Download Solutions for Midterm I - Stony Brook Math Department

Stony Brook University
Mathematics Department
Julia Viro
Introduction to Linear Algebra
MAT 211
March 6, 2008
Solutions for Midterm I
Problem 1. Solve the system using the Gauss-Jordan elimination and verify your answer

+ x4 = 0
 x1 + 3x2
x2 + x3
=−1

− x2 − x3 + x4 = 3.
Solution. Elementary row transformations of the augmented matrix of the system give rise to
the reduced row-echelon form:





1 3 0 1
1 0 −3 0
1
3
0 1
0
0
1
 0
 0 1
1
1 0 −1  ∼  0 1 1 0 −1 
1 0 −1  .
∼
R3+R2
R1−3R2−R3
3
2
2
0 −1 −1 1
0 0 0 1
0 0
0 1

Obviously, the rank of the matrix is 3. The number of free variables is 4 − 3 = 1 (the number
of unknowns minus the rank). We write down the solution starting from the back:
x4 = 2 (that is what the last row in the rref says),
x3 = t (choose x3 as a free variable),
x2 = −1 − t (from the second row in the rref),
x1 = 1 + 3t (from the first row in the rref).
So the solution is
(x1 , x2 , x3 , x4 ) = (1 + 3t, −1 − t, t, 2) = (1, −1, 0, 2) + t(3, −1, 1, 0),
where t is an arbitrary real number. Geometrically, the solution is a line in R4 .
To verify the solution, we substitute it into the three equations of the system:


1 + 3t + 3(−1 − t) + 2 = 0
−1 − t + t
= −1 .

−(−1 − t) − t + 2
=3
Since all the equations are satisfied, our solution is correct.
Answer:
(x1 , x2 , x3 , x4 ) = (1 + 3t, −1 − t, t, 2), t ∈ R.
2
Problem 2. Let T : R2 → R2 be an orthogonal projection onto the line x − 2y = 0 followed
by a counterclockwise rotation by 45◦ . Find a matrix of T (with respect to the standard basis).
Describe geometrically and show on a picture the kernel and the image of T . Is T invertible?
Explain!
Solution. Denote by P the orthogonal projection onto the line x − 2y = 0 and by R the
counterclockwise rotation by 45◦ . Then T = R ◦ P . Let A and B be standard matrices of P
and R respectively. Then the standard matrix of T is BA. Let us evaluate matrices A and B.
Matrix A consists of two columns representing the coordinates of the images of the standard
basis vectors e1 , e2 under the transformation P :


|
|
A = P e1 P e2  .
|
|
To evaluate the images we use a formula defining the orthogonal projection P :
x·u
Px =
u,
kuk2
where x is an arbitrary vector in R2 , u is a vector along the line of projection, kuk is its length
and · is a dot product. Take u = (2, 1). (Note that one can take any vector u√= (x, y) whose
√
coordinate satisfy the equation of the line x − 2y = 0.) Its length is kuk = 22 + 12 = 5.
Calculate the images of e1 = (1, 0) and e2 = (0, 1) under projection P :
(1, 0) · (2, 1)
2
4 2
P e1 =
(2, 1) = (2, 1) =
,
,
5
5
5 5
2 1
1
(0, 1) · (2, 1)
(2, 1) = (2, 1) =
,
.
P e2 =
5
5
5 5
It gives us the matrix A:
1 4 2
4/5 2/5
A=
=
.
2/5 1/5
5 2 1
The standard matrix of a counterclockwise rotation by 45◦ is
√ √ √
2 1 −1
cos 45◦ − sin 45◦
2/2
−
2/2
√
B=
= √
=
.
sin 45◦ cos 45◦
2/2
2/2
2 1 1
The standard matrix of T is
√ √ 2 1 −1 1 4 2
2 2 1
BA =
=
.
2 1 1 5 2 1
10 6 3
The kernel of T is a subspace which is annihilated by T . This is a line passing through the
origin which is orthogonal to the line x − 2y = 0. The equation of this line is 2x + y = 0. Note
that Ker T is spanned by vector (1, −2) which is annihilated by T .
The image of T is the line x − 2y = 0 rotated counterclockwise by 45◦ around the origin.
The
1
equation of this line is 3x − y = 0. Note that Im T is spanned by a column vector
of the
3
matrix of T .
3
Transformation T is not invertible. It can be explained in many different ways. For example,
Ker T 6= 0 or Im T 6= R2 or the determinant of the matrix of T is 0.
The picture makes all calculations crystal clear:
Ker T
Im T
2y=
x−
3x−
0
y=
Answer:
2x+
y=0
45
√ 2 2 1
the standard matrix of T is
,
10 6 3
Ker T = span {(1, −2)},
Im T = span {(1, 3)},
T is not invertible.
0
4
Problem 3.
A secret agent has got an encoded message
7, 12, −12, 16, 24, −24
representing the date of a secret operation. He knows that the coding
matrix is


1 0 2
 0 1 3 ,
−2 0 −3
but nevertheless cannot decode since he is not good in Linear Algebra.
Help him to decode the secret date!
Solution. Oh, boy! First, invert the matix:






