Determinants - Uniqueness and Properties

In order to show that there's only one determinant function on $M(n,R)$ , I'm going to derive another formula for the determinant. It involves permutations of the rows, so I'll give a brief introduction to permutations first.

Definition. A permutation of the set $\{1, 2, \ldots, n\}$ is an invertible function $\sigma: \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$ . The set of all permutations of $\{1, 2, \ldots, n\}$ is denoted $S_n$ and is called the symmetric group on n letters.

A transposition is a permutation in which two elements are swapped, and everything else doesn't move.

Recall that a function has an inverse if and only if it is bijective: Different inputs produce different outputs ( injective), and everything in the target set is an output of the function ( surjective). So a permutation of $\{1, 2, \ldots, n\}$ is just a way of "scrambling" the numbers from 1 to n.

Example. There are 6 permutations of $\{1, 2, 3\}$ :

$$\hbox{\epsfysize=1.75in \epsffile{det-unique1.eps}}$$

The last two in the first row and the first one in the second row are transpositions. In each case, two elements are swapped, and the other element doesn't move.

For example, the one in the bottom right corner of the diagram is the function

$$\sigma(1) = 3, \quad \sigma(2) = 1, \quad \sigma(3) = 2.$$

Now many permutations are there of $\{1, 2, \ldots, n\}$ ?

There are n places that 1 can be mapped to. Once you know where 1 goes, there are $n - 1$ places remaining where 2 can be mapped to. (1 and 2 can't go to the same place because permutations are injective.) Continuing in this way, you find that the total numbers of possibilities is

$$n(n - 1)(n - 2)\cdots 2\cdot 1 = n!.$$

Therefore, there are $n!$ permutations of $\{1, 2, \ldots, n\}$ .

Permutations are multiplied by composing them as functions, i.e. by doing one permutation followed by another. The following result says that every permutation can be built out of transpositions.

Theorem. (a) Every permutation can written as a product of transpositions.

(b) No permutation can be written as a product of both an even and an odd number of transpositions.

You'll probably see these results proved in a course in abstract algebra. You can see that the first part reasonable: It says that any way of "scrambling" the elements in $\{1, 2, \ldots, n\}$ could be accomplished by swapping two elements at a time, one swap after another.

The second part of the theorem is important, because it justifies the following definition.

Definition. A permutation is even if it can be written as a product of an even number of transpositions. A permutation is odd if it can be written as a product of an odd number of transpositions.

If $\sigma$ is a permutation, the sign of $\sigma$ is

$$\sgn(\sigma) = \cases{+1 & if $\sigma$ is even \cr -1 & if $\sigma$ is odd \cr}.$$

Theorem. If $D: M(n,R) \rightarrow R$ is linear in each row and is 0 on matrices with equal rows, then

$$D(A) = \sum_{\sigma \in S_n} \sgn(\sigma) A_{1\sigma(1)} \cdots A_{n\sigma(n)} D(I).$$

Proof. Write

$$A = \left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

Observe that

$$r_i = \sum_j A_{ij} e_j,$$

where $e_j$ is the j-th standard basis vector (equivalently, the j-th row of the identity matrix).

By linearity,

$$D(A) = D \left[\matrix{\leftarrow & \sum_j A_{1j} e_j & \rightarrow \cr \leftarrow & r_2 & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = \sum_j A_{1j} D\left[\matrix{\leftarrow & e_j & \rightarrow \cr \leftarrow & r_2 & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

Now do the same with row 2:

$$D(A) = \sum_j \sum_k A_{1j} A_{2k} D\left[\matrix{\leftarrow & e_j & \rightarrow \cr \leftarrow & e_k & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

Do this for all n rows, using $j_1, \ldots, j_n$ as the summation indices:

$$D(A) = \sum_{j_1, \ldots, j_n} A_{1j_1} A_{2j_2} \cdots A_{nj_n} D\left[\matrix{ \leftarrow & e_{j_1} & \rightarrow \cr \leftarrow & e_{j_2} & \rightarrow \cr & \vdots & \cr \leftarrow & e_{j_n} & \rightarrow \cr}\right].$$

