Row reduce: Since we only did type (3) elementary row operations, det(a) = 3. Following a different path to REF, let D = det(a).

Math 20580

March 22, 2021

1. Elementary row operations

Here is what each elementary row operation does to the

determinant.

(1) Permute the rows by τ : multiply the determinant by

(−1)

τ

.

(2) Multiply a row by c: multiply the determinant by c.

(3) Add a multiple of row j to row i: no change in the

determinant.

We will verify (3) later but first, find the determinant of

Row reduce:

           1 2 −3 2 2 5 −1 7 5 12 −4 17 −1 −3 3 3            ∼            1 2 −3 2 0 1 5 3 0 2 11 7 0 −1 0 5            ∼            1 2 −3 2 0 1 5 3 0 0 1 1 0 0 5 8            ∼            1 2 −3 2 0 1 5 3 0 0 1 1 0 0 0 3           

Since we only did type (3) elementary row operations, det(A) = 3

Following a different path to REF, let D = det(A).

2. Laplace expansion

The Laplace expansion of a determinant gives another way to calculate

which has theoretical consequences and can be a fast way to compute

if your matrix has a lot of 0’s. It comes about by asking for the matrix

of the linear transformation d

j

: R

n

→ R. From general theory there

exists a row vector with entries [C

1 j

· · · C

n j

] such that

det



Ah ~

Ai

_j



= [C

_{1 j}

· · · C

_{n j}

] ~

A

As usual C

_{i j}

= d

_j

(~

e

_i

) = det (Ah~

e

_i

i

_j

).

Appendix B checks that Ci j = (−1)i+jAi j so in general (ce) det(A) = [(−1)j+1|A_{1 j}| · · · (−1)j+n|A_{n j}|A~j = n X k=1 Ck j xk j

If you work out the matrix for the row i linear transformation you will get (re) det(A) = R_i     (−1)i+1|A_{i 1}| ... (−1)i+n|A_{i n}|     = n X k=1 x_{i k} C_{i k} The book shows you the checkerboard pattern of signs:

3. Determinants multiply

det(BA) = det(B) det(A) See Appendix C for a proof.

4. The adjugate of a matrix

The adjugate of a square matrix A is the matrix adj(A) whose (i j)th entry is C_{j i}, backwards to what you might expect. In other words adj(A) = [C_{i j}]T.

A adj(A) = det(A)1_n×n adj(A)A = det(A)1_n×n

See Appendix D for a proof. Notice that the adjugate is a pretty good substitute for the inverse because the adjugate always exists and is a ma-trix of polynomials in the xi j. If A is a matrix with det(A) 6= 0 then

A−1 = 1

det(A)adj(A).

5. Properties of the adjugate

It is easy to compute directly that adj(1n×n) = 1n×n and adj(0) = 0.

adj(cA) = cn−1adj(A) det adj(A)
= det(A)
n−1

adj(AB) = adj(B)adj(A) adj adj(A)
= det(A)
n−2A

These formulas can be guessed by pretending that adj(A) = A−1 and then adjusting by some determinants. Details may be found in Appendix E.

6. Cramer’s rule

Cramer’s rule gives a solution to the equation A~x = ~b if A is an invertible square matrix. It is very inefficient except for small n.

The formula comes from adj(A)A~x = adj(A)~b =⇒ det(A)~x = adj(A)~b. Hence det(A)x_i is the ith row of adj(A)~b.

Appendix A. Type (3) elementary row operations So why does the formula for type (3) elementary row operations work?

Suppose we have A and we wish to add c times row j to row i. That is, if the rows of A are Rk, the new matrix C has rows Rk, k 6= i and row i is Ri+ cRj.

Look at the matrix B which has rows R_k, k 6= i and row i is cR_j. Then one row of B is a multiple of another so det(B) = 0. By the row version of (P1) det(C) = det(A) + det(B).

Appendix B. The Laplace expansion

The point of Laplace expansion is that the C_{i j} can be worked out in terms of (n − 1) × (n − 1) determinants instead of the n × n ones which follow from the definition of the matrix associated to dj.

