Dr. Sarah's Sample Problem Set Write-Ups

1.1 Number 65

Notice that cos(theta) and sin(theta) are numbers. I'll use t for web convenience. The augmented matrix for this system is _rtable[199295356] I want to use the cos(t) to eliminate the -sin(t) below it.

First, in the case when cos(t)=0, then the matrix is already in reduced form:
row 1 = [0 sin(t) 1]
row 2 = [-sin(t) 0 0].
In addition, we know that sin(t) in this case is plus or minus 1 (as cos(t)=0 and this is the unit circle where the pythagorean theorem holds for points [cos(t),sin(t)]). Namely sin(t) is nonzero. Then, using row 2 and dividing by -sin(t), we see that x =0/-sin(t) = 0. In addition, from row 1, sin(t)y = 1, so y = 1/sin(t) = plus or minus 1, whichever sin(t) is. Notice that the solution is then the same as (cos(t), sin(t)) for this t value.

So it remains to examine the case when cos(t) is non-zero. So I will do r2' = sin(t)/cos(t) r1 + r2 to obtain

Now we examine the a22 spot and obtain a common denominator in order to obtain

a22 := sin(t)^2/cos(t)+cos(t) = (sin(t)^2+cos(t)^2)...

Notice that this is equal to 1/cos(t) after using use the trig identity sin^2 +cos^2 = 1.

Thus row 3 now becomes: [0 1/cos(t) sin(t)/cos(t)]. In other words, 1/cos(t)*y = sin(t)/cos(t)

Hence y=sin(t).

Substitute into the equation corresponding to the first row and use the standard trig identity again to obtain x=cos(t)

Hence the solution is x=cost(t), y=sin(t), ie (cos(t), sin(t)), just like in the other case.

1.2 Number 53 Parts A, B, and C

Part A : A 6x3 matrix has 6 rows

This statement is true. As listed on page 14," an mxn matrix (read "m x n") has n rows (horizontal lines) and n columns (verticle lines)". Hence, a 6x3 matrix does indeed have 6 rows.

Part B : Every matrix is row-equivalent to a matrix in row-echelon form.

This statement is true. Just after Example 4, on page 18, the book says that, "It can be shown that every matrix is row-equivalent to a matrix in row-echelon form, which matches the statement exactly.

Part C : If the row-echelon form of the augmented matrix of a system of linear equations contains the row

[1 0 0 0 0], then the original system is inconsistent.

This statement is false. We must produce a counterexample to show this, ie a matrix which has the row [1 0 0 0 0] in row-echelon form, but is consistent instead of inconsistent. Look at the augmented matrix

> M:=Matrix([[1,0,0,0,0],[0,1,0,0,1],[0,0,1,0,2],[0,0,0,1,3]]);

M := _rtable[137346484]

We see that this satisfies the definition of row-echelon form on page 18 of the book - there are no rows consisting entirely of zeros, the first nonzero entry of each row is a 1, and for each two successive nonzero rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. Hence we have produced a row-echelon form with the desired row as its first row. Yet this system is not inconsistent since it has the solution x1=0, x2=1, x3=2, and x4=3.

Caution from Dr. Sarah: We have to be careful about true-false statements. Just because the statement in Part C is false and we have produced a counterexample whose system is consistent, does not mean that such a system is always consistent:

> M:=Matrix([[1,0,0,0,0],[0,1,0,0,1],[0,0,1,0,2],[0,0,0,0,1]]); M := _rtable[136470352]

M would be a counterexample. A true complete statement could read:

If the row-echelon form of the augmented matrix of a system of linear equations contains the row [1 0 0 0 0], then the original system is sometimes but not always consistent.

2.1 Number 54

Part A : The book tells us that the matrix C gives the total cost for manufacturing each product in each location,

and that labor accounts for 2/3 of the total costs. Since C gives the total costs, then 2/3 C

will give us the labor costs. Hence scalar multiplication by 2/3 on C (which is defined as 2/3 times each and every entry), will give us the matrix L that gives the labor costs for each product in each location:

> L:=(2/3*Matrix([[627,681],[135,150]]));

L := _rtable[135943016]

Part B: The book tells us that there are only matrial and labor costs, and asks us to find the material costs. Since there are only two costs, then material costs + labor costs = total costs, and so the matrix C (total costs) -

the matrix L (labor costs) will give us the material costs. Recall that matrix subtraction is defined by substracting corresponding entries.

> M:=Matrix([[627,681],[135,150]])-L;

M := _rtable[136462316]