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4.3. Properties of determinants

In this subsection, we state the basic properties of determinants that you should know. The proofs of these properties are omitted; in most cases they are proved by induction. It will be convenient to illustrate some of these properties using a $3\times 3$ matrix, in which case we will consider the arbitrary matrix:

A=\begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}\quad\hbox{with $a,b,c,d,e,f,g,h,i\in{\mathbb{R}}$.}\quad

As usual, we write $R_{i}$ and $C_{i}$ for the $i$ -th row and the $i$ -th column of a given matrix. Notice that the letter $i$ is being used in two different ways: a coefficient of the matrix $A$ , and an index for the rows and columns. The context should make clear which one we mean.

DET1.

$\det B=\det(B^{t})$ .

One can prove this by induction on the size of $B$ . A sketch of the argument is as follows. Let $B=(b_{ij})\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , and write $B^{t}=(b^{\prime}_{ij})$ for the coefficients of the transpose of $B$ . The relationship between the coefficients and the cofactors of $B$ and of $B^{t}$ can be inductively assumed to be

b_{ij}=b^{\prime}_{ji}\quad\hbox{and}\quad B_{ij}=B^{t}_{ji}\quad\hbox{for all% $1\leq i,j\leq n$.}\quad

By expanding $\det B$ about the first row and $\det B^{t}$ about the first column:

\det B=b_{11}B_{11}+\dots+b_{1n}B_{1n}=b^{\prime}_{11}B^{t}_{11}+\dots+b^{% \prime}_{n1}B^{t}_{n1}=\det B^{t}.

DET2.

Swapping two rows (or columns) multiplies the determinant by $-1$ .

DET3.

If two rows (or two columns) are identical then the determinant is zero.

DET4.

Multiplying every coefficient in a row (or column) by a scalar $\lambda\in{\mathbb{R}}$ multiplies the value of the determinant by $\lambda$ .

For example, suppose that a given number $\lambda\in{\mathbb{R}}$ multiplies $R_{2}$ in the above $3\times 3$ matrix $A$ . Then

\det\hat{A}=\begin{vmatrix}a&b&c\\ \lambda d&\lambda e&\lambda f\\ g&h&i\end{vmatrix}=-\lambda d\begin{vmatrix}b&c\\ h&i\end{vmatrix}+\lambda e\begin{vmatrix}a&c\\ g&i\end{vmatrix}-\lambda f\begin{vmatrix}a&b\\ g&h\end{vmatrix}=\lambda\begin{vmatrix}a&b&c\\ d&e&f\\ g&h&i\end{vmatrix}=\lambda\det A.

Combining the last two properties, we get the following.

DET5.

If one row (or column) is a multiple of another row (or column) the determinant has the value zero.

Assume that $C\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ has its $i$ -th row equal to $\lambda$ times its $j$ -th row, for some $i\neq j$ . That is, there is a matrix $B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , with $C$ obtained from $B$ by multiplying all the coefficients in the $i$ -th row by some $\lambda\in{\mathbb{R}}$ , and with $B$ having $R_{i}=R_{j}$ . By property DET4, we have that $\det C=\lambda\det B$ . By property DET3, we also have that $\det B=0$ , because $R_{i}=R_{j}$ . So it follows that

\det C=\lambda\det B=\lambda 0=0.

DET6.

Adding a multiple of one row (or column) to another row (or column) does not alter the value of the determinant.