Since D is 0 on matrices with equal rows, I only need to consider terms where all the rows are different --- i.e. terms where $j_1, \ldots, j_n$ are distinct. This means that $\{ j_1, \ldots, j_n\}$ is a permutation of $\{ 1, \ldots, n\}$ , and I'm summing over $\sigma \in S_n$ :

$$D(A) = \sum_{\sigma \in S_n} A_{1\sigma(1)} A_{2\sigma(2)} \cdots A_{n\sigma(n)} D\left[\matrix{\leftarrow & e_{\sigma(1)} & \rightarrow \cr \leftarrow & e_{\sigma(2)} & \rightarrow \cr & \vdots & \cr \leftarrow & e_{\sigma(n)} & \rightarrow \cr}\right].$$

Now

$$\left[\matrix{ \leftarrow & e_{\sigma(1)} & \rightarrow \cr \leftarrow & e_{\sigma(2)} & \rightarrow \cr & \vdots & \cr \leftarrow & e_{\sigma(n)} & \rightarrow \cr}\right]$$

is just the identity matrix with its rows permuted by $\sigma$ . Every permutation is a product of transpositions. By an earlier lemma, since D is linear on the rows and is 0 for matrices with equal rows, a transposition (i.e. a row swap) multiplies the value of D by -1. Hence,

$$D\left[\matrix{ \leftarrow & e_{\sigma(1)} & \rightarrow \cr \leftarrow & e_{\sigma(2)} & \rightarrow \cr & \vdots & \cr \leftarrow & e_{\sigma(n)} & \rightarrow \cr}\right] = \sgn(\sigma) D(I).$$

Therefore,

$$D(A) = \sum_{\sigma \in S_n} A_{1\sigma(1)} A_{2\sigma(2)} \cdots A_{n\sigma(n)}\sgn(\sigma) D(I).\quad\halmos$$

Since the determinant function defined by expansion by cofactors is alternating and linear in each row, it satisfies the conditions of the theorem. Since $\det I = 1$ , the formula in the theorem becomes

$$\det A = \sum_{\sigma \in S_n} \sgn(\sigma) A_{1\sigma(1)} \cdots A_{n\sigma(n)}.$$

I'll call this the permutation formula for the determinant.

Example. Compute the following real determinant

$$\det \left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6 \cr 7 & 8 & 9 \cr}\right]$$

using the permutation formula.

I have to choose 3 entries from the matrix at a time, in such a way that each choice involves exactly one element from each row and each column. For each such choice, I take the product of the three elements and multiply by the sign of the permutation of the elements, which I'll describe below. Finally, I add up the results.

In order to do this systematically, focus on the first column. I can choose 1, 4, or 7 from column 1.

If I choose 1 from column 1, I can either choose 5 from column 2 and 9 from column 3, or 8 from column 2 and 6 from column 3. (Remember that I can't have two elements from the same row or column.)

$$\left[\matrix{1 & \ast & \ast \cr \ast & 5 & \ast \cr \ast & \ast & 9 \cr}\right]\quad \left[\matrix{1 & \ast & \ast \cr \ast & \ast & 6 \cr \ast & 8 & \ast \cr}\right]$$

If I choose 4 from column 1, I can either choose 2 from column 2 and 9 from column 3, or 8 from column 2 and 3 from column 3.

$$\left[\matrix{\ast & 2 & \ast \cr 4 & \ast & \ast \cr \ast & \ast & 9 \cr}\right]\quad \left[\matrix{\ast & \ast & 3 \cr 4 & \ast & \ast \cr \ast & 8 & \ast \cr}\right]$$

Finally, if I choose 7 from column 1, I can either choose 2 from column 2 and 6 from column 3, or 5 from column 2 and 3 from column 3.

