Math. 311 (Fulling) 18 February 2004

Test A – Solutions

Name:

Calculators may be used for simple arithmetic operations only!

1. (15 pts.) Find all solutions (w, x, y, z) of the system \{ \begin{matrix}w + 2x + y + z & = 1 , \\ 3w + 2x − y − 2z & = 0 . \end{matrix}\}

Form the augmented matrix and reduce:

( \begin{matrix}1 & 2 & 1 & 1 & 1 \\ 3 & 2 & −1 & −2 & 0 \end{matrix} ) ⟶^{(2)→(2)−3(1)} ( \begin{matrix}1 & 2 & 1 & 1 & 1 \\ 0 & −4 & −4 & −5 & −3 \end{matrix} ) ⟶^{(2)→− \frac{1}{4} (2)}

( \begin{matrix}1 & 2 & 1 & 1 & 1 \\ 0 & 1 & 1 & \frac{5}{4} & \frac{3}{4} \end{matrix} ) ⟶^{(1)→(1)−2(2)} ( \begin{matrix}1 & 0 & −1 & − \frac{3}{2} & − \frac{1}{2} \\ 0 & 1 & 1 & \frac{5}{4} & \frac{3}{4} \end{matrix} ) .

Thus

w − y − \frac{3}{2} z = − \frac{1}{2} ,
x + y + \frac{5}{4} z = \frac{3}{4} .

Let

y = s, z = t s and t arbitrary).

Then

w = s + \frac{3}{2} t − \frac{1}{2} , x = −s − \frac{5}{4} t + \frac{3}{4}

is the general solution. It is quickly checked by substituting back into the original equations.

2. (10 pts.) Define a mapping T of the function space C2(−∞, ∞) into the function space

C(−∞, ∞) by

[T (f)](z) ≡ f′′(z) + z2f(z) + ∫{z \atop 0} e−uf(u) du .

Is T a linear function? Explain.

YES. Let λ be an arbitrary real number and f and g be arbitrary functions in C^{2} . Then

T (λf + g)(z) = (λf + g)′′(z) + z^{2}(λf + g)(z) + ∫_{0}^{z} e^{−u}(λf(u) + g(u)) du
= λf′′(z) + g′′(z) + λz^{2}f(z) + z^{2}g(z) + λ ∫_{0}^{z} e^{−u}f(u) du + ∫_{0}^{z} e^{−u}g(u) du
= λ[T f](z) + [T g](z)

(where the known linearity of differentiation and integration have been used). Thus T satisfies the definition of linearity.

311A-S04 Page 2

3. (30 pts.) Define \{ \begin{matrix}u & = x3 − 2y + z , \\ v & = 4x + ey + z2, \end{matrix}\}  r⃗ = ( \begin{matrix}x \\ y \\ z \end{matrix} ) ,  r⃗0 = ( \begin{matrix}2 \\ 1 \\ 2 \end{matrix} ).

(a) Find the direction of most rapid increase of u at the point r0 .

The direction of fastest increase is the direction of the gradient.

∇u(r⃗_{0}) = (3x^{2}, −2, 1)|r⃗_{0} = (12, −2, 1).

The unit vector in that direction (for one point extra credit) is

\frac{1}{\sqrt{144 + 4 + 1}} ( \begin{matrix}12 \\ −2 \\ 1 \end{matrix} ) = \frac{1}{\sqrt{149}} ( \begin{matrix}12 \\ −2 \\ 1 \end{matrix} ) .

(b) Define F: R3 → R2 by ( \begin{matrix}u \\ v \end{matrix} ) = F (r⃗) . Construct the best affine approximation (a.k.a.

the first-order approximation) to F around r⃗0 .

The Jacobian matrix of this function is

JF = ( \begin{matrix}\frac{∂u}{∂x} & \frac{∂u}{∂y} & \frac{∂u}{∂z} \\ \frac{∂v}{∂x} & \frac{∂v}{∂y} & \frac{∂v}{∂z} \end{matrix} ) = ( \begin{matrix}3x^{2} & −2 & 1 \\ 4 & e^{y} & 2z \end{matrix} ) .

Evaluate it at r⃗_{0} :

Jr⃗_{0}F = ( \begin{matrix}12 & −2 & 1 \\ 4 & e & 4 \end{matrix} ) .

Now

F (r⃗) ≈ F (r⃗_{0}) + dr⃗_{0}F (r⃗ − r⃗_{0}),

where the matrix of the differential dr⃗_{0}F is Jr⃗_{0}F . That is,

F (r) ≈ ( \begin{matrix}8 \\ 12 + e \end{matrix} ) + ( \begin{matrix}12 & −2 & 1 \\ 4 & e & 4 \end{matrix} ) ( \begin{matrix}x − 2 \\ y − 1 \\ z − 2 \end{matrix} ) .

Further simplification is optional (cf. part (c)).

(c) Define a curve by r⃗(t) = ( \begin{matrix}2t2 \\ t \\ −2 cos(πt) \end{matrix} ). Note that r⃗(1) = r⃗0 . Find the tangent

vector to the curve at that point, and the parametrized equation of the tangent line.

The tangent vector is

r⃗′(1) = ( \begin{matrix}4t \\ 1 \\ 2π sin(πt) \end{matrix} )|_{t=1} = ( \begin{matrix}4 \\ 1 \\ 0 \end{matrix} ) .

So the tangent line is

r⃗ = r⃗(1) + r⃗′(1)(t − 1)

= ( \begin{matrix}2 \\ 1 \\ 2 \end{matrix} ) + ( \begin{matrix}4 \\ 1 \\ 0 \end{matrix} ) (t − 1)

= ( \begin{matrix}2 + 4(t − 1) \\ 1 + (t − 1) \\ 2 \end{matrix} ) .