For example, let $\hat{A}$ be obtained from $A$ by adding $\lambda R_{2}$ to $R_{1}$ , and expand about the first row:

	$\displaystyle\det\hat{A}$	$\displaystyle=\begin{vmatrix}a+\lambda d&b+\lambda e&c+\lambda f\\ d&e&f\\ g&h&i\end{vmatrix}$
		$\displaystyle=(a+\lambda d)\begin{vmatrix}e&f\\ h&i\end{vmatrix}-(b+\lambda e)\begin{vmatrix}d&f\\ g&i\end{vmatrix}+(c+\lambda f)\begin{vmatrix}d&e\\ g&h\end{vmatrix}$
		$\displaystyle=\left(a\begin{vmatrix}e&f\\ h&i\end{vmatrix}-b\begin{vmatrix}d&f\\ g&i\end{vmatrix}+c\begin{vmatrix}d&e\\ g&h\end{vmatrix}\right)+\lambda\left(d\begin{vmatrix}e&f\\ h&i\end{vmatrix}-e\begin{vmatrix}d&f\\ g&i\end{vmatrix}+f\begin{vmatrix}d&e\\ g&h\end{vmatrix}\right)$
		$\displaystyle=\det A+\lambda\underbrace{\begin{vmatrix}d&e&f\\ d&e&f\\ g&h&i\end{vmatrix}}_{D}=\det A$

because $D=0$ by property DET3.

For the next property, we need to explain the terminology of triangular matrices.

Definition 4.3.1.

Let $A=(a_{ij})\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ .

(i)

$A$ is upper-triangular if $a_{ij}=0$ whenever $i>j$ . That is, all the coefficients “below” the diagonal are zero.
(ii)

$A$ is lower-triangular if $a_{ij}=0$ whenever $i<j$ . That is, all the coefficients “above” the diagonal are zero.
(iii)

$A$ is diagonal if $a_{ij}=0$ whenever $i\neq j$ . That is all the coefficients outside of the diagonal are zero.

Example 4.3.2.

The following matrices are upper-triangular, but not lower-triangular:

$\begin{pmatrix}1&2&3&4\\ 0&1&2&3\\ 0&0&0&2\\ 0&0&0&8\end{pmatrix},\begin{pmatrix}1&0&5\\ 0&-1&2\\ 0&0&3\\ \end{pmatrix},\begin{pmatrix}1&-1&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}.$

The following matrices are lower-triangular, but not upper-triangular:

$\begin{pmatrix}2&0\\ 1&1\end{pmatrix},\begin{pmatrix}2&0&0\\ 1&1&0\\ 0&0&0\end{pmatrix}.$

The following matrices are both upper- and lower-triangular, and therefore diagonal:

$\begin{pmatrix}2&0\\ 0&1\end{pmatrix},\begin{pmatrix}1&0&0\\ 0&5&0\\ 0&0&-1\end{pmatrix}.$

In particular, we note that a square matrix is diagonal if and only if it is at the same time upper- and lower-triangular. For short, we call a matrix triangular if its either lower- or upper-triangular.

DET7.

For a triangular or diagonal matrix $A=(a_{ij})\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ we have $\det A=a_{11}a_{22}\cdots a_{nn}$ .

Example 4.3.3.

Now let’s use these tricks to speed up the calculation of the determinant of the matrix

$A=\begin{pmatrix}1&2&5\\ -3&1&1\\ 0&2&4\end{pmatrix}$

We could directly expand the computations to obtain the value, but there is a smarter move. Namely, as we see from the computations of determinants by expanding about a row or column of the matrix, it is convenient to pick the simplest row or column, i.e. with the more zero terms. We also can use the above rules in order to create zeros in a row or a column and then expand about that simplified row or column. That is, we use elementary operations to make all but one entry in a row or column equal to zero. The most useful property to do this is DET6. That is, we add a multiple of a row or column to another row or column, and the determinant remains the same.