$$\left[\matrix{\ast & 2 & \ast \cr \ast & \ast & 6 \cr 7 & \ast & \ast \cr}\right]\quad \left[\matrix{\ast & \ast & 3 \cr \ast & 5 & \ast \cr 7 & \ast & \ast \cr}\right]$$

This gives me 6 products:

$$(1)(5)(9) \quad (1)(6)(8) \quad (4)(2)(9) \quad (4)(8)(3) \quad (7)(2)(6) \quad (7)(5)(3)$$

Next, I have to attach a sign to each product. To do this, I count the number of row swaps I need to move the 1's in the identity matrix into the same positions as the numbers in the product. I'll illustrate with two examples.

$$\left[\matrix{\ast & \ast & 3 \cr 4 & \ast & \ast \cr \ast & 8 & \ast \cr}\right]: \qquad \left[\matrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr}\right] \matrix{\to \cr r_1 \leftrightarrow r_2 \cr} \left[\matrix{0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr}\right] \matrix{\to \cr r_1 \leftrightarrow r_3 \cr} \left[\matrix{0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr}\right]$$

It took 2 row swaps to move the 1's into the same positions as 4, 8, and 3. Since 2 is even, the sign of $(4)(8)(3)$ is $+1$ .

$$\left[\matrix{\ast & \ast & 3 \cr \ast & 5 & \ast \cr 7 & \ast & \ast \cr}\right]: \qquad \left[\matrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr}\right] \matrix{\to \cr r_1 \leftrightarrow r_3 \cr} \left[\matrix{0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr}\right]$$

It took 1 row swap to move the 1's into the same positions as 7, 5, and 3. Since 1 is odd, the sign of $(7)(5)(3)$ is -1.

Continuing in this fashion, I get

$$\det \left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6 \cr 7 & 8 & 9 \cr}\right] = (1)(5)(9) - (1)(6)(8) - (4)(2)(9) + (4)(8)(3) + (7)(2)(6) - (7)(5)(3) = 0.\quad\halmos$$

Notice how ugly the computation was! While the permutation formula can be used for computations, it's easier to use row or column operations or expansion by cofactors.

Corollary. There is a unique determinant function $|\cdot|: M(n,R) \rightarrow R$ .

Proof. A determinant function D is linear in each row, is 0 on matrices with equal rows, and satisfies $D(I) = 1$ . The theorem then shows that $D(A)$ is completely determined by A.

In other words, the determinant function I defined by cofactor expansion is the only determinant function on $n \times n$ matrices.

Remark. Another way to express the proof of the Corollary is: If $D: M(n,R) \rightarrow R$ is alternating and linear in each row, then

$$D(A) = |A| D(I),$$

where $|\cdot|$ denotes the determinant function on $M(n,R)$ .

Corollary. If $A \in M(n,R)$ , then $|A| = |A^T|$ .

Proof.

$$|A^T| = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^n A_{i\sigma(i)}^T = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^n A_{\sigma(i)i} =$$

$$\sum_{\sigma \in S_n} \sgn(\sigma) \prod_{j=1}^n A_{j\sigma^{-1}(j)} = \sum_{\sigma \in S_n} \sgn\left(\sigma^{-1}\right) \prod_{j=1}^n A_{j\sigma^{-1}(j)} =$$

$$\sum_{\sigma^{-1} \in S_n} \sgn\left(\sigma^{-1}\right) \prod_{j=1}^n A_{j\sigma^{-1}(j)} = \sum_{\tau \in S_n} \sgn\left(\tau\right) \prod_{j=1}^n A_{j\tau(j)} = |A|.$$

I got the next-to-the-last equality by letting $\tau = \sigma^{-1}$ .

Remark. Since the rows of A are the columns of $A^T$ and vice versa, the Corollary implies that you can also use column operations to compute determinants. The allowable operations are swapping two columns, multiplying a column by a number, and adding a multiple of a column to another column. They have the same effects on the determinant as the corresponding row operations.