Further simplification is possible but not recommended.

(d) Find \frac{d}{dt}F (r⃗(t)) at t = 1 .

We can use parts of (b) and (c) in the chain rule:

\frac{d}{dt} F(r⃗(t))|_{t=1} = dr⃗_{0}F (r⃗′(1)) [also written (Jr⃗_{0}F)(r⃗′(1))]

= ( \begin{matrix}12 & −2 & 1 \\ 4 & e & 4 \end{matrix} ) ( \begin{matrix}4 \\ 1 \\ 0 \end{matrix} ) = ( \begin{matrix}46 \\ 16 + e \end{matrix} ) .

4. (15 pts.) Producing a yacht requires 1 ton of steel and 1 ton of aluminum. Producing an airplane requires 3 tons of steel and 2 tons of aluminum. Producing a ton of steel consumes 1 ton of coal and 2 tons of hematite. Producing a ton of aluminum consumes 4 tons of coal and 2 tons of bauxite. Organize these facts into matrices, and find the matrix that tells you how much coal ( c), hematite ( h), and bauxite ( b) is needed to make y yachts and p airplanes.

Translate the given information into equations:

( \begin{matrix}s \\ a \end{matrix} ) = ( \begin{matrix}1 & 3 \\ 1 & 2 \end{matrix} ) ( \begin{matrix}y \\ p \end{matrix} ) , ( \begin{matrix}c \\ h \\ b \end{matrix} ) = ( \begin{matrix}1 & 4 \\ 2 & 0 \\ 0 & 2 \end{matrix} ) ( \begin{matrix}s \\ a \end{matrix} ) .

Give the matrices names:

A = ( \begin{matrix}1 & 3 \\ 1 & 2 \end{matrix} ) , B = ( \begin{matrix}1 & 4 \\ 2 & 0 \\ 0 & 2 \end{matrix} ) .

Then

( \begin{matrix}c \\ h \\ b \end{matrix} ) = BA ( \begin{matrix}y \\ p \end{matrix} )

where

BA = ( \begin{matrix}5 & 11 \\ 2 & 6 \\ 2 & 4 \end{matrix} ) .

5. (10 pts.) The commutator of two matrices is defined as [A, B] = AB − BA. The trace of a matrix is the sum of its diagonal elements:

tr M = ∑ ; j Mjj .

(a) What condition must the matrices A and B satisfy in order for their commutator to

be defined?

They must be square matrices of the same size ( n × n).

(b) Prove that the trace of any commutator is equal to zero.

tr(AB) = ∑_{j=1}^{n}(AB)_{jj} = ∑_{j=1}^{n} ∑_{k=1}^{n} A_{jk}B_{kj}

by definition of matrix multiplication. Therefore, by interchanging of the matrices, then renaming of indices, then commuting the multiplication of numbers,

tr(BA) == ∑_{j=1}^{n} ∑_{k=1}^{n} B_{jk}A_{kj} = ∑_{k=1}^{n} ∑_{j=1}^{n} B_{kj}A_{jk} = tr(AB).

Thus

tr(AB − BA) = 0,

since tr is obviously a linear function of M .

6. (20 pts.) Find the inverse (if it exists) of the matrix M = ( \begin{matrix}3 & 1 & 1 \\ 1 & 0 & 1 \\ 2 & 4 & 2 \end{matrix} ).

( \begin{matrix}3 & 1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 2 & 4 & 2 & 0 & 0 & 1 \end{matrix} ) ⟶^{(1)↔(2)} ( \begin{matrix}1 & 0 & 1 & 0 & 1 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 \\ 2 & 4 & 2 & 0 & 0 & 1 \end{matrix} ) ⟶^{{(2)→(2)−3(1) \atop (3)→(3)−2(1)}}

( \begin{matrix}1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & −2 & 1 & −3 & 0 \\ 0 & 4 & 0 & 0 & −2 & 1 \end{matrix} ) ⟶^{(3)→(3)−4(2)} ( \begin{matrix}1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & −2 & 1 & −3 & 0 \\ 0 & 0 & 8 & −4 & 10 & 1 \end{matrix} ) ⟶^{{(2)→(2)+ \frac{1}{4} (3) \atop (3)→ \frac{1}{8} (3)}}

( \begin{matrix}1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & − \frac{1}{2} & \frac{1}{4} \\ 0 & 0 & 1 & − \frac{1}{2} & \frac{5}{4} & \frac{1}{8} \end{matrix} ) ⟶^{(1)→(1)−(3)} ( \begin{matrix}1 & 0 & 0 & \frac{1}{2} & − \frac{1}{4} & − \frac{1}{8} \\ 0 & 1 & 0 & 0 & − \frac{1}{2} & \frac{1}{4} \\ 0 & 0 & 1 & − \frac{1}{2} & \frac{5}{4} & \frac{1}{8} \end{matrix} ) .

Therefore,

M^{−1} = ( \begin{matrix}\frac{1}{2} & − \frac{1}{4} & − \frac{1}{8} \\ 0 & − \frac{1}{2} & \frac{1}{4} \\ − \frac{1}{2} & \frac{5}{4} & \frac{1}{8} \end{matrix} ) .

Check:

MM^{−1} = ( \begin{matrix}3 & 1 & 1 \\ 1 & 0 & 1 \\ 2 & 4 & 2 \end{matrix} ) ( \begin{matrix}\frac{1}{2} & − \frac{1}{4} & − \frac{1}{8} \\ 0 & − \frac{1}{2} & \frac{1}{4} \\ − \frac{1}{2} & \frac{5}{4} & \frac{1}{8} \end{matrix} ) = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) .