With the given matrix $A$ , we do the row operation $R_{2}=r_{2}+3r_{1}$ , which does not change the value of $\det A$ . The outcome is

$\det A=\begin{vmatrix}1&2&5\\ -3&1&1\\ 0&2&4\end{vmatrix}=\begin{vmatrix}1&2&5\\ 0&7&16\\ 0&2&4\end{vmatrix}$

Now we use the cofactor expansion along the first column $C_{1}$ , and we get:

$\det A=(-1)^{1+1}(1)\begin{vmatrix}7&16\\ 2&4\end{vmatrix}+0+0=7\cdot 4-16\cdot 2=-4.$

Example 4.3.4. (Vandermonde determinant)

$\quad\hbox{We evaluate the determinant}\quad\begin{vmatrix}1&1&1\\ a&b&c\\ a^{2}&b^{2}&c^{2}\end{vmatrix},\quad\hbox{where $a,b,c\in{\mathbb{R}}$.}\quad$

Use the elementary row operations $R_{2}=r_{2}-ar_{1}$ and $R_{3}=r_{3}-a^{2}r_{1}$ , which don’t change the determinant by DET6, and therefore give

$\displaystyle\begin{vmatrix}1&1&1\\ a&b&c\\ a^{2}&b^{2}&c^{2}\end{vmatrix}$ $\displaystyle=\begin{vmatrix}1&1&1\\ 0&b-a&c-a\\ 0&b^{2}-a^{2}&c^{2}-a^{2}\end{vmatrix}=\begin{vmatrix}b-a&c-a\\ b^{2}-a^{2}&c^{2}-a^{2}\end{vmatrix}$

$\displaystyle=(b-a)(c^{2}-a^{2})-(c-a)(b^{2}-a^{2})$

$\displaystyle=(b-a)(c-a)\big{(}(c+a)-(b+a)\big{)}$

$\displaystyle=(b-a)(c-a)(c-b).$

Example 4.3.5.

In order to illustrate the efficiency of the elementary operations, let us take again the matrix $A\in\operatorname{M}_{{4}}({{\mathbb{R}}})$ of Example 4.2 and compare the difficulties of both methods:

$A=\begin{pmatrix}2&0&3&5\\ 0&4&-1&0\\ 1&0&0&1\\ 0&2&1&1\end{pmatrix}\;.$

If we do $C_{4}=c_{4}-c_{1}$ , which doesn’t change the determinant by DET6, we get

$|A|=\begin{vmatrix}2&0&3&3\\ 0&4&-1&0\\ 1&0&0&0\\ 0&2&1&1\end{vmatrix}.$

As the third row has a unique non-zero term, we perform the cofactor expansion about it and find that

$|A|=(-1)^{3+1}(1)\begin{vmatrix}0&3&3\\ 4&-1&0\\ 2&1&1\end{vmatrix}+0+0+0.$

Now, we perform $C_{2}=c_{2}-c_{3}$ with the resulting $3\times 3$ determinant, and then perform the cofactor expansion about $C_{2}$ , since it has only one non-zero coefficient:

$|A|=\begin{vmatrix}0&0&3\\ 4&-1&0\\ 2&0&1\end{vmatrix}=0+(-1)^{2+2}(-1)\begin{vmatrix}0&3\\ 2&1\end{vmatrix}+0=(-1)(-6)=6.$

So we found the same answer as in Example 4.2, but with less work. Since it is advantageous to be able to quickly evaluate determinants in exams, you would be wise to know the shortcuts!

Remark 4.3.6.

Up to now, for pedagogical reasons, we have always written the signs $(-1)^{\hbox{\tiny row+column}}$ of the cofactors in the expansion of a determinant. However, this is not necessary.

	$\displaystyle\begin{vmatrix}1&1&1\\ a&b&c\\ a^{2}&b^{2}&c^{2}\end{vmatrix}$	$\displaystyle=\begin{vmatrix}1&1&1\\ 0&b-a&c-a\\ 0&b^{2}-a^{2}&c^{2}-a^{2}\end{vmatrix}=\begin{vmatrix}b-a&c-a\\ b^{2}-a^{2}&c^{2}-a^{2}\end{vmatrix}$
		$\displaystyle=(b-a)(c^{2}-a^{2})-(c-a)(b^{2}-a^{2})$
		$\displaystyle=(b-a)(c-a)\big{(}(c+a)-(b+a)\big{)}$
		$\displaystyle=(b-a)(c-a)(c-b).$