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Determinants and Invertibility

We began Section 3.6 with the remark that a $2 \times 2$ $2 \times 2$ matrix A is invertible if and only if its determinant is nonzero: $| A | \neq 0.$ $| A | \neq 0.$ Now we want to show that this result also holds for $n \times n$ $n \times n$ matrices. This connection between determinants and invertibility is closely related to the fact that the determinant function “respects” matrix multiplication in the sense that

| A B | = | A | | B |

$| A B | = | A | | B |$ (9)

if A and B are $n \times n$ $n \times n$ matrices. Our first step is to show that Eq. (9) holds if A is an elementary matrix obtained from the $n \times n$ $n \times n$ identity matrix I by performing a single elementary row operation.

Proof:

This is really just a restatement of Theorem 1. For instance, suppose that E is obtained from I by multiplying the pth row by c, so that $| E | = c .$ $| E | = c .$ Then Theorem 5 in Section 3.5 tells us that the product EB is the result of multiplying the pth row of B by c. Therefore,

| E B | = c | B | = | E | | B |,

$| E B | = c | B | = | E | | B |,$

and so we have verified Eq. (10) for the first of the three types of elementary matrices. The verifications for the other two types are similar.

Now, let A be an $n \times n$ $n \times n$ matrix whose invertibility we want to discuss, and let R be the reduced echelon form of A. If the elementary matrices $F_{1}, F_{2}, \dots, F_{k}$ $F_{1}, F_{2}, \dots, F_{k}$ correspond to the elementary row operations that reduce A to R, then

F_{k} \dots F_{2} F_{1} A = R

$F_{k} \dots F_{2} F_{1} A = R$ (11)

by Theorem 5 in Section 3.5. Recalling that every elementary matrix is invertible (Section 3.5), we can rewrite Eq. (11) as

A = E_{1} E_{2} \dots E_{k} R,

$A = E_{1} E_{2} \dots E_{k} R,$ (12)

where each $E_{i} = {(F_{i})}^{- 1}$ $E_{i} = {(F_{i})}^{- 1}$ is an elementary matrix. It now follows, by k applications of the lemma, that

A = | E_{1} | | E_{2} | \dots | E_{k} | | R | .

$A = | E_{1} | | E_{2} | \dots | E_{k} | | R | .$ (13)

This relation is the key both to the proof of (9) and to the proof of the following theorem.

Proof:

If (as previously) R is the reduced echelon form of A, then Theorem 6 in Section 3.5 implies that

\begin{array}{l} A & is invertible & if and only if & R = I . \end{array}

$\begin{array}{l} A & is invertible & if and only if & R = I . \end{array}$ (14)

Because R is a square reduced echelon matrix, we see that either R is the identity matrix I and $| R | = 1,$ $| R | = 1,$ or R has an all-zero row and, consequently, $| R | = 0.$ $| R | = 0.$ Therefore,

\begin{array}{l} R = I & if and only if & | R | \end{array} \neq 0.

$\begin{array}{l} R = I & if and only if & | R | \end{array} \neq 0.$ (15)

Finally, because $| E | \neq 0$ $| E | \neq 0$ if E is an elementary matrix, it follows immediately from Eq. (13) that

\begin{array}{l} | R | \neq 0 & if and only if & | A | \neq 0. \end{array}

$\begin{array}{l} | R | \neq 0 & if and only if & | A | \neq 0. \end{array}$ (16)

Combining the statements in (14), (15), and (16), we see that A is invertible if and only if $| A | \neq 0.$ $| A | \neq 0.$

So now we can add the statement det $A \neq 0$ $A \neq 0$ to the list of equivalent properties of nonsingular matrices stated in Theorem 7 of Section 3.5. Indeed, some texts define the square matrix A to be nonsingular if and only if det $A \neq 0$ $A \neq 0$

Proof:

If R is the reduced echelon form of A, then we see that

A = E_{1} E_{2} \dots E_{k} R,

$A = E_{1} E_{2} \dots E_{k} R,$ (12)

where $E_{1}, E_{2}, \dots,$ $E_{1}, E_{2}, \dots,$ and $E_{k}$ $E_{k}$ are elementary matrices. Hence

A B = E_{1} E_{2} \dots E_{k} R B .

$A B = E_{1} E_{2} \dots E_{k} R B .$

We now take the determinant of both sides, using the lemma stated earlier to “split off” the elementary matrices:

\begin{array}{rcl} | A B | & = & | E_{1} | | E_{2} E_{3} \dots E_{k} R B | & (lemma once) \\ = & | E_{1} | | E_{2} | | E_{3} \dots E_{k} R B | & (lemma twice) \\ = & | E_{1} E_{2} | | E_{3} \dots E_{k} R B | . & (lemma thrice) \end{array}

$\begin{array}{rcl} | A B | & = & | E_{1} | | E_{2} E_{3} \dots E_{k} R B | & (lemma once) \\ = & | E_{1} | | E_{2} | | E_{3} \dots E_{k} R B | & (lemma twice) \\ = & | E_{1} E_{2} | | E_{3} \dots E_{k} R B | . & (lemma thrice) \end{array}$

After $2 k - 1$ $2 k - 1$ steps, we get

| A B | = | E_{1} E_{2} \dots E_{k} | | R B | .

$| A B | = | E_{1} E_{2} \dots E_{k} | | R B | .$ (17)

The remainder of the proof depends on whether or not A is invertible.

If A is invertible, then $R = I,$ $R = I,$ so Eq. (12) yields

A = E_{1} E_{2} \dots E_{k}

$A = E_{1} E_{2} \dots E_{k}$

and also that $R B = I B = B .$ $R B = I B = B .$ In this case the meaning of Eq. (17) is precisely that $| A B | = | A | | B | .$ $| A B | = | A | | B | .$

If A is not invertible, then $| A | = 0$ $| A | = 0$ by Theorem 2. Also, as we noted previously, the reduced echelon form R of A has an all-zero row in this case. Hence it follows from the definition of matrix multiplication that the product RB has an all-zero row and, therefore, that $| R B | = 0.$ $| R B | = 0.$ In this case Eq. (17) implies that $| A B | = 0.$ $| A B | = 0.$ Because both $| A | = 0$ $| A | = 0$ and $| A B | = 0,$ $| A B | = 0,$ the equation $| A B | = | A | | B |$ $| A B | = | A | | B |$ holds, and the proof is complete.

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Table of Contents for
Determinants and Invertibility