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4.4. More on elementary matrices

There are two main results of this subsection: The first is that a matrix is invertible if and only if its determinant is non-zero (Theorem 4.4.4). If you were to choose a matrix at random, its determinant would almost certainly be non-zero; therefore “most” matrices are invertible! The second result is that the determinant is multiplicative, in other words, $\det(AB)=(\det A)(\det B)$ , see Theorem 4.4.5. These will be proved by studying the determinants of elementary matrices, and building from there. We will begin by translating the above properties into statements about the elementary matrices (see Section 2 for an introduction to elementary matrices).

Theorem 4.4.1 (Elementary operations).

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , let $\lambda\in{\mathbb{R}}$ be non-zero, and let $1\leq i,j\leq n$ with $i\neq j$ . The three types of elementary row operations change the determinant as follows.

DET4

tells us that $\det(E_{i}(\lambda)\cdot A)=\lambda\det A$ ,
DET6

tells us that $\det(E_{ij}(\lambda)\cdot A)=\det A$ ,
DET2

tells us that $\det(E_{ij}\cdot A)=-\det A$ .

In view of this theorem, the “easiest” row operation to use for evaluating determinants is $E_{ij}(\lambda)$ , which adds a multiple of one row to another. This is the only row operation that doesn’t change the determinant. Let us give a few more results on elementary matrices which are immediate consequences of the above Theorem.

Lemma 4.4.2.

Let $\lambda\in{\mathbb{R}}$ be non-zero, and let $i,j\in{\mathbb{N}}$ be distinct. Then,

\det E_{i}(\lambda)=\lambda,\quad\det E_{ij}(\lambda)=1\quad\hbox{and}\quad% \det E_{ij}=-1.

Proof.

This follows from Theorem 4.4.1 since each elementary matrix is obtained by applying the corresponding elementary row operation to the identity matrix $I_{n}$ , whatever its size $n$ . ∎

Lemma 4.4.3.

Let $A,L\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ and assume that $L$ is an elementary matrix. Then,

\det(LA)=(\det L)(\det A).

Proof.

The equality follows from Theorem 4.4.1 and Lemma 4.4.2. ∎

Next, an invertibility criterion for square matrices.

Theorem 4.4.4.

Let $A$ be an $n\times n$ matrix. Then,

A

is invertible if and only if

\det A\neq 0

Proof.

Let $A_{r}$ be the reduced echelon form of $A$ , and let $L_{1},\dots,L_{k}$ be elementary matrices such that

A_{r}=L_{k}\ldots L_{1}A.

By using Lemma 4.4.3 over and over, we have that

\det A_{r}=\det L_{k}\dots\det L_{1}\det A,

with $\det L_{i}$ all non-zero. So $\det A\neq 0$ if and only if $\det A_{r}\neq 0$ . Since $A_{r}$ is in reduced echelon form (and hence upper-triangular), by DET7, $\det A_{r}\neq 0$ if and only if $A_{r}=I_{n}$ . By Lemma 3.3.3, $A_{r}=I_{n}$ if and only if $A$ is invertible. Putting these facts together, we see $\det A\neq 0$ if and only if $A$ is invertible, as required. ∎

Now we can prove that the determinant is multiplicative, as follows.

Theorem 4.4.5.

Let $A,B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ . Then,

\det(AB)=(\det A)(\det B).

Proof.

If $A$ is not invertible then $A B$ cannot be invertible, by Theorem 3.1.5. So, by Theorem 4.4.4, both sides are zero, and therefore equal. If $A$ is invertible then $A$ is a product of elementary matrices, by Theorem 3.3.4, and so the result follows from repeated applications of Lemma 4.4.3. ∎

Here are a few explicit examples which use the above results.

Example 4.4.6.

Let $x\in{\mathbb{R}}$ be a parameter, and let $A$ be the matrix

$A=\begin{pmatrix}1&2&-1\\ x&1&3\\ 3&3&x\end{pmatrix}.$

We want to find the values of $x$ for which is $A$ invertible, without using the algorithm for calculating the inverse.

