11.1.

1
 1
−1

1 1
 0 0
0 0


1
1 1 −2
0 → 0 0
4
2
0 0 −1


2
1
1
3 → 0
5
−1 −1
0


1
−2
2
1 1 −2
1 −5
1
−3  →  0 0
0
1 11/19
0 0
0
1 −2 2
1
2 1
−1
1 3

1 1 −2
 0 0 −1
0 0
4

2
1
−1 −1  →
3
5
1
0
0

2 1
5 3 →
19 11


0 −3/19
1
0 −2/19  →  0
1 11/19
0
We can take x2 as the free variable. Then

 
x1
−7/19 − x2
 x2  
x2

 
 x3  = 
−2/19
x4
11/19
11.2.

1
 0

 0
0

1
 0

 0
0

1
 0

 0
0

1
 0

 0
0

1
 0

 0
0
1
2
1
3
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
0
0
0
1
0
0
0
1

−7/19
−2/19 
11/19


.




1 1 1 1 0 0 0 1
0


0 
→ 0 2 1 0 1 0 0 0 →
 3 4 2 1 0 1 0 0 
0 
1
1 4 3 2 0 0 1 0



1 1
1
1 0 0 0
1
1 0 0 0
1
1


1
0 1 0 0
0 
 →  0 1 −1 −2 0 1 0 −3  →


−1 −2 0 1 0 −3
0 2
1
0 1 0 0
0 
2
1 0 0 1 −1
0 3
2
1 0 0 1 −1



1
1 0
0 0
1
1 1
1
1 0
0 0
1

−1 −2 0
1 0 −3 
1 0 −3 
 →  0 1 −1 −2 0
→


3
4 1 −2 0
0 0
6
8 2 −4 0 12 
6
5
7 0 −3 1
8
0 0
5
7 0 −3 1
8


1
1 0
1 1
1
1
0
0
1
0
0
0
1
 0 1 −1 −2
1
0 −3 
0
1
0
−3
−1 −2 0
→
 0 0
1
1 2 −1 −1
4 
1
1
2 −1 −1
4
5
7 0 −3
1
8
0 0
0
2 −10
2
6 −12


1
1
0
0
0
1
1 1
1 0
5 −1 −3
7
 0 1 −1 0 −10
−1 −2
0
1
0 −3 
3
6
−15
→
 0 0
1
1
2 −1 −1
4 
1 0
7 −2 −4
10
0
1 −5
1
3 −6
0 0
0 1 −5
1
3 −6



1 0 0 0
0 0 −2
1
1 −3
1
0 −1
2

0 0 −3
1
2 −5 
1
2 −5 
 →  0 1 0 0 −3
→


1 0
7 −2 −4 10
0 0 1 0
7 −2 −4 10 
0 1 −5
1
3 −6
0 0 0 1 −5
1
3 −6
0 2 1 0
 3 4 2 1

 1 4 3 2
1 1 1 1

−2
−1
0
1
0
0
0
0
1
0
0
0
0
1
0
1


→



→

It follows that the inverse is

1
 −3

 7
−5

0 −1
2
1
2 −5 

−2 −4 10 
1
3 −6
11.3.
Find the inverse of the matrix using the adjoint matrix


0 2 1
 −1 3 2  .
0 1 1
The adjoint is

3
1



  2
 −
 1


  2
3
−1
2 −
0
1
2 1 −1
0
1 1 0
0
1 1 0
− 0
0
1 −
−1
2
1 2 0
−1
>
3 1 




1

2 
 −1
=
1 

1


2 
3 The determinant of the matrix is equal
−1 3
−1 2 3 2 + 1 − 2
0·
0 1
0 1 1 1 > 
1 −1
1
0
0  = 1
−1
2
−1

−1
1
0 −1 
0
2
to
= 0 · 1 + 2 · 1 + 1 · (−1) = 2 − 1 = 1.

1 −1
1
0 −1 .
It follows that the inverse matrix is  1
−1
0
2
11.4. If f, g are polynomials from the set, then f (2) + 3f (1) = 0 and g(2) +
3g(1) = 0, hence f (2) + g(2) + 3(f (1) + g(1)) = 0 and cf (2) + 3cf (1) = 0, so
f + g and cf belong to the set. Consequently, it is a subspace.
11.5.








1 −2 1
1 −2 1
1 −2 1
1 0 3
 −1




3 0 
1 1 
1 1 

→ 0
→ 0
→ 0 1 1 
 2






1 7
0
5 5
0
0 0
0 0 0 
3 −2 7
0
4 4
0
0 0
0 0 0

 

1
−2 




 3 
−1 




We see that the rank of the matrix is 2, and that the first two columns 
,
2   1 





3
−2
form a basis of the column space. We can choose the third variable free, then

2
the solution of the homogeneous system is

 



x1
−3x3
−3
 x2  =  −x3  = x3  −1  ,
x3
x3
1


 −3 
hence a basis of the null space consists of one vector  −1  .