1 0
2 1 0 0
1 0 2 1 0 0
1 0 0 −3 0 −2
 0 1
3 0 1 0  ∼  0 1 3 0 1 0  ∼  0 1 0 −6 1 −3 
R3+2R1
R1−2R3
−2 0 −3 0 0 1
0 0 1 2 0 1
0 0 1
2 0
1
R2−3R3
Second, multiply the inverse matrix by the two vectors (7, 12, −12) and (16, 24, −24) from the
encoded message:
  

3
7
−3 0 2
−6 1 −3  12  = 6 ,
2
−12
2 0 1

  

0
16
−3 0 2
−6 1 −3  24  = 0 .
8
−24
2 0 1

Third, read the secret date: 3, 6, 2, 0, 0, 8 or 3/6/2008. (The secret operation is Midterm I : -)
Answer:
3/6/2008
5
Problem 4. Let V be a subspace of R4 given by
V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0}.
Find a basis in V . Show that the vector v = (0, 1, −1, 2) belongs to V and find the coordinates
of v with respect to the basis you have found.
Solution. The subspace V consists of all the solutions of the homogeneous system
x1 + 2x2
− x4
x2 + x3
= 0
= 0.
En elementary row transformation of the augmented matrix gives rise to the reduced rowechelon form:
1 2 0 −1 0
0 1 1
0 0
∼
R1−2R2
1 0 −2 −1 0
0 1
1
0 0
.
The number of free variables in the solution is 4 − 2 = 2 (the number of unknowns minus
the rank of the matrix). We put x4 = t and x3 = s ( t, s are arbitrary real numbers) and get
x2 = −s and x1 = t + 2s. It gives us the solution in the form
(x1 , x2 , x3 , x4 ) = (t + 2s, −s, s, t) = t(1, 0, 0, 1) + s(2, −1, 1, 0),
t, s ∈ R.
Now we see that the subspace V is spanned by two vectors, (1, 0, 0, 1) and (2, −1, 1, 0). They
comprise a basis of V . Note that a basis is not unique and other answers are possible.
One can get a basis of V in a short cut:
V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0}
= {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 = −2x2 + x4 , x2 = −x3 }
= {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 = 2x3 + x4 , x2 = −x3 }
= {(2x3 + x4 , −x3 , x3 , x4 ) | x3 , x4 ∈ R}
= {x3 (2, −1, 1, 0) + x4 (1, 0, 0, 1) | x3 , x4 ∈ R} = span{(2, −1, 1, 0), (1, 0, 0, 1)}.
Now let us check that the vector v = (0, 1, −1, 2) belongs to V . Its components x1 = 0, x2 = 1,
x3 = −1, x4 = 2 satisfy the two equations defining V : 0 + 2 · 1 − 2 = 0 and 1 + (−1) = 0.
In order to find the coordinates of v with respect to the basis B = {(1, 0, 0, 1), (2, −1, 1, 0)}, we
organize a linear combination
v = (0, 1, −1, 2) = a(1, 0, 0, 1) + b(2, −1, 1, 0)
and find the coefficients a and b: a = 2, b = −1.
Answer:
A basis of V is B = {(1, 0, 0, 1), (2, −1, 1, 0)}.
Coordinates of v = (0, 1, −1, 2) with respect to B are (2, −1).
6
Problem 5. A linear transformation T : R3 → R4 is defined by
T (x, y, z) = (y + 2z, −y + z, 3y + z, 4y + 2z).
a) Find the matrix of T with respect to the standard bases.
b) Find a basis in the kernel of T and a basis in the image of T .
c) Find the dimensions of the kernel and the image.
d) Find the rank of T .
e) Verify the Kernel-Image theorem for T .
Solution. The matrix of T with respect to the standard bases in R3 and R4 is


0 1 2
0 −1 1

A=
0 3 1
0 4 2 4×3
We perfom elementary row transformations to get the reduced row-echelon form of A:

 
 

0 1 0
0 1 2
0 1 2
0 −1 1 0 0 3  0 0 1

 
 
A=
0 3 1 ∼ 0 0 −5 ∼ 0 0 0 = rref(A).
0 0 0
0 0 −6
0 4 2
Obviously, the rank of A is 2. It is the dimension of the image of T . The image of T is
generated by the second and the third columns of A, since the leading ones in the rref(A) stay
in the second and the third columns: Im T = span{(1, −1, 3, 4), (2, 1, 1, 2)}. Since the spanning
vectors are linearly independent they comprise a basis of T .
The Kernel-Image Theorem says that
dim Ker T + dim Im T = dim R3
or
dim Ker T + 2 = 3.
So dim Ker T = 1. A basis of Ker T can be found by solving a homogenous linear system with
coefficient matrix A. It is easy to read the solution from the rref(A): z = 0, y = 0, x = t, where
t is an arbitrary real number. Hence
Ker T = {(t, 0, 0) | t ∈ R} = span{(1, 0, 0)}.

Answer:

0 1 2
0 −1 1

The standard matrix of T is 
0 3 1 ,
0 4 2
a basis of Ker T is {(1, 0, 0)},
a basis of Im T is {(1, −1, 3, 4), (2, 1, 1, 2)},
dim Ker T = 1,
dim Im T = rk T = 2.