Math. 311 (Fulling) 17 February 1999

Test A – Solutions

Calculators may be used for simple arithmetic operations only!

1. (15 pts.) Find a parametric representation of the plane (in R3 with coordinates (x, y, z) ) whose equation is 3x − 2z = −2 .

Let’s do this by the formal row-reduction method of Chapter 2. The augmented matrix and its reduced form are

( 3 0 −2 −2 ) → ( 1 0 − \frac{2}{3} − \frac{2}{3} ) .

Working from the end of the variable list back to the beginning, we see that we must take

z = t, y = s (arbitrary parameters),

and then

x = \frac{2}{3} t − \frac{2}{3} .

In vector form,

x⃗ = s ( \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ) + t ( \frac{2}{3} 0 1 ) + ( \begin{matrix}− \frac{2}{3} \\ 0 \\ 0 \end{matrix} ) .

It should be noted that there are other correct answers, such as

x⃗ = s ( 1 0 \frac{3}{2} ) + t ( \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ) + ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ) .

Also, there are other ways of presenting an answer, such as

x⃗ = ( \begin{matrix}\frac{2}{3} (t − 1) \\ s \\ t \end{matrix} ) .

Remarks: The answer to a problem like this is easily checked: The constant term must by itself satisfy the equation, since it corresponds to s = t = 0 . The vectors multiplied by s and t must satisfy the corresponding homogeneous equation, 3x − 2z = 0 . Unfortunately, passing these tests does not guarantee that your solution is complete: The main lesson of this problem is the necessity of including the term proportional to (0, 1, 0) (or saying something about what the coordinate y does!). If you leave out that term, you have constructed just a line, not a parametrized plane.

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2. (15 pts.) An A2X30 module contains 1000 cubic inches of steel and 20 cubic inches of titanium. A B62W subassembly contains 10 cubic inches of steel and 1 cubic inch of titanium. A supertanker is built from 10 A2X30s and 8 B62Ws. A minesweeper is built from 5 A2X30s and 3 B62Ws.

Organize these facts into matrices, and find the matrix that should be used to calculate the quantities of metals needed to make k tankers and m minesweepers.

Let A , B , s . t have the obvious meanings. Translating the sentences into equations, we have

( \begin{matrix}s \\ t \end{matrix} ) = ( \begin{matrix}1000 & 10 \\ 20 & 1 \end{matrix} ) ( \begin{matrix}A \\ B \end{matrix} ) (let’s call this matrix M ),

( \begin{matrix}A \\ B \end{matrix} ) = ( \begin{matrix}10 & 5 \\ 8 & 3 \end{matrix} ) ( \begin{matrix}k \\ m \end{matrix} ) (let’s call this matrix N ).

(Be sure that your matrices express the correct relationships, calculating the required inputs from the desired outputs. For example, the top line of the first matrix equation expresses the formula s = 1000A + 10B , saying that we need 1000 units of steel for each A module and 10 units of steel for each B module.) Therefore,

( \begin{matrix}s \\ t \end{matrix} ) = MN ( \begin{matrix}k \\ m \end{matrix} ) ,

where

MN = ( \begin{matrix}1000 & 10 \\ 20 & 1 \end{matrix} ) ( \begin{matrix}10 & 5 \\ 8 & 3 \end{matrix} ) = ( \begin{matrix}10080 & 5030 \\ 208 & 103 \end{matrix} ) .

3. (20 pts.) Atoms near the point x⃗0 ≡ ( \begin{matrix}3 \\ 2 \\ 1 \end{matrix} ) sit in an electric field E⃗ = ( \begin{matrix}x2 − y \\ x2 + y2 \\ z3 \end{matrix} ) .

(a) Find the first-order (best affine) approximation to E⃗(x⃗) for x⃗ near x⃗0 .

The Jacobian matrix is

\frac{dE⃗}{dx⃗} = ( \begin{matrix}2x & −1 & 0 \\ 2x & 2y & 0 \\ 0 & 0 & 3z^{2} \end{matrix} ) = ( \begin{matrix}6 & −1 & 0 \\ 6 & 4 & 0 \\ 0 & 0 & 3 \end{matrix} ) at x⃗_{0} .

Therefore,

E⃗(x⃗) ≈ E⃗(x⃗_{0}) + \frac{dE⃗}{dx⃗} (x⃗ − x⃗_{0}) = ( \begin{matrix}7 \\ 13 \\ 1 \end{matrix} ) + ( \begin{matrix}6 & −1 & 0 \\ 6 & 4 & 0 \\ 0 & 0 & 3 \end{matrix} ) ( \begin{matrix}x − 3 \\ y − 2 \\ z − 1 \end{matrix} ) .

