2,3}\, A'$$

where,

$$A' =( I \, a'_4 \, a'_5 \, a'_6)$$

such that $$I$$ is a $$3\times 3$$ unit matrix, $$a'_4=(a_1 \, a_2\, a_3)^{-1} a_4$$ and similar expressions for $$a'_5$$ and $$a'_6$$.

Now, we note that

$$A_{i,j,k}=A_{1,2,3} \, A'_{i,j,k}$$ Therefore, $$|A_{4,2,3}|=|A_{1,2,3}|\,\,|A'_{4,2,3}|=|A_{1,2,3}| (a'_4)_{11}$$ Similar identity gives us, $$\left| \begin{array}{ccccc} |A_{4,2,3}| & |A_{5,2,3}| & |A_{6,2,3}| \\ |A_{1,4,3}| & |A_{1,5,3}| & |A_{1,6,3}| \\ |A_{1,2,4}| & |A_{1,2,5}| & |A_{1,2,6}| \\ \end{array} \right|=|A_{1,2,3}|^{3} |a'_4 \, a'_5 \, a'_6|=|A_{1,2,3}|^{3} A'_{4,5,6}=|A_{1,2,3}|^{2} A_{4,5,6}$$

Hence, proved!