Problem. Let \(D\colon M_{n\times n}(\mathbb{F})\to\mathbb{F}\) be a volume function, and let \(A\in M_{n\times n}(\mathbb{F})\).
We will denote by \(\varepsilon\) the elementary row operation \(R_{i}\to R_{i}+cR_{j}\), where \(1\le i,j\le n\), \(i\ne j\), and \(c\in\mathbb{F}\).
Prove that \(D(\varepsilon(A))=D(A)\).
Proof. Let us mark \(A\) with \(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\) (we assume without loss of generality that \(j<i\)).
Then the matrix \(A\) after operation \(\varepsilon\) will be
\[\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}+cR_{j}\\ \vdots\\ R_{n} \end{pmatrix}\]
Since a volume function is in particular a multi-linear function,
\[D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}+cR_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)=D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\right)+D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ cR_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)=D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\right)+cD\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)\]
Recall that for a matrix \(B\) and for the elementary row operation \(R_{w}\leftrightarrow R_{k}\) which we mark with \(\varepsilon'\), \(D\left(\varepsilon'\left(B\right)\right)=-D\left(B\right)\).
\[\therefore D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\right)+cD\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)=D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\right)+c\cdot-D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ R_{j}\\ \vdots\\ \vdots\\ R_{n} \end{pmatrix}\right)\]
And since \(D\) is a volume function, it’s an alternating function in particular, which means it returns zero on a matrix that has two adjacent identical rows: \[c\cdot-D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ R_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)=c\cdot-0=0\]
Hence
\[D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}+cR_{j}\\ \vdots\\ R_{n} \end{pmatrix}\right)=D\left(\begin{pmatrix}R_{1}\\ \vdots\\ R_{j}\\ \vdots\\ R_{i}\\ \vdots\\ R_{n} \end{pmatrix}\right)\]
Which means \(D(\varepsilon(A))=D(A)\) as required. ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.