# 25. Properties of determinants

## 25. Properties of determinants

Recall that $\det(M)=\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)M_{1\sigma(1)}M_{2\sigma(2)}% \cdots M_{n\sigma(n)}.$ We'll use this formula to prove some nice properties of the determinant.

Lemma:

If two rows coincide then $\det(M)=0$ . For example $\det\begin{pmatrix}1&1&0\\ 1&1&0\\ 2&4&5\end{pmatrix}=0$ .

If row $i$ and row $j$ coincide then $M_{ik}=M_{jk}$ for all $k$ . Consider one of the terms $\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(i)}\cdots M_{j\sigma(j)}% \cdots M_{n\sigma(n)}$ in the determinant. Since $M_{ik}=M_{jk}$ for all $k$ , we have $M_{i\sigma(i)}=M_{j\sigma(i)},\qquad M_{j\sigma(j)}=M_{i\sigma(j)}.$ Therefore $\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(i)}\cdots M_{j\sigma(j)}% \cdots M_{n\sigma(n)}=\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(j)}% \cdots M_{j\sigma(i)}\cdots M_{n\sigma(n)}$ This is another term in our determinant, corresponding to the permutation $\sigma^{\prime}$ , which is defined by $\sigma^{\prime}(i)=\sigma(j),\qquad\sigma^{\prime}(j)=\sigma(i),\qquad\sigma^{% \prime}(k)=\sigma(k)\mbox{ for all }k\neq i,j.$ In other words: $M_{1\sigma(1)}\cdots M_{i\sigma(j)}\cdots M_{j\sigma(i)}\cdots M_{n\sigma(n)}=% M_{1\sigma^{\prime}(1)}\cdots M_{i\sigma^{\prime}(i)}\cdots M_{j\sigma^{\prime% }(j)}\cdots M_{n\sigma^{\prime}(n)}$ Since $\sigma^{\prime}$ differs from $\sigma$ by a single transposition, $\mathrm{sgn}(\sigma^{\prime})=-\mathrm{sgn}(\sigma)$ . Therefore these two terms contribute with opposite signs to $\det(M)$ , and cancel out $\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(j)}\cdots M_{j\sigma(i)}% \cdots M_{n\sigma(n)}+\mathrm{sgn}(\sigma^{\prime})M_{1\sigma^{\prime}(1)}% \cdots M_{i\sigma^{\prime}(j)}\cdots M_{j\sigma^{\prime}(i)}\cdots M_{n\sigma^% {\prime}(n)}=0$ Therefore all terms cancel in pairs, and the determinant is zero.

### Type I row operations

Lemma:

If $M^{\prime}$ is obtained from $M$ by the row operation $R_{i}\mapsto R_{i}+\lambda R_{j}$ then $\det(M^{\prime})=\det(M)$ .

We have $\det(M^{\prime})=\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)M^{\prime}_{1\sigma% (1)}\cdots M^{\prime}_{i\sigma(i)}\cdots M^{\prime}_{n\sigma(n)}$ $=\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots(M_{i\sigma(i)}% +\lambda M_{j\sigma(i)})\cdots M_{n\sigma(n)}$ $=\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(i)}% \cdots M_{n\sigma(n)}+\lambda\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)M_{1% \sigma(1)}\cdots M_{j\sigma(i)}\cdots M_{n\sigma(n)}.$ The first term here is $\det(M)$ . The second term is the determinant of the matrix obtained from $M$ by replacing the $i$ th row with the $j$ th row; in other words, the $j$ th row now appears twice. By the previous lemma, this second term vanishes, so $\det(M^{\prime})=\det(M)$ .

We can put any matrix into echelon form using type I row operations without changing the determinant. Once you're in echelon form (since you're then upper triangular), the determinant is just the product of diagonal entries.

### Type III row operations

Lemma:

If $M^{\prime}$ is obtained from $M$ by switching two rows (say $R_{i}\leftrightarrow R_{j}$ then $\det(M^{\prime})=-\det(M)$ .

Every term in $\det(M)$ has the form $\mathrm{sgn}(\sigma)M^{\prime}_{1\sigma(1)}\cdots M^{\prime}_{i\sigma(i)}% \cdots M^{\prime}_{j\sigma(j)}\cdots M^{\prime}_{n\sigma(n)}.$ Since $M^{\prime}$ and $M$ are related by switching rows $i$ and $j$ , this is just $\mathrm{sgn}(\sigma)M_{1\sigma(1)}\cdots M_{i\sigma(j)}\cdots M_{j\sigma(i)}% \cdots M_{n\sigma(n)},$ which is equal to $\mathrm{sgn}(\sigma)M_{1\sigma^{\prime}(1)}\cdots M_{i\sigma^{\prime}(i)}% \cdots M_{j\sigma^{\prime}(j)}\cdots M_{n\sigma^{\prime}(n)},$ where $\sigma^{\prime}$ is (as before): $\sigma^{\prime}(i)=\sigma(j),\qquad\sigma^{\prime}(j)=\sigma(i),\qquad\sigma^{% \prime}(k)=\sigma(k)\mbox{ for all }k\neq i,j.$ Since $\mathrm{sgn}(\sigma^{\prime})=-\mathrm{sgn}(\sigma)$ , this is equal to $-\mathrm{sgn}(\sigma^{\prime})M_{1\sigma^{\prime}(1)}\cdots M_{i\sigma^{\prime% }(i)}\cdots M_{j\sigma^{\prime}(j)}\cdots M_{n\sigma^{\prime}(n)},$ so for every term in $\det(M^{\prime})$ there's a corresponding term in $\det(M)$ which differs only by a sign. Therefore $\det(M^{\prime})=-\det(M)$ .

### Type II row operations

Lemma:

If $M^{\prime}$ is obtained from $M$ by $R_{i}\mapsto\lambda R_{i}$ then $\det(M^{\prime})=\lambda\det(M)$ .

Exercise.