Calculus of One and Several Variables

A.2.10 Cramer’s Rule

In case you are solving a system of equations, Ax = y for x, it follows that if A⁻¹ exists,

x = (A− 1A)x = A−1 (Ax ) = A −1y

thus solving the system. Now in the case that A⁻¹ exists, there is a formula for A⁻¹ given above. Using this formula,

∑n ∑n 1 xi = a−ij1 yj = ------cof (A )jiyj. j=1 j=1 det(A)

By the formula for the expansion of a determinant along a column,

( ∗ ⋅⋅⋅ y ⋅⋅⋅ ∗ ) 1 | . .1 . | xi = det(A) det( .. .. .. ) , ∗ ⋅⋅⋅ yn ⋅⋅⋅ ∗

where here the i^th column of A is replaced with the column vector

^T, and the determinant of this modified matrix is taken and divided by det. This formula is known as Cramer’s rule.

Appendix B
Implicit Function Theorem*

The implicit function theorem is one of the greatest theorems in mathematics. There are many versions of this theorem which are of far greater generality than the one given here. The proof given here is like one found in one of Caratheodory’s books on the calculus of variations. It is not as elegant as some of the others which are based on a contraction mapping principle but it may be more accessible. However, it is an advanced topic. Don’t waste your time with it unless you have first read and understood the material on rank and determinants found in the chapter on the mathematical theory of determinants. You will also need to use the extreme value theorem for a function of n variables and the chain rule of multivariable calculus as well as everything about matrix multiplication.

Proof: Let

( ) f1(x,y ) || f2(x,y )|| f (x,y ) = |( .. |) . . fn (x,y)

Define for

∈B

(x0,δ)

ⁿ and y ∈ B the following matrix.

( ( 1 ) ( 1 ) ) ( 1 n ) | f1,x1.x ,y ⋅⋅⋅ f1,xn x. ,y | J x ,⋅⋅⋅,x ,y ≡ ( .. .. ) . (*) fn,x1 (xn,y) ⋅⋅⋅ fn,xn (xn,y )

(*)

Then by the assumption of continuity of all the partial derivatives and the extreme value theorem, there exists r > 0 and δ₀,η₀ > 0 such that if δ ≤ δ₀ and η ≤ η₀, it follows that for all

∈B

(x0,δ)

ⁿ and y ∈B

(y0,η)

,

| ( ( 1 n ))| |det J x ,⋅⋅⋅,x ,y | > r > 0. (2.3)

(2.3)

and B

×B

(y0,η0)

⊆ U. By continuity of all the partial derivatives and the extreme value theorem, it can also be assumed there exists a constant, K such that for all ∈B

(x0,δ0)

×B

(y0,η0)

and i = 1,2,,n, the i^th row of D₂f, given by D₂f_i satisfies

|D2fi(x,y)| < K, (2.4)

(2.4)

and for all

∈B

(x0,δ0)

ⁿ and y ∈B

(y0,η0)

the i^th row of the matrix,

( ) J x1,⋅⋅⋅,xn,y −1

which equals e_i^T

satisfies

| ( ( )−1)| ||eTi J x1,⋅⋅⋅,xn,y || < K. (2.5)

(2.5)

(Recall that e_i is the column vector consisting of all zeros except for a 1 in the i^th position.)

To begin with it is shown that for a given y ∈ B

there is at most one x ∈ B such that f = 0.

Pick y ∈ B

and suppose there exist x,z ∈B

(x0,δ)

such that f = f = 0. Consider f_i and let

h(t) ≡ fi(x + t(z− x),y).

Then h

= h and so by the mean value theorem, h^′ = 0 for some t_i ∈. Therefore, from the chain rule and for this value of t_i,

′ ∑n -∂-- h (ti) = ∂xjfi(x + ti(z − x),y)(zj − xj) = 0. (2.6) j=1

(2.6)

Then denote by xⁱ the vector, x + t_i

. It follows from 2.6 that  

( ) J x1,⋅⋅⋅,xn,y (z − x ) = 0

and so from 2.3 z − x = 0. (The matrix, in the above is invertible since its determinant is nonzero.) Now it will be shown that if η is chosen sufficiently small, then for all y ∈ B

, there exists a unique x ∈ B such that f = 0.