1
0
11.6. We have L(f + g) = x(f + g) + (f + g)(x) − (f + g)(0) = xf 0 + xg 0 +
f (x) + g(x) − f (0) − g(0) = xf 0 + f (x) − f (0) + xg 0 + g(x) − g(0) = L(f ) + L(g).
Similarly, L(cf ) = x(cf )0 + (cf )(x) − (cf )(0) = cxf 0 + cf (x) − cf (0) = c(xf 0 +
f (x) − f (0)) = cL(f ). It follows that L is a linear transformation. We have
L(1) = 0 + 1 − 1 = 0, L(x) = x(x)0 + x − 0 = 2x, L(x2 ) = x(x2 )0 + x2 − 0 = 3x2 ,
L(x3 ) = x(x3 )0 + x3 − 0 = 4x3 , hence the matrix of the operator is


0 0 0 0
 0 2 0 0 


 0 0 3 0 .
0 0 0 4
Its rank is 3.
11.7. Denote v1 = (1, −1, −1, 1)> , v2 = (5, 1, 1, 5)> , v3 = (6, 2, 0, 4)> . We
apply the Gram-Schmidt orthogonalization process to these vectors. We take
u1 =
(1, −1, −1, 1)>
v1
√
=
= (1/2, −1/2, −1/2, 1/2)> .
kv1 k
4
Subtracting projection of v2 onto u1 from v2 we get
v2 − hv2 , u1 iu1 =
5−1−1+5
(5, 1, 1, 5)> −
(1/2, −1/2, −1/2, 1/2)> = (5, 1, 1, 5)> −(2, −2, −2, 2)> = (3, 3, 3, 3)> .
2
Normalizing this vector (dividing by its length 6), we get
u2 = (1/2, 1/2, 1/2, 1/2)> .
Let us subtract the projection of the third vector onto the span of u1 , u2 :
v3 − hv3 , u1 iu1 − hv3 , u2 iu2 =
6−2−0+4
6+2+0+4
(6, 2, 0, 4)> −
(1/2, −1/2, −1/2, 1/2)> −
(1/2, 1/2, 1/2, 1/2)> =
2
2
(6, 2, 0, 4)> − (2, −2, −2, 2)> − (3, 3, 3, 3)> = (1, 1, −1, −1)> .
Dividing by its length, we get u3 = (1/2, 1/2, −1/2, −1/2)> .
Consequently, an orthonormal basis of the span of v1 , v2 , v3 is
(1/2, −1/2, −1/2, 1/2)> , (1/2, 1/2, 1/2, 1/2)> , (1/2, 1/2, −1/2, −1/2)> .
3
(b) Let us project the vector r = (1, −1, 0, 0)> (the radius vector of the point)
onto the span of v1 , v2 , v3 . Passing to the orthonormal basis of the subspace,
we get that the projection is
hr, v1 iv1 +hr, v2 iv2 +hr, v3 iv3 = 1·v1 +0·v2 +0·v3 = v1 = (1/2, −1/2, −1/2, 1/2)> .
The distance from r to the projection (which is the same as the distance to the
span) is
k(1, −1, 0, 0)> − (1/2, −1/2, −1/2, 1/2)> k = k(1/2, −1/2, 1/2, −1/2)k = 1.
11.8. The characteristic polynomial is
1−x
−1
2
=
3
−3 − x
6
−2
2
−4 − x (1 − x)(−3 − x)(−4 − x) + 12 + 12 + 4(−3 − x) + 3(−4 − x) − 12(1 − x) =
− x3 − 6x2 = −x2 (x + 6).
We have two eigenvalues: 0 (double) and -6 (single).
The eigenvectors for 0:




1 −1 2
1 −1
2
 3 −3
0 0 .
6 → 0
0
0 0
−2
2 −4
Take the last two variables x2 , x3 as free. Then solution of the homogeneous
system is

 

 


x1
x2 − 2x3
1
−2
 x2  = 
 = x2  1  + x3  0  .
x2
x3
x3
0
1
  

−2 
 1
It follows that a basis of the eigenspace is  1  ,  0  .


1
0
The eigenvectors for -6:






−1 2
−2
2 2
1 −1 −1
3 6  →  7 −1 2  →  7 −1
2 →
2 2
3
3 6
3
3
6






1 −1 −1
1 −1 −1
1 0 1/2
 0
6
9 → 0
1 3/2  →  0 1 3/2 
0
6
9
0
0
0
0 0
0


−1
Setting the third variable equal to 2, we get a solution  −3 .
2
7
 3
−2
4
It is diagonalizable, since we have three independent eigenvectors.
11.9. Since the matrix is block-triangular, we have
1−x
2
−1 2 2
= (1 − x) 1 − x
0
1
−
x
2
= (1 − x)(x − 2x − 3)
2
1
−
x
0
2
1−x 11.10. Let us find the
4−x
2
−2
2
1
−
x
−1
−2
−1
1−x
characteristic polynomial:
= (4−x)(1−x)2 +4+4−4(1−x)−4(1−x)−2(4−x) = −x2 (x−6).
For the double eigenvalue x = 0 we find a basis of the eigenspace by solving the
homogeneous system




2 1 −1
4
2 −2
 2
0 .
1 −1  →  0 0
0 0
0
−2 −1
1
The solution is

 





x1
(−x2 + x3 )/2
−1/2
1/2
 x2  = 
 = x2 
x2
1  + x3 
0 .
x3
x3
0
1

  
1 
 −1
We can take a basis  2  ,  0  .


