Exercise 1
Let f : R3→ R3be the function defined by:
f (x , y , z) = (−x + kz, y + kz, z + y ); prove that it is a linear map. Then find, for any k ∈ R, Imf , ker f , and a basis for each of them.
Exercise 1, answer
I Given any (x , y , z), (x0, y0, z0) ∈ R3:
f [(x , y , z) + (x0, y0, z0)] = f (x + x0, y + y0, z + z0) =
= (−(x + x0) + k(z + z0), y + y0+ k(z + z0), z + z0+ y + y0) =
= (−x + kz − x0+ kz0, y + kz + y0+ kz0, z + y + z0+ y0) =
= (−x + kz, y + kz, z + y ) + (−x0+ kz0, y0+ kz0, z0+ y0) =
= f (x , y , z) + f (x0, y0, z0)
I Given any λ ∈ R, (x, y , z) ∈ R3: f [λ(x , y , z)] =f (λx , λy , λz) =
=(−λx + kλz, λy + kλz, λz + λy ) =
=(λ(−x + kz), λ(y + kz), λ(z + y )) =
=λ(−x + kz, y + kz, z + y ) =
=λf (x , y , z) Thus f is linear.
Exercise 1, answer (cont.)
I
Imf =hf (1, 0, 0), f (0, 1, 0), f (0, 0, 1)i =
=h(−1, 0, 0), (0, 1, 1), (k, k, 1)i =
={(−a + kc, b + kc, b + c) | a, b, c ∈ R}
Since
−1 0 0
0 1 1
k k 1
= −(1 − k), it follows that
dim(Imf ) = 3 ⇔ k 6= 1
while, if k = 1, then dim(Imf ) = 2. Consequently, if k 6= 1, then ((−1, 0, 0), (0, 1, 1), (k, k, 1)) is a basis of Imf .
If k = 1, then ((−1, 0, 0), (0, 1, 1)) is a basis of Imf .
Exercise 1, answer (cont.)
I By the above, if k 6= 1 then dim(ker f ) = 0, so ker f = {(0, 0, 0)}
and ∅ is a basis of ker f .
If k = 1, then dim(ker f ) = 1, and ker f is the set of all (x , y , z) ∈ R3satisfying
−1 0 0
0 1 1
x y z
=
0 0 0
that is
−x = 0
y + z = 0 which means
x = 0
z = −y
Thus ker f = {(0, y , −y ) | y ∈ R}, and ((0, 1, −1)) is a basis of ker f .
Exercise 2
Reduce and reduce totally the following matrices in Mat(R, m, n) using Gauss’ algorithm and, if it is possible, find the inverse:
0 3 0 2 1 0 1 0 1
,
0 23 0 0 1 0 5 0 3
,
0 23 0 0
0 1 0 3
5 0 3 1
Exercise 2, answer
0 3 0 | 1 0 0 2 1 0 | 0 1 0 1 0 1 | 0 0 1
→
1 0 1 | 0 0 1 2 1 0 | 0 1 0 0 3 0 | 1 0 0
The matrix
1 0 1 2 1 0 0 3 0
is a reduction of
0 3 0 2 1 0 1 0 1
.
Exercise 2, answer (cont.)
To get a strong reduction with Gauss’ algorithm:
1 0 1 | 0 0 1 2 1 0 | 0 1 0 0 3 0 | 1 0 0
→
1 0 1 | 0 0 1 0 3 0 | 1 0 0 2 1 0 | 0 1 0
→
→
1 0 1 | 0 0 1
0 3 0 | 1 0 0
0 1 −2 | 0 1 −2
→
→
1 0 1 | 0 0 1
0 3 0 | 1 0 0
0 0 −2 | −13 1 −2
→
→
1 0 1 | 0 0 1
0 1 0 | 13 0 0 0 0 1 | 16 −12 1
→
→
1 0 0 | −16 12 0
0 1 0 | 13 0 0
0 0 1 | 16 −12 1
Exercise 2, answer (cont.)
Thus the identity matrix
1 0 0 0 1 0 0 0 1
is a strong reduction of
0 3 0 2 1 0 1 0 1
, and
0 3 0 2 1 0 1 0 1
−1
=
−16 12 0
1
3 0 0
1
6 −12 1
Esercise 2, answer (cont.)
Since the first two rows of the matrix
0 23 0 0 1 0 5 0 3
are proportional, the matrix does not have an inverse.
0 23 0 0 1 0 5 0 3
→
5 0 3 0 1 0 0 23 0
→
5 0 3 0 1 0 0 0 0
The matrix
5 0 3 0 1 0 0 0 0
is a reduction of
0 23 0 0 1 0 5 0 3
. To obtain a strong reduction with Gauss’ algorithm:
5 0 3 0 1 0 0 0 0
→
1 0 35 0 1 0 0 0 0
Exercise 2, answer (cont.)