Claim: If η is small enough, then the function, x → h_y

≡² achieves its minimum value on B

(x0,δ)

at a point of B. (The existence of a point in B

(x0,δ)

at which h_y achieves its minimum follows from the extreme value theorem.)

Proof of claim: Suppose this is not the case. Then there exists a sequence η_k → 0 and for some y_k having

< η_k, the minimum of h_ykon B

(x0,δ)

occurs on a point x_k such that = δ. Now taking a subsequence, still denoted by k, it can be assumed that x_k → x with = δ and y_k → y₀. This follows from the fact that is a closed and bounded set and is therefore sequentially compact. Let ε > 0. Then for k large enough, the continuity of y → h_y implies h_yk < ε because h_y0 = 0 since f = 0. Therefore, from the definition of x_k, it is also the case that h_yk < ε. Passing to the limit yields h_y0 ≤ ε. Since ε > 0 is arbitrary, it follows that h_y0 = 0 which contradicts the first part of the argument in which it was shown that for y ∈ B there is at most one point, x of B

(x0,δ)

where f = 0. Here two have been obtained, x₀ and x. This proves the claim.

Choose η < η₀ and also small enough that the above claim holds and let x

denote a point of B at which the minimum of h_y on B

(x0,δ)

is achieved. Since x is an interior point, you can consider h_y for small and conclude this function of t has a zero derivative at t = 0. Now

∑n 2 hy(x (y )+ tv ) = fi (x(y) +tv,y) i=1

and so from the chain rule,

∑n ∑n d-hy(x(y)+ tv) = 2fi(x (y) + tv,y) ∂fi(x-(y)+-tv,y)vj. dt i=1j=1 ∂xj

Therefore, letting t = 0, it is required that for every v,

∑n ∑n 2fi(x(y),y) ∂fi(x-(y-),y)vj = 0. i=1j=1 ∂xj

In terms of matrices this reduces to

0 = 2f (x (y),y)T D1f (x(y),y)v

for every vector v. Therefore,

0 = f (x(y),y)T D f (x(y),y) 1

From 2.3, it follows f

= 0. This proves the existence of the function y → x such that f = 0 for all y ∈ B.

It remains to verify this function is a C¹ function. To do this, let y₁ and y₂ be points of B

. Then as before, consider the i^th component of f and consider the same argument using the mean value theorem to write

0 = fi(x(y1),y1)− fi(x(y2),y2) = fi(x (y1(),yi1 ))− fi(x (y2),y1) +fi(x (y2(),y1)− fi)i(x(y2),y2) (2.7) = D1fi x ,y1 (x(y1)− x (y2))+ D2fi x (y2),y (y1 − y2).

(2.7)

where yⁱ is a point on the line segment joining y₁ and y₂. Thus from 2.4 and the Cauchy-Schwarz inequality,

||D2fi(x(y2),yi)(y1 − y2)|| ≤ K |y1 − y2|.

Therefore, letting M

≡ M denote the matrix having the i^th row equal to D₂f_i, it follows

( )1∕2 |M (y − y )| ≤ ∑ K2 |y − y |2 = √mK- |y − y |. (2.8) 1 2 i 1 2 1 2

(2.8)

Also, from 2.7,

( ) J x1,⋅⋅⋅,xn,y1 (x (y1) − x(y2)) = − M (y1 − y2) (2.9)

(2.9)

and so from 2.8, 2.5,

=

Now let y₂ = y,y₁ = y + he_k for small h. Then M depends on h and

lim M (h) = D2f (x (y) ,y ) h→0

thanks to the continuity of y → x

just shown. Also,

x-(y-+-hek)−-x-(y) = − J (x1(h),⋅⋅⋅,xn (h),y+ hek)−1M (h )ek h

Passing to a limit and using the formula for the inverse of a matrix in terms of the cofactor matrix, and the continuity of y → x

shown above, this yields

-∂x = − D1f (x (y),y )−1 D2fi(x(y),y)ek ∂yk

Then continuity of y → x

and the assumed continuity of the partial derivatives of f shows that each partial derivative of y → x exists and is continuous. ■

This implies the inverse function theorem given next.

Proof: Apply the implicit function theorem to the function

F (x,y) ≡ f (x)− y

where y₀ ≡ f

. Thus the function y → x defined in that theorem is f⁻¹. Now let

W ≡ B (x ,δ)∩ f−1(B (y ,η)) 0 0

and

V ≡ B (y0,η).

This proves the theorem. ■