As I noted earlier, this result also means that you can compute determinants using cofactors of rows as well as columns.

The uniqueness of determinant functions is useful in deriving properties of determinants.

Theorem. Let $A, B \in M(n,R)$ . Then $|AB| = |A||B|$ .

Proof. Fix B, and define

$$D(A) = |AB|.$$

Let $r_i$ denote the i-th row of A. Then

$$D(A) = \left|\matrix{\leftarrow & r_1B & \rightarrow \cr \leftarrow & r_2B & \rightarrow \cr & \vdots & \cr \leftarrow & r_nB & \rightarrow \cr}\right|.$$

Now $|\cdot|$ is alternating, so interchanging two rows in the determinant above multiplies $D(A)$ by -1. Hence, D is alternating.

Next, I'll show that D is linear:

$$D\left[\matrix{& \vdots & \cr \leftarrow & kx + y & \rightarrow \cr & \vdots & \cr}\right] = \left|\matrix{& \vdots & \cr \leftarrow & (kx + y)B & \rightarrow \cr & \vdots & \cr}\right| =$$

$$k\cdot \left|\matrix{& \vdots & \cr \leftarrow & xB & \rightarrow \cr & \vdots & \cr}\right| + \left|\matrix{& \vdots & \cr \leftarrow & yB & \rightarrow \cr & \vdots & \cr}\right| = k\cdot D\left[\matrix{& \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr}\right] + D\left[\matrix{& \vdots & \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right].$$

This proves that D is linear in each row.

By the remark following the uniqueness theorem,

$$|AB| = D(A) = |A| D(I) = |A||B|.\quad\halmos$$

In words, this important result says that the determinant of a product is the product of the determinants. Note that the same thing does not work for sums.

Example. It's not true that $\det (A + B) = \det A + \det B$ . For example, in $M(2,\real)$ ,

$$\det \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] = 1 \quad\hbox{and}\quad \det \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right] = 1.$$

But

$$\det\left(\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] + \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right]\right) = \det \left[\matrix{0 & 0 \cr 0 & 0 \cr}\right] = 0.\quad\halmos$$

Theorem. Let F be a field, and let $A \in M(n, F)$ . A is invertible if and only if $|A| \ne 0$ .

Proof. If A is invertible, then

$$|A| |A^{-1}| = |AA^{-1}| = |I| = 1.$$

This equation implies that $|A| \ne 0$ (since $|A| = 0$ would yield "$0 = 1$ ").

Conversely, suppose that $|A| \ne 0$ . Suppose that A row reduces to the row reduced echelon matrix R, and consider the effect of elementary row operations on $|A|$ . Swapping two rows multiplies the determinant by -1. Adding a multiple of a row to another row leaves the determinant unchanged. And multiplying a row by a nonzero number multiplies the determinant by that nonzero number. Clearly, no row operation will make the determinant 0 if it was nonzero to begin with. Since $|A| \ne 0$ , it follows that $|R| \ne 0$ .

Since R is a row reduced echelon matrix with nonzero determinant, it can't have any all-zero rows. An $n \times n$ row reduced echelon matrix with no all-zero rows must be the identity, so $R = I$ . Since A row reduces to the identity, A is invertible.

Corollary. Let F be a field, and let $A \in M(n, F)$ . If A is invertible, then

$$|A^{-1}| = \dfrac{1}{|A|}.$$

Proof. I showed in proving the theorem that $|A||A^{-1}| = 1$ , so $|A^{-1}| = \dfrac{1}{|A|}$ .

Here's an easy application of this result.

Corollary. Let F be a field, and let $A, P \in M(n, F)$ . If P is invertible, then $|PAP^{-1}| = |A|$ .

Proof.

$$|PAP^{-1}| = |P||A||P^{-1}| = |P||P^{-1}||A| = |A|.\quad\halmos$$

(The matrices $PAP^{-1}$ and A are said to be similar. This will come up later when I talk about changing coordinates in vector spaces.)