By Theorem 4.4.4, we can calculate $\det A$ and find the values of $x$ for which $\det A\neq 0$ . Note that $\det A$ is a real-valued function of $x$ (in fact, it is a polynomial). Hence, we look for the values of $x$ for which the function $\det A$ is non-zero. We calculate $\det A$ by simplifying first the computations using elementary row operations which do not change the value of $\det A$ .

$\begin{vmatrix}1&2&-1\\ x&1&3\\ 3&3&x\end{vmatrix}\xrightarrow{\begin{array}[]{c}\\ R_{2}=r_{2}-xr_{1}\\ R_{3}=r_{3}-3r_{1}\end{array}}\begin{vmatrix}1&2&-1\\ 0&1-2x&3+x\\ 0&-3&x+3\end{vmatrix}.$

Expanding about the first column gives

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

$\quad\hbox{So,}\quad\det A=(1-2x)(x+3)-(-3)(x+3)=(x+3)(4-2x).$

Hence, by Theorem 4.4.4, the matrix $A$ is invertible if and only if $\det A\neq 0$ , which is the same as $x\neq-3,2$ .

Example 4.4.7.

We want to determine whether the following matrices are invertible or not, without using the algorithm for calculating the inverse.

$A=\begin{pmatrix}1&2&3&4\\ 2&3&4&1\\ 3&4&1&2\\ 4&1&2&3\end{pmatrix}\quad\hbox{and}\quad B=\begin{pmatrix}a&b&c&d\\ b&c&d&a\\ c&d&a&b\\ d&a&b&c\end{pmatrix}\;,$

where $a,b,c,d\in{\mathbb{R}}$ are subject to $a+b+c+d=0$ .

Again, using Theorem 4.4.4, this exercise consists of evaluating the determinants of $A$ and $B$ . We simplify first the computations using elementary row operations which do not change the value of the determinant. Let us start with $A$ .

$\det A=\begin{vmatrix}1&2&3&4\\ 2&3&4&1\\ 3&4&1&2\\ 4&1&2&3\end{vmatrix}\xrightarrow{\begin{array}[]{l}R_{4}=r_{4}+r_{1}\\ R_{4}=r_{4}+r_{2}\\ R_{4}=r_{4}+r_{3}\end{array}}\begin{vmatrix}1&2&3&4\\ 2&3&4&1\\ 3&4&1&2\\ 10&10&10&10\end{vmatrix}.$

Now, we do the column operations: $C_{2}=c_{2}-c_{1}~{},~{}C_{3}=c_{3}-c_{1}$ and $C_{4}=c_{4}-c_{1}$ , and expand the determinant by using the cofactors of the bottom row:

$\det A=\begin{vmatrix}1&1&2&3\\ 2&1&2&-1\\ 3&1&-2&-1\\ 10&0&0&0\end{vmatrix}=(-1)^{4+1}10\begin{vmatrix}1&2&3\\ 1&2&-1\\ 1&-2&-1\end{vmatrix}+0+0+0.$

Finally, we do $R_{1}=r_{1}-r_{2}$ and $R_{3}=r_{3}-r_{2}$ and expand the determinant using the cofactors from the first row, we get

$\det A=-10\begin{vmatrix}0&0&4\\ 1&2&-1\\ 0&-4&0\end{vmatrix}=-10\left(0+0+(-1)^{1+3}4\begin{vmatrix}1&2\\ 0&-4\end{vmatrix}\right)=-10\cdot 4(-4)=160.$

Since $\det A$ is non-zero, we conclude that $A$ is invertible, by Theorem 4.4.4.

For $B$ , since $a+b+c+d=0$ , we see that if we add all the rows, then we get a zero row:

$\det B=\begin{vmatrix}a&b&c&d\\ b&c&d&a\\ c&d&a&b\\ d&a&b&c\end{vmatrix}\xrightarrow{\begin{array}[]{l}R_{4}=r_{4}+r_{1}\\ R_{4}=r_{4}+r_{2}\\ R_{4}=r_{4}+r_{3}\end{array}}\begin{vmatrix}a&b&c&d\\ b&c&d&a\\ c&d&a&b\\ 0&0&0&0\end{vmatrix}=0.$

Thus, $B$ is not invertible by Theorem 4.4.4.