MATH 223
Assignment #3
due Friday September 26.
I like to see the solutions to a system of equations in Parametric Vector Form (or Vector Parametric
Form if you wish). If we find a set of solutions:
x1
x2
x3
x4
x5
x6
=
−3r
=
r
=
=
=
= 1/3
−4s −2t
−2s
s
t
for all choices r, s, t ∈ R then we can write the set of solutions in parametric vector form as follows:
−2
−4
−3
0
x1










 0 
 0 
 1 
 0 
 x2 


















 x3 
 =  0  + r  0  + s  −2  + t  0  ,

 0 
 1 
 0 
 0 
x 








 4










 1 
 0 
 0 
 0 
 x5 
0
0
0
1/3
x6










r, s, t ∈ R
1. Give the solutions in vector parametric form for the two planes:
a) π1 = {(x, y, z) : 2x − 2y + 3z = 5}.
b) π2 = {(x, y, z) : 3y = 6}.
2. Give the vector parametric form of all solutions to the following system of equations:
2x1
2x1
3x1
3x1
+4x4
+5x4
+8x4
+13x4
+2x2
+4x2
+6x5
+7x5
+9x5
+12x5
=
=
=
=
14
16
27
39
3. Express the line given by the vectors

 



1
1
x


 
 
y  = 1 + s 2 ,
z
1
−1
s∈R
as the set of solutions to a system of equations in x, y, z (two equations suffice). Then use Gaussian
Elimination on this system to re-express the solutions in vector parametric form.
4. Express the solutions to


x1


 

0 1 0 1 3 
7
 x2 



 
 0 0 0 1 2   x3  =  5 


0 0 0 0 0  x4 
0
x5
in vector parametric form as x = a + sb + tc + ud. Show that the only solution to


s
 
[bcd]  t  = 0
u

is

s
 
 t  = 0.
u
5. Express the inverse of the following matrix A as a product of elementary matrices and use this
product to express the matrix A as a product of elementary matrices.


1 1 0

A = 2 3 1

0 0 1
6. Let A be a 2×2 matrix with first column x and second column y. We wish to show that | det(A)|
is the area of the parallelogram formed by the two vectors x, y. Using assignment 2, question 1
(a), we readily deduce that det(A) = 0 if and only if the parallelogram is degenerate with no area.
So assume det(A) 6= 0.
a) There exists an angle θ such that R(θ)x = x0 points in the direction of the x-axis. Let y 0 = R(θ)y.
Show that the area of the parallelogram formed by x0 , y0 is the same as the area of the parallelogram
formed by the two vectors x, y.
#
"
1 s
b) We can choose a value s so that if we set S =
, then this shear matrix has Sx0 = x0
0 1
while Sy0 = y00 has y00 in the direction of the y-axis or its opposite. Show that the area of the
parallelogram formed by x0 , y00 is the same as the area of the parallelogram formed by the two
vectors x0 , y0 .
c) If we let B be a 2 × 2 matrix with first column x0 and second column y00 , show that | det(B)| is
the area of the rectangular box formed by x0 , y00 .
d) Using the product rule for determinants and verifying that det(R(θ)) = 1 and det(S) = 1, show
that | det(A)| is the area of the parallelogram formed by the two vectors x, y.
7. Show that the 2×2 matrix of 0’s is diagonalizable (I mention this because many students confuse
invertibility and diagonalizability; they are not related).
8. This is harder than question 7 from assignment 2 and essentially is the reverse observation. Let
A be a 2 × 2 matrix (not necessarily diagonalizable) and define
Assume limn→∞
xn
yn
1
xn
= An
.
2
yn
1
is an eigenvector of A.
1
= 1. Show that
1
,
1
Hint: This is one possible approach. Rewrite the matrix with respect to the two vectors
1
.. Explain why we can find c1 , c2 , c3 , c4 with
2
1
1
1
A
= c1
+ c2
;
1
1
2
1
1
1
A
= c3
+ c4
.
2
1
2
1
1
What do you need to show about c1 , c2 , c3 , c4 so that
An
1
1
1
with An+1
= A An
2
2
2
.
is an eigenvector of A? Now compare