Definition. Let R be a commutative ring with identity, and let $A \in M(n, R)$ . The classical adjoint $\adj A$ is the matrix whose i-j-th entry is

$$(\adj A)_{ij} = (-1)^{i+j} |A(j\mid i)|.$$

In other words, $\adj A$ is the transpose of the matrix of cofactors.

Example. Compute the adjoint of

$$A = \left[\matrix{1 & 0 & 3 \cr 0 & 1 & 1 \cr 1 & -1 & 2 \cr}\right].$$

First, I'll compute the cofactors. The first line shows the cofactors of the first row, the second line the cofactors of the second row, and the third line the cofactors of the third row.

$$+\left|\matrix{1 & 1 \cr -1 & 2 \cr}\right| = 3, \quad -\left|\matrix{0 & 1 \cr 1 & 2 \cr}\right| = 1, \quad +\left|\matrix{0 & 1 \cr 1 & -1 \cr}\right| = -1,$$

$$-\left|\matrix{0 & 3 \cr -1 & 2 \cr}\right| = -3, \quad +\left|\matrix{1 & 3 \cr 1 & 2 \cr}\right| = -1, \quad -\left|\matrix{1 & 0 \cr 1 & -1 \cr}\right| = 1,$$

$$+\left|\matrix{0 & 3 \cr 1 & 1 \cr}\right| = -3, \quad -\left|\matrix{1 & 3 \cr 0 & 1 \cr}\right| = -1, \quad +\left|\matrix{1 & 0 \cr 0 & 1 \cr}\right| = 1.$$

The adjoint is the transpose of the matrix of cofactors:

$$\adj A = \left[\matrix{3 & -3 & -3 \cr 1 & -1 & -1 \cr -1 & 1 & 1 \cr}\right].\quad\halmos$$

I'll need the following little lemma later on.

Lemma. Let R be a commutative ring with identity, and let $A \in M(n, R)$ . Then

$$(\adj A)^T = \adj A^T.$$

Proof. Consider the $(i,j)^{\rm th}$ elements of the matrices on the two sides of the equation.

$$[(\adj A)^T]_{ij} = (\adj A)_{ji} = (-1)^{j+i} |A(i \mid j)|,$$

$$[\adj A^T]_{ij} = (-1)^{i+j} |A^T(j \mid i)|.$$

The signs $(-1)^{j+i}$ and $(-1)^{i+j}$ are the same; what about the other terms? $|A(i \mid j)|$ is the determinant of the matrix formed by deleting the $i^{\rm th}$ row and the $j^{\rm th}$ column from A. $|A^T(j \mid i)|$ is the determinant of the matrix formed by deleting the $j^{\rm th}$ row and $i^{\rm th}$ column from $A^T$ . But the $i^{\rm th}$ row of A is the $i^{\rm th}$ column of $A^T$ , and the $j^{\rm th}$ column of A is the $j^{\rm th}$ row of $A^T$ . So the two matrices that remain after these deletions are transposes of one another, and hence they have the same determinant. Thus, $|A(i \mid j)| = |A^T(j \mid i)|$ . Hence, $[(\adj A)^T]_{ij} = [\adj A^T]_{ij}$ .

Theorem. Let R be a commutative ring with identity, and let $A \in M(n, R)$ . Then

$$|A|\cdot I = A\cdot \adj A.$$

Proof. Expand $|A|$ by cofactors of row i:

$$|A| = \sum_j (-1)^{i+j} A_{ij} |A(i\mid j)|.$$

Take $k \ne i$ , and construct a new matrix by replacing row k of A with row i. Since the new matrix has two equal rows, its determinant is 0. On the other hand, expanding by cofactors of row k,

$$0 = \sum_j (-1)^{k+j} A_{kj} |A(k\mid j)| = \sum_j (-1)^{k+j} A_{ij} |A(k\mid j)|.$$

