# Prove that the determinants are equal

$$Let\ A= \begin{bmatrix} 0 & a^2 & b^2 & c^2\\ a^2 & 0 & z^2 & y^2\\ b^2 & z^2 & 0 & x^2\\ c^2 & y^2 & x^2 & 0\\ \end{bmatrix}$$ $$Let\ B= \begin{bmatrix} 0 & ax & by & cz\\ ax & 0 & cz & by\\ by & cz & 0 & ax\\ cz & by & ax & 0\\ \end{bmatrix}$$

Show that $det(A)=det(B)$ I have tried by multiplying and dividing xyz and abc to symmetric rows and columns however was unable to take out the common.Please do not write any method of expanding the determinants and comparing them.

• +1 because this thing is nice and I am wondering whether it generalizes. Aug 7, 2014 at 13:59

I think this question is not suitable here. Anyway here is the solution.

Act on the matrix $A$ as follows:

1. Multiply the first row by $xyz$
2. Multiply the second row by $xbc$
3. Multiply the third row by $ayc$
4. Multiply the fourth row by $abz$

You get a matrix $C$ with $\det C =a^2b^2c^2x^2y^2z^2$. Now act on $C$ as follows

1. Divide the first column by $abc$
2. Divide the second column by $ayz$
3. Divide the third column by $xbz$
4. Divide the fourth column by $xyc$

You get $B$ and $\det C= a^2b^2c^2x^2y^2z^2\det B$. So, $\det A=\det B$ at least when all $a,b,c,x,y,z$ are not $0$, and therefore (by continuity) for all values of them.

• (by "here", the author means "on MathOverflow" FYI). Aug 7, 2014 at 15:24

Another approach, which might be easier to generalize.

Step 1 In the Leibniz expansion of the determinant of a $3\times 3$ matrix $M$, each term contains at least one among $\{m_{11},m_{12},m_{21},m_{22}\}$.

Proof: you have to take three terms from different rows and columns. If you wish to avoid the topmost $2\times 2$ square, you can take at most one term from the last row and one term from the last column.

Step 2 Let $N$ be any $2\times 2$ submatrix of a $3\times 3$ matrix $M$. In the Leibniz expansion of the determinant, each term contains one element of $N$.

Proof: it's really the same as Step 1, just on two arbitrary rows/columns instead of the first one.

Step 3 In the Leibniz expansion of $\det A$, each nonzero term containing $c^2$ contains $z^2$ as well.

Proof: after taking one of the two $c^2$ factors and eliminating its row and column, the remaining $3\times 3$ submatrix contains a $2\times 2$ submatrix $\begin{bmatrix}z^2 & 0\\ 0 & z^2\end{bmatrix}$. So we have to take either $0$ or $z^2$.

Step 4 In each nonzero term of the Leibniz expansion of $\det A$, the term $z^2$ is taken at least as many times as $c^2$.

Proof: if you take both factors $c^2$, the remaining two terms have to be taken from $\begin{bmatrix}z^2 & 0\\ 0 & z^2\end{bmatrix}$, so the result holds. If you take only one factor $c^2$, then the result holds by Step 3. If you take no factors $c^2$, there is nothing to prove.

Step 5 In each nonzero term of the Leibniz expansion of $\det A$, $c^2$ and $z^2$ are taken the same number of times.

Proof: Call $P(c,z)$ the statement of Step 4. All we used to prove it is that the submatrix complementary to $\begin{bmatrix}c^2 & 0\\ 0 & c^2\end{bmatrix}$ is $\begin{bmatrix}z^2 & 0\\ 0 & z^2\end{bmatrix}$, so the same reasoning can be used to prove $P(z,c)$. But then this means that we take $z^2$ and $c^2$ the same number of times.

Step 6 In each nonzero term of the Leibniz expansion of $\det A$, ($a^2$ and $x^2$) and ($b^2$ and $y^2$) are taken the same number of times.

Proof: Analogous to Step 5.

Step 7 $\det A=\det B$ Call $A=F(a^2,b^2,c^2,x^2,y^2,z^2)$. Then $B=F(ax,by,cz,ax,by,cz)$. Since the terms $c^2$ and $z^2$ always appear in pairs, if we replace them both with $cz$ the determinant is unchanged, because $(c^2)(z^2)=(cz)(cz)$. The same holds for the other two pairs of letters.