0
2
For the single eigenvalue x = 6, we have to solve the system

−2
 2
−2


1
2 −2
−5 −1  →  0
0
−1 −5

−2
A solution is  −1 .
1
We get

−1
−3
−3

1
−3  →
−3

1
 0
0


−1 1
1
1 1 → 0
0 0
0

−1
S= 2
0
1
0
2

−2
−1  ,
1
5

0
D= 0
0
0
0
0

0
0 .
6
0
1
0

2
1 
0
11.11. We have to find S −1 :




−1 1 −2 1 0 0
1 −1
2 −1 0 0
 2 0 −1 0 1 0  →  0
2 −5
2 1 0 →
0 0 1
0 2
1 0 0 1
0
2
1




1 −1
2 −1
0 0
1 −1
2
−1
0
0
 0
2
1 0 → 0
2
1
0 →
2 −5
2 −5
0
0
6 −2 −1 1
0
0
1 −1/3 −1/6 1/6




1 −1 0 −1/3
1/3 −1/3
1/3 −1/3
1 −1 0 −1/3
 0
2 0
1/3
1/6
5/6  →  0
1 0
1/6 1/12 5/12  →
0
0 1 −1/3 −1/6
1/6
0
0 1 −1/3 −1/6
1/6


1 0 0 −1/6 5/12 1/12
 0 1 0
1/6 1/12 5/12  ,
0 0 1 −1/3 −1/6 1/6
so

5/12 1/12
1/12 5/12  .
−1/6 1/6

S −1
−1/6
=  1/6
−1/3
Therefore,



1 0 0
−1/6 5/12 1/12
−1 1 −2
eA = SeD S −1 =  2 0 −1   0 1 0   1/6 1/12 5/12  =
0 0 e6
−1/3 −1/6 1/6
0 2
1




6
−1 1 −2e
−1/6 5/12 1/12
 2 0 −e6   1/6 1/12 5/12  =
0 2
e6
−1/3 −1/6 1/6


(1 + 2e6 )/3 (−1 + e6 )/3 (1 − e6 )/3
 (−1 + e6 )/3 (5 + e6 )/6 (1 − e6 )/6  .
(1 − e6 )/3
(1 − e6 )/6 (5 + e6 )/6

As determinant of a product of matrices is equal to the product of determinants,
we have
det(eA ) = det(S) det eD det(S −1 ) = det(S) det(S −1 ) det eD = det eD = e6 .
We have T r(A) = 4 + 1 + 1 = 6.
11.12. When a point goes along the curve, x goes from
R 1 1 to 0 and then back
1
from 0 to 1. It follows that the integral is equal to 2 0 sin x dx = −2 cos x|0 =
−2 cos 1 + 2 cos 0 = 2 − 2 cos 1.
2
∂(x2 +y)
11.13. We have ∂x
= 1. The results are different, so the
∂x = 2x and
∂y
field is not conservative.
11.14. The triangle can be parametrized by x and z, then we have (x, y, z) =
∂
(x, 1 − x − z, z). The derivative ∂x
of the parametrization is (1, −1, 0). The
6
derivative
∂
∂z
hence dS =
is (0, −1, 1). Their
i
j
~
dS = 1 −1
0 −1
√
3 dx dz, and
√ Z
xz dS = 3
ZZ
cross-product is
k 0 dx dz = (1, 1, 1) dx dz,
1 1
0
S
√ Z 1 x(1 − x)2
1−x
xz /2 z=0 dx = 3
dx =
2
0
0
0
√ Z 1
√ √
3
3 1 2 1
3
x − 2x2 + x3 dx =
− +
=
2 0
2
2 3 4
24
Z
1−x
√ Z
xz dz dx = 3
1
2
11.15.
i
~ F~ = ∂/∂x
∇×
z2
j
∂/∂y
y2
k
∂/∂z
xy
2
= ∂xy − ∂y ,
∂y
∂z
∂z 2
∂xy
−
,
∂z
∂x
∂y 2
∂z 2
−
∂x
∂y
= (x, 2z−y, 0).
Let us parametrize the 2D triangle by x and y. Then z = 2−2x−2y, and the
parametrization is (x, y, 2 − 2x − 2y). Its derivatives over x and y are (1, 0, −2)
and (0, 1, −2). We take them in this order, if we choose the upward orientation,
i.e., the counterclockwise, if looking from above. Then
i j
k ~ = 1 0 −2 = (2, 2, 1)
dS
0 1 −2 The integral is then, by Stockes’ Theorem equal to
1
Z
Z
1−x
Z
1
Z
(x, 2(2−2x−2y)−y, 0)·(2, 2, 1) dy dx =
0
Z
0
8−6x−10y dy dx =
0
1
8(1 − x) − 6x(1 − x) − 5(1 − x)2 dx =
0
Z
0
7
1−x
0
1
x2 − 4x + 3 dx =
4
1
−2+3 = .
3
3