In words, this means that if you expand by cofactors of a row using the cofactors from another row, you get 0. You only get $|A|$ if you use the "right" cofactors --- i.e. the cofactors for the given row. In terms of the Kronecker delta function,

$$\sum_j (-1)^{k+j} A_{ij} |A(k\mid j)| = \delta_{ik} |A| \quad \hbox{for all} \quad i, k.$$

Interpret this equation as a matrix equation, where the two sides represent the i-k-th entries of their respective matrices. What {\it are} the respective matrices? The right side is obviously the i-k-th entry of $|A|\cdot I$ . For the left side,

$$(A\cdot \adj A)_{ik} = \sum_j A_{ij} (\adj A)_{jk} = \sum_j A_{ij} (-1)^{j+k} |A(k\mid j)|.$$

Therefore,

$$|A|\cdot I = A\cdot \adj A.\quad\halmos.$$

I can use the theorem to obtain an important corollary. I already know that a matrix over a field is invertible if and only if its determinant is nonzero. The next result explains what happens over a commutative ring with identity, and also provides a formula for the inverse of a matrix.

Corollary. Let R be a commutative ring with identity, and let $A \in M(n, R)$ . A is invertible if and only if $|A|$ is invertible in R, in which case

$$A^{-1} =\dfrac{1}{|A|} \adj A.$$

Proof. First, suppose A is invertible. Then $AA^{-1} = I$ , so

$$|A||A^{-1}| = |AA^{-1}| = |I| = 1.$$

Therefore, $|A|$ is invertible in R.

Conversely, suppose $|A|$ is invertible in R. I have

$$|A|\cdot I = A\cdot \adj A, \quad\hbox{so}\quad I = A\cdot |A|^{-1} \adj A. \eqno{(1)}$$

Take the last equation and replace A with $A^T$ :

$$I = A^T\cdot |A^T|^{-1} \adj A^T.$$

Use the facts that $|A^T| = |A|$ and (from the earlier lemma) that $(\adj A)^T = \adj A^T$ :

$$I = A^T\cdot |A|^{-1} (\adj A)^T.$$

Use the transpose rule $(MN)^T = N^TM^T$ :

$$I = |A|^{-1} \left(\adj A\cdot A\right)^T.$$

Finally, transpose both sides, using the transpose rules $I^T = I$ and $(M^T)^T = M$ :

$$I = |A|^{-1} \adj A\cdot A. \eqno{(2)}$$

Equations (1) and (2) show that $|A|^{-1} \adj A$ multiplies A to I on both the left and right. Therefore, A is invertible, and

$$A^{-1} = |A|^{-1} \adj A.\quad\halmos$$

The last result can be used to find the inverse of a matrix. It's not very good for big matrices from a computational point of view: The usual row reduction algorithm uses fewer steps. However, it's not too bad for small matrices --- say $3 \times 3$ or smaller.

Example. Compute the inverse of the following real matrix using cofactors:

$$A = \left[\matrix{1 & -2 & -2 \cr 3 & -2 & 0 \cr 1 & 1 & 1 \cr}\right].$$

First, I'll compute the cofactors. The first line shows the cofactors of the first row, the second line the cofactors of the second row, and the third line the cofactors of the third row.

$$+\left|\matrix{-2 & 0 \cr 1 & 1 \cr}\right| = -2, \quad -\left|\matrix{3 & 0 \cr 1 & 1 \cr}\right| = -3, \quad +\left|\matrix{3 & -2 \cr 1 & 1 \cr}\right| = 5,$$

$$-\left|\matrix{-2 & -2 \cr 1 & 1 \cr}\right| = 0, \quad +\left|\matrix{1 & -2 \cr 1 & 1 \cr}\right| = 3, \quad -\left|\matrix{3 & -2 \cr 1 & 1 \cr}\right| = -3,$$

$$+\left|\matrix{-2 & -2 \cr 2 & 0 \cr}\right| = -4, \quad -\left|\matrix{1 & -2 \cr 3 & 0 \cr}\right| = -6, \quad +\left|\matrix{1 & -2 \cr 3 & -2 \cr}\right| = 4.$$