Here is one way to proceed. Let A_{i j} be the (i, j)th minor of A. That is the (n − 1) × (n − 1) matrix you get by deleting the ith row and jth column from A. Let Y be the matrix you get by replacing the jth column and the ith row by 0’s except for yi j = 1.

Looking at the matrix Ah~eiij we see that all terms in the Leibniz expansion

are 0 except for ones contain x_{i j}. Look at x_{i k}. Since x_{i j} is present in any non-zero term, x_{i k} is not present in any non-zero term unless k = j. Hence |Y| = |Ah~e_ii_j| and Y_{i j} = A_{i j}

To compute |Y_{i j}| move the jth _{column to the far left. This introduces a}

The new matrix looks like   Yi j ~0 0 · · · 0 1   Hence Ci j = (−1)i+j       Ai j ~0 0 · · · 0 1       = (−1)i+jAi j  .

Then the Laplace expansion along the jth column is (ce) det(A) = [(−1)j+1|A_{1 j}| · · · (−1)j+n|A_{n j}|A~_j =

n

X

k=1

C_{k j} x_{k j} where ~A_j is the jth column of A.

Either repeating the above argument using rows or playing with transposes yields the Laplace expansion along the ith row :

(re) det(A) = Ri     (−1)i+1|A_{i 1}| ... (−1)i+n|A_{i n}|     = n X k=1 xi k Ci k

Appendix C. Multiplicative properties of the determinant We discussed elementary matrices earlier in the course but here is a quick review.

Given any elementary row operation, one can apply it to the m × m identity matrix to get a matrix E which has the property that EA does the same elementary row operation on any m × n matrix A.

(1) If τ permutes the rows of 1 then multiplication by E(τ ) permutes the rows of A by τ : det E(τ )
= (−1)τ_.

(2) If c is scalar and i is a row index, then multiplication by E(c, i) multiplies the ith row of A by c: det E(c, i)
= c.

(3) If c is scalar and i and j are row indices, then multiplication by E(c, i, j) adds c times the jth row to the ith of A: det E(c, i, j)
= 1.

Let E be any of the three kinds of m × m elementary matrices above. Then det(EA) = det(E) det(A)

Given any square matrix A there exists a sequence of elementary row oper-ations which reduce it to REF, R. Hence there is an E with det(EA) = det(E) det(A) = det(R). Now det(E) is not zero and det(R) is the product of the diagonal entries. All the diagonal entries of R are non-zero if and only if det(R) 6= 0 if and only if R is invertible if and only if A is invertible.

Hence the determinant determines if a square matrix is invertible. Moreover

det(BA) = det(B) det(A)

To see this observe that if det(B) 6= 0 B is a product of elementary matrices since there is such a product which reduces B to RREF which must be the identity matrix. Hence if det(B) 6= 0 we already known the equation holds. If det(B) = 0 then B is not invertible so there is a non-zero row vector R such that RB is the zero row vector. But then RBA is also the zero row vector and BA is not invertible so det(BA) = 0 and the equation holds in this case as well.

Appendix D. Basic property of the adjugate of a matrix To see the first formula, let A adj(A) = [s_{i j}] Then

si j = n

X

k=1

xi kCj k

Laplace expansion along the ith row is given by (re): det(A) =

n

X

k=1

xi k Ci k,

If i 6= j let Y have kth row the same as the kth row of A except that the ith row of Y is the jth row of A. Hence the entries of Y are y_{r s} = x_{r s} except y_{i k} = x_{j k}.

Let Cr s(A) be the indicated cofactor of A with Cr s(Y ) be the indicated

cofactor of Y. Since the rows of A and Y are the same except for row i, Ci k(A) = Ci k(Y ). det(Y) = n X k=1 y_{i k} C(Y )_{i k} = n X k=1

x_{j k} C(A)_{i k} = s_{j i}. But Y has two equal rows so det(Y) = 0 and hence sj i = 0 and

A adj(A) = det(A)1_n×n A similar proof using columns shows

adj(A) A = det(A)1_n×n (1) If rank(A) = n, rank adj(A)
= n.

(2) If rank(A) = n − 1 then rank adj(A)
= 1.

