Raising the bar

Axioms 1-3 and Properties I-III

Axioms

Axiom 1. Homogeneity. If $latex A^\prime$ denotes the matrix obtained from $latex A$ by multiplying one of its rows by a number $latex k,$ then $latex \det A^\prime=k\det A.$

This implies that for $latex k\neq 0,$ $latex A^\prime$ and $latex A$ are invertible simultaneously.

Axiom 2. If $latex A^\prime$ denotes the matrix obtained from $latex A$ by adding one of the rows of $latex A$ to another, then $latex \det A^\prime=\det A$.

We remember that adding one of the rows of $latex A$ to another corresponds to adding one equation of the system $latex Ax=y$ to another and so this operation should not impact solvability of the system.

Axiom 3. $latex \det I=1.$

Adding this axiom on top of the previous two makes the determinant a unique function of a matrix.

Properties

Assuming that $latex A$ is of size $latex n\times n,$ it is convenient to partition it into rows:

$latex A=\left(\begin{array}{c}A_1 \\... \\A_n\end{array}\right).$

Property I. If one of the rows of $latex A$ is zero, then $latex \det A=0.$

Indeed, if $latex A_i=0,$ then $latex A_i=0\times A_i$ and by Axiom 1 $latex \det A=0\times\det A=0.$

Property II. If we add to one of the rows of $latex A$ another row multiplied by some number $latex k,$ the determinant does not change.

Proof. If $latex k=0,$ there is nothing to prove. Suppose $latex k\neq 0$ and we want to add $latex kA_j$ to $latex A_i.$ This result can be achieved in three steps.

a) Multiply $latex A_j$ by $latex k.$ By Axiom 1, $latex \det A$ gets multiplied by $latex k.$

b) In the new matrix, add row $latex kA_j$ to $latex A_i.$ By Axiom 2, this does not change the determinant.

c) In the resulting matrix, divide row numbered $latex j$ by $latex k.$ The determinant gets divided by $latex k.$

The determinant of the very first matrix will be the same as that of the very last matrix, while the $latex i$th row of the last matrix will be $latex A_i+kA_j.$

Property III. If rows of $latex A$ are linearly dependent, then $latex \det A=0.$

Proof. Suppose rows of $latex A$ are linearly dependent. Then one of the rows can be expressed as a linear combination of the others. Suppose, for instance, that $latex A_1=k_2A_2+...+k_nA_n.$ Multiply the $latex i$th row by $latex -k_i$ and add the result to the first row, for $latex i=2,...,n.$ Thereby we make the first row equal to $latex 0,$ while maintaining the determinant the same by Property II. Then by Property I $latex \det A=0.$

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