The adjoint is the transpose of the matrix of cofactors:

$$\adj A = \left[\matrix{-2 & 0 & -4 \cr -3 & 3 & -6 \cr 5 & -3 & 4 \cr}\right].$$

Since $\det A = -6$ , I have

$$A^{-1} = -\dfrac{1}{6} \left[\matrix{-2 & 0 & -4 \cr -3 & 3 & -6 \cr 5 & -3 & 4 \cr}\right].\quad\halmos$$

Another consequence of the formula $|A|\cdot I = A\cdot \adj A$ is Cramer's rule, which gives a formula for the solution of a system of linear equations. As with the adjoint formula for the inverse of a matrix, Cramer's rule is not computationally efficient: It's better to use row reduction to solve systems.

Corollary. ( Cramer's rule) If A is an invertible $n \times n$ matrix, the unique solution to $Ax = y$ is given by

$$x_i = \dfrac{|B_i|}{|A|},$$

where $B_i$ is the matrix obtained from A by replacing its i-th column by y.

Proof.

$$Ax = y,$$

$$(\adj A)Ax = (\adj A)y,$$

$$|A|x_i = \sum_j (\adj A)_{ij}y_j = \sum_j (-1)^{i+j} y_j |A(j\mid i)|.$$

What is the last sum? It is a cofactor expansion of A along column i, where instead of the elements of A's column i I'm using the components of y. This is exactly $|B_i|$ .

Example. Use Cramer's Rule to solve the following system over $\real$ :

$$\matrix{2x & + & y & + & z & = & 1 \cr x & + & y & - & z & = & 5 \cr 3x & - & y & + & 2z & = & -2 \cr}$$

In matrix form, this is

$$\left[\matrix{2 & 1 & 1 \cr 1 & 1 & -1 \cr 3 & -1 & 2 \cr}\right] \left[\matrix{x \cr y \cr z \cr}\right] = \left[\matrix{1 \cr 5 \cr -2 \cr}\right].$$

I replace the successive columns of the coefficient matrix with $\langle 1, 5, -2\rangle$ , in each case computing the determinant of the resulting matrix and dividing by the determinant of the coefficient matrix:

$$x = \dfrac{\left|\matrix{{\bf 1} & 1 & 1 \cr {\bf 5} & 1 & -1 \cr {\bf -2} & -1 & 2 \cr}\right|} {\left|\matrix{2 & 1 & 1 \cr 1 & 1 & -1 \cr 3 & -1 & 2 \cr}\right|} = \dfrac{-10}{-7} = \dfrac{10}{7}, \quad y = \dfrac{\left|\matrix{2 & {\bf 1} & 1 \cr 1 & {\bf 5} & -1 \cr 3 & {\bf -2} & 2 \cr}\right|} {\left|\matrix{2 & 1 & 1 \cr 1 & 1 & -1 \cr 3 & -1 & 2 \cr}\right|} = \dfrac{-6}{-7} = \dfrac{6}{7}, \quad z = \dfrac{\left|\matrix{2 & 1 & {\bf 1} \cr 1 & 1 & {\bf 5} \cr 3 & -1 & {\bf -2} \cr}\right|} {\left|\matrix{2 & 1 & 1 \cr 1 & 1 & -1 \cr 3 & -1 & 2 \cr}\right|} = \dfrac{19}{-7} = -\dfrac{19}{7}.$$

This looks pretty simple, doesn't it? But notice that you need to compute four $3 \times 3$ determinants to do this. It becomes more expensive to solve systems this way as the matrices get larger.

Send comments about this page to: Bruce.Ikenaga@millersville.edu.

Bruce Ikenaga's Home Page

Copyright 2008 by Bruce Ikenaga