(3) If rank(A) < n−1 then rank adj(A)
= 0 which means adj(A) = 0_n×n. (1): If rank(A) = n, A is invertible, so det(A) 6= 0 so det adj(A)
6= 0 and hence rank adj(A)
= n.

(3): If rank(A) < n − 1, there are (n − 2) columns of A, k and j such that k and j are linear combinations of the remaining vectors. Hence if you remove one column, the remaining columns are still linearly dependent. Hence the columns of each Ai j are dependent and therefore det(Ai j) = 0; Cj = 0 and

(2): If rank(A) = n − 1, (n − 1) columns are linearly independent and one column, say j, is a unique linear combination of the others. Let Y be the n × (n − 1) matrix which is A with the jth column deleted. The columns of Y are linearly independent or, equivalently, rank(Y) = n − 1. This means that (n − 1) rows of Y are independent. But there are n rows so row must be a linear combination of the independent rows. Suppose row i is such a row. Then Ai j has linearly independent rows and hence det(Ai j) 6= 0 as is Ci j

and so rank adj(A)
> 1.

Then adj(A)A = det(A)1_n×n = 0_n×n. There are (n − 1) linearly independent columns of A and each of them is a null vector for adj(A). By the Rank theorem, rank adj(A)
+ dim null space(adj(A)
= n so rank adj(A)
6 1 and therefore rank adj(A)
= 1

Appendix E. Properties of the adjugate It is easy to compute directly that adj(cA) = cn−1adj(A).

det adj(A)
= det(A)
n−1

It follows from adj(A) A = det(A)1_n×n that det(A) det adj(A)
= det(A)
n. It is a theorem that cancellation holds with multi-nomials so our formula holds. Our formula implies A is invertible if and only if adj(A) is invertible. Moreover adj(A) = det(A)A−1. Hence the adjugate behaves a lot like the inverse except the adjugate always exists.

Start with AYadj(AY) = det(AY)1n×n =⇒ adj(A)AYadj(AY) =

det(AY)adj(A) =⇒ det(A)1n×nYadj(AY) = det(AY)adj(A) =⇒

det(A)adj(Y)Yadj(AY) = det(AY)adj(Y)adj(A) =⇒

det(A) det(Y)adj(AY) = det(AY)adj(Y)adj(A). Cancel det(AY) to get the formula.

adj adj(A)
= det(A)
n−2A

A adj(A) = det(A)1_n×n =⇒ adj A adj(A)
= det(A)
n−11_n×n =⇒ adj adj(A)
adj(A) = det(A)
n−11_n×n =⇒

adj adj(A)
adj(A)A = det(A)
n−11_n×nA =⇒ adj adj(A)
det(A) = det(A)
n−1A. Cancel a copy of det(A) to get our formula.

adj(A)
T = adj(AT)

A adj(A) = det(A)1_n×n =⇒ A adj(A)
T = det(A)1_n×n =⇒ adj(A)
TAT = det(A)1_n×n =⇒ adj(A)
TATadj(AT) =

det(A)1n×nadj(AT) =⇒ adj(A)

T

det(AT)1n×n = det(A)adj(AT) =⇒

det(A) adj(A)
T = det(A)adj(AT). Cancel a copy of det(A) to get our formula.

Appendix F. Cramer’s Rule

The formula comes from adj(A)A~x = adj(A)~b =⇒ det(A)~x = adj(A)~b. Hence det(A)x_i is the ith row of adj(A)~b.

det(A)x_i =

n

X

k=1

det A_i(~b)
=

n

X

k=1

C_{k i} A_i(~b)
b_k

but because we have removed the ith column in forming Ck i Ai(~b)
,

Ck i Ai(~b)
= Ck i A)
so det(A)x_i = n X k=1 C_{k i}b_k = n X k=1 C_{k i} A_i(~b))
b_k = det A_i(~b)

Cramer’s Rule: If A~x = ~b has a solution, then det(A)x_i = det A_i(~b)

If det(A) 6= 0 Cramer’s rule computes each xi. But it does say more. If

det(A) = 0 and if A~x = ~b has a solution, then each det Ai(~b)
= 0.