(b) Suppose that the index of refraction of a crystal depends on the electric field according

to the law

n = 1 + 0.01Ex2 + 0.04Ey2 + 0.02Ez2.

Use the multidimensional chain rule to find \frac{∂n}{∂y} at x⃗0 .

Method 1: ∇n = \frac{dn}{dE⃗} \frac{dE⃗}{dx⃗} = (0.02E_{x}, 0.08E_{y}, 0.04E_{z})|x⃗_{0} \frac{dE⃗}{dx⃗} =

(0.14, 1.04, 0.04) ( \begin{matrix}6 & −1 & 0 \\ 6 & 4 & 0 \\ 0 & 0 & 3 \end{matrix} ) = ( ∗ , 4.02, ∗),

where the numbers ∗ are irrelevant and 4.02 = \frac{∂n}{∂y} .

Method 2: (This is really the same method, but in “classical” partial-derivative notation instead of vectors and matrices.)

\frac{∂n}{∂y} = \frac{∂n}{∂E_{x}} \frac{∂E_{x}}{∂y} + \frac{∂n}{∂E_{y}} \frac{∂E_{y}}{∂y} + \frac{∂n}{∂E_{z}} \frac{∂E_{z}}{∂y} = 0.02E_{x} (−1) + 0.08E_{y} (2y) + 0.04E_{z} (0).

When all this is evaluated at (x, y, z) = (3, 2, 1) , we again get 4.02.

4. (15 pts.) Let’s define a mapping Q of the function space C1(0, ∞) into the function space C(0, ∞) by

Q(f)(t) ≡ \frac{df}{dt} + (2t + 1)f(t)2.

(Here t is the independent variable of the functions in C1(0, ∞) , and f is a generic element of C1(0, ∞) .) Is Q linear, affine, or nonlinear? Justify your answer.

Nonlinear. It is enough to show that either of the linearity conditions is violated; for instance,

Q(2f) = 2 \frac{df}{dt} + 4(2t + 1)f^{2} ≠ 2 \frac{df}{dt} + 2(2t + 1)f^{2} = 2Q(f).

5. (15 pts.)

(a) Solve the system \{ \begin{matrix}x + y & = 2, \\ x − by & = 0 \end{matrix}\} for x and y (with b as a parameter).

Set up an augmented matrix and reduce:

( \begin{matrix}1 & 1 & 2 \\ 1 & −b & 0 \end{matrix} ) → ( \begin{matrix}1 & 1 & 2 \\ 0 & −b − 1 & −2 \end{matrix} ) → ( \begin{matrix}1 & 0 & 2 − \frac{2}{b+1} \\ 0 & 1 & \frac{2}{b+1} \end{matrix} ) .

We note that the last step assumes b + 1 ≠ 0 . Therefore, if b ≠ −1 , we have

x = 2 − \frac{2}{b + 1} = \frac{2b}{b + 1} , y = \frac{2}{b + 1} .

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(b) Point out any values of b that are “special” with regard to existence and uniqueness of solutions. (Explain.)

If b = −1 , the equations are inconsistent; no solutions exist.

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

( \begin{matrix}3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 1 & −2 & 0 & 1 & 0 \\ 2 & 1 & 0 & 0 & 0 & 1 \end{matrix} )  \begin{matrix}(1) ↔ (2) \\ (2) ← (2) − 3(1) \\ (3) ← (3) − 2(1) \\ ⟶ \end{matrix}
( \begin{matrix}1 & 1 & −2 & 0 & 1 & 0 \\ 0 & −1 & 7 & 1 & −3 & 0 \\ 0 & −1 & 4 & 0 & −2 & 1 \end{matrix} )  \begin{matrix}(2) ← −(2) \\ (1) ← (1) − (2) \\ (3) ← (3) + (2) \\ ⟶ \end{matrix}
( \begin{matrix}1 & 0 & 5 & 1 & −2 & 0 \\ 0 & 1 & −7 & −1 & 3 & 0 \\ 0 & 0 & 3 & 1 & −1 & −1 \end{matrix} )  \begin{matrix}(3) ← \frac{1}{3} (3) \\ (1) ← (1) − 5(3) \\ (2) ← (2) + 7(3) \\ ⟶ \end{matrix}

( \begin{matrix}1 & 0 & 0 & − \frac{2}{3} & − \frac{1}{3} & \frac{5}{3} \\ 0 & 1 & 0 & \frac{4}{3} & \frac{2}{3} & − \frac{7}{3} \\ 0 & 0 & 1 & \frac{1}{3} & − \frac{1}{3} & − \frac{1}{3} \end{matrix} ) .

Therefore,

M^{−1} = \frac{1}{3} ( \begin{matrix}−2 & −1 & 5 \\ 4 & 2 & −7 \\ 1 & −1 & −1 \end{matrix} ) .