See and Observe!

$\large \Delta = \left | \begin{array}{ccc} x & y & z \\ p & q & r \\ a & b & c \\ \end{array} \right | \times \left | \begin{array}{ccc} 0 & m & n \\ -m & 0 & k \\ -n & k & 0 \\ \end{array} \right | = 0$

Relevant wiki: Determinants

Intuition : we can notice much symmetry if we multiply the third line by $m$ and the first line by $k$ and the second line by $n$ , we can then proceed to sum the second line with the third line to get two same lines, thus, giving us the cancellation of the determinant.

Key points of the proof :

We will suppose all variables are in some $\mathbb{K}$ domain which is a field where determinants exist.

• (1) we can multiply each line by a scalar $X \in \mathbb{K}$ which will multiply the value of the determinant by $X \in \mathbb{K}$ ;
• (2) summing lines in a determinant will not change the value of the determinant ;
• (3) if two lines or two columns are equal in a determinant, then the determinant is zero ;
• (4) $\mathbb{K}$ is an integral domain, i.e. $\forall (a, b) \in \mathbb{K}, ab = 0 \implies a = 0 \text{ or } b = 0$ .

An approach :

By (1) :

$\begin{aligned} (mkn)\Delta & = \begin{vmatrix} mky + nkz & mkq + nkr & mkb + knc \\ nkz - mnx & nkr - mnp & nkc - mna \\ mnx + mky & mnp + mkq & mna + mkb \end{vmatrix} \text{ (notice how we get chunks of terms everywhere)} \\ & = \begin{vmatrix} mky + nkz & mkq + nkr & mkb + knc \\ nkz - mnx & nkr - mnp & nkc - mna \\ mky + nkz & mkq + nkr & mkb + nkc \end{vmatrix} \text{ (we sum line 2 and line 3 and replace it by line 3 by property (2))} \\ & = 0 \text{ (by property (3))} \end{aligned}$

Then, by (4) , either $m = 0$ , either $k = 0$ , either $n = 0$ , either $\Delta = 0$ .

Here we do some casework :

• If $m, n, k$ are not zero, then $\Delta = 0$ .

• If $m = 0$ , then $\Delta = \begin{vmatrix} nz & nr & nc \\ kz & kr & kc \\ nx + ky & np + kq & na + kb \end{vmatrix} = 0$ because of (3) and (1), we can factor out $k$ and $n$ and first line and second line are the same.

• If $k = 0$ , then $\Delta = \begin{vmatrix} my + nz & mq + nr & mb + nc \\ -mx & -mp & -ma \\ nx & np & na \end{vmatrix} = 0$ , same for $m = 0$ by factoring $-m$ and $n$ .

• Easily, we deduce the same for $n = 0$ for the same reasons. Left as an exercise to the reader.