The matrix
0 23 0 0
0 1 0 3
5 0 3 1
is not a square matrix, so it does not have an inverse.
As for a reduction using Gauss’ algorithm:
0 23 0 0
0 1 0 3
5 0 3 1
→
5 0 3 1
0 1 0 3
0 23 0 0
Esercizio 2, answer (cont.)
To reach a strong reduction:
5 0 3 1
0 1 0 3
0 23 0 0
→
5 0 3 1
0 1 0 3
0 0 0 −2
→
5 0 3 1 0 1 0 3 0 0 0 1
→
→
5 0 3 1
0 1 0 0
0 0 0 1
→
5 0 3 0 0 1 0 0 0 0 0 1
→
1 0 35 0
0 1 0 0
0 0 0 1
Exercise 3
Compute the determinant of the following matrices in Mat(R, m, n)
A =
1 2 3 4 0 6 0 8 9
B =
11
49 0 11233
4 0 6
0 81 9
C = 1 −2
−1 0
D =
1 2 0 0
4 0 6 1
0 8 −1 1 1 1 −1 1
Exercise 3, answer
det A = − 4
2 3 8 9
− 6
1 2 0 8
= −4(18 − 24) − 6 · 8 = 24 − 48 = −24
det B =11 7 2 · 9
1
7 0 163
2 0 3
0 9 1
= 198 7
−2
0 163 9 1
− 3
1
7 0
0 9
=
=198 7
−2
−3 169
− 31 79
= 198 7
27 8 −27
7
=198 7
−27 56
=
= −2673 196 det C = − 2
Exercise 3, answer (cont.)
det D =
0 6 1
8 −1 1 1 −1 1
− 2
4 6 1
0 −1 1 1 −1 1
= 6 − 8 + 1 − 48 − 2(−4 + 6 + 1 + 4) =
= − 49 − 2 · 7 = −63
Exercise 4
Given the matrix
A =
k − 2 0 6
−1 4 k + 3
0 −2 0
∈ Mat(R, 3, 3)
compute, when it exists, its inverse A−1.
Exercise 4, answer
det A = 2
k − 2 6
−1 k + 3
= 2[(k − 2)(k + 3) + 6] = 2(k2+ k) so
det A = 0 ⇔ k = 0 or k = −1 Thus A−1exists if and only if k 6= 0 and k 6= 1.
The transposition of A is the matrix
k − 2 −1 0
0 4 −2
6 k + 3 0
Exercise 4, answer (cont.)
Thus, for k /∈ {0, 1},
A−1= 2(k21+k)
4 −2
k + 3 0
−
0 −2
6 0
0 4
6 k + 3
−
−1 0
k + 3 0
k − 2 0
6 0
−
k − 2 −1 6 k + 3
−1 0 4 −2
−
k − 2 0
0 −2
k − 2 −1
0 4
=
= 2(k21+k)
2(k + 3) −12 −24
0 0 −k2− k
2 2(k − 2) 4(k − 2)
=
=
k+3
k2+k −k26+k −k212+k
0 0 −12
1 k2+k
k−2 k2+k
2k−4 k2+k
Exercise 5
Solve the linear system
x −y +z = 0 2x −y −2z = 0
Exercise 5, answer
As
1 −1 2 −1
= −1 + 2 = 1 6= 0
the solution depends on a free variable. which can be taken to be z:
x − y = −z 2x − y = 2z whence
x =
−z −1 2z −1
= z + 2z = 3z
y =
1 −z 2 2z
= 2z + 2z = 4z
Exercise 6
Solve the linear system
x −2y +z = 0
y +z +t = 0
Exercise 6, answer
The system is reduced, and it can be solved with respect to the free variables z and t. From the second equation:
y = −z − t Substituting this value in the first equation
x = 2y − z = −2z − 2t − z = −3z − 2t Thus:
x = −3z − 2t y = −z − t is the solution of the system
Exercise 7
Solve the linear system
3x −2y = 1
x +y +z = 3
2x −3y −z = −2
Exercise 7, answer
As the first equation is the sum of the other two, the system is equivalent to any pair of equations (since they are linearly independent), for instance
x + y + z = 3 3x − 2y = 1
This can be solved by taking x as free variable. The second equation yields
y = 3 2x −1
2 substituting this value in the first equation,
z = −x −3 2x +1
2 + 3 = −5 2x +7
2
Thus
y = 32x − 12 z = −52x + 72 is the solution of the system.
Exercise 8
Solve the linear system
2x +2y = 4
x +y = 1
Answer. Since the matrix of coefficients2 2 1 1
has rank 1, while the complete matix2 2 | 4
1 1 | 1
has rank 2, the system has no solution.
