# algebre exam blanc .pdf

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Prop´e blanc d’alg`ebre lin´eaire I 1
R`
egles applicables :
- Le test 2 dure 3 heures, aucun document n’est autoris´e
- Il y a 20 exercices, 34 questions.
- Chaque r´eponse correcte rapporte 1 point.
- Pas de r´eponse ou une r´eponse fausse ne fait pas perdre de point.
Bon travail !
Exercices :
1. The linear system

 x + 2y + 2z = 4
−x + 3y + 3z = −2

2x + y + z = 0

has
(A)
(B)
(C)
(D)

no solution
a unique solution
a line of solutions
a plane of solutions

2. Let R be the reduced echelon form

1 3
 2 4
3 5

of the matrix

5 7
6 8 
7 9

Then
(A)
(B)
(C)
(D)

R13
R13
R13
R13

= −1
=0
=1
=2

c
1. °EPFL
2002 - Facult´e d’Informatique et Communications (I&amp;C)
2. Toute ressemblance avec un test existant serait purement fortuite.

1

3. Let M be the vector space of 2 × 2 matrices and define
¸
¸
·
·
1 0
1 0
,B=
A=
0 −1
0 0
a) a base for M is given by
¸¾
¸ ·
½·
0 1
1 0
,
(A)
1 0
0 1
½·
¸ ·
¸ ·
¸ ·
¸¾
1 3
2 4
3 1
1 5
(B)
,
,
,
2 4
3 1
4 2
1 −7
¸¾
¸ ·
¸ ·
½
·
1 0
0 0
0 1
,
,
(C) A, B,
0 1
−1 0
0 0
½
·
¸ ·
¸ ¾
0 2
0 2
(D) A, A − B,
,
,
−1 0
1 0
b) If a sub-space of M contains A and B, then it must also contain
(A) all of M
·

¸
1 1
(B) the matrix
0 −1
· ¸
¤
1 £
2 0
(C) the matrix
0
· ¸
¤
1 £
0 2
(D) the matrix
0
4. Let

2
4 −6
2
1
3  , b =  5 .
A= 0
−3 −5 7
−3
Then the solution x = (x1 ,x2 ,x3 ) of Ax = b has
(A)
(B)
(C)
(D)

x1
x1
x1
x1

=0
=1
= −1
=2

2

5. Let {v1 ,v2 ,v3 }, and {w1 ,w2 ,w3 } be two bases of R3 .
(a) Define the linear transformation T : R3 → R3 by T (v1 ) = w2 ,
T (v2 ) = T (v3 ) = w1 + w3 and let A be the matrix associated
with T. Then
(A)
(B)
(C)
(D)

A32
A32
A23
A33

=0
=1
=1
=0

(b) For the same transformation T as above one has:
(A)
(B)
(C)
(D)

Ker(T ) = span[v2 − v3 ]
Ker(T ) = {0}
Ker(T ) = span[v1 ,v2 ,v3 ]
Ker(T ) = span[w1 + w3 ]

(c) Define the linear transformation S : R3 → R3 by S(v1 ) = w1 +
w2 + w3 , S(v2 ) = w2 + w3 , S(v3 ) = w3 . Then the vector v such
that S(v) = w1 is :
(A)
(B)
(C)
(D)

v
v
v
v

= v 2 − v1
= v1 − v2 − 2v3
= v 1 − v2
= v 1 − v2 − v3

(d) The inverse of the matrix B associated with the transformation
S above has :
(A)
(B)
(C)
(D)

(B −1 )21
(B −1 )21
(B −1 )21
(B −1 )21

=1
= −1
=0
=2

3

6. Let

A=

22 4
−6 −14
−3 41 −16 −2
2
4
−6
13
−3 −5
7
−9
−2 0
4
−3
32 14 −7 −11

12
72
−7
−3
−14

 , and B =  −24 −31 4 −16 −27 

 13
29
4
−6
10 

−8 −3 −5
7
−9

Then AB has
(A)
(B)
(C)
(D)

(AB)53
(AB)53
(AB)53
(AB)53

= 45
= 54
= −36
= −54

7. Let

0 1 2
A =  1 0 3 .
4 −3 8
Then A−1 has
(A)
(B)
(C)
(D)

(A−1 )31
(A−1 )31
(A−1 )31
(A−1 )31

= 1/2
= 3/2
= 5/2
= 7/2

8. Given·3 n × n matrices
A,B and C, construct the 2n × 2n matrix
¸
A B
M=
,
C 0n
where 0n is the n × n zero matrix.
(a) Then:
(A)
(B)
(C)
(D)

·

¸
AA + CC AB
=
,
BA
0n
· T
¸
A A + C T C AT B
MT M =
BT A
BT B
· T
¸
A A + C T C BAT
T
M M=
AB T
BB T
· T
¸
A A BT B
MT M =
CT C
0n
MT M

4

(b) M is invertible if and only if
(A)
(B)
(C)
(D)

A and B have rank n
A and C have rank n
B and C have rank n
A, B and C have rank n

(c) Supposing that M is invertible one has :
· −1
¸
A
B −1
(A) M −1 =
,
C −1 0n
· −1
¸
A
In
−1
(B) M =
0n C −1
¸
·
0n
C −1
(C) M −1 =
B −1 −B −1 AC −1
¸
·
A−1
−A−1 BC −1
−1
(D) M =
A−1 BC −1
C −1
9. Define the matrix

2 −4 4 −2
A =  6 −9 7 −3  .
−1 −4 8 0
Calculate the LU decomposition of A (using only the elementay row
operation of adding a multiple of a line to a line below itself). The
resulting matrix L has:
(A)
(B)
(C)
(D)

L32
L32
L32
L32

=1
=0
=2
= −2

5

10. Given that A is a 3 × 3 matrix with eigenvalues −1,1,2, then
(a) the
(A)
(B)
(C)
(D)

rank of A is
1
2
0
3

(b) The determinant of AT A is
(A) 2
(B) 4
(C) 0
(D) 3
(b) The determinant of A + I is
(A)
(B)
(C)
(D)

1
6
0
-1

(b) The determinant of A−1 is
(A) -2
(B) -1
(C) 1
(D) -1/2
11. Let A be an m × n matrix whose columns generate Rm . Then
(a) (A) The system Ax = 0 has a nontrivial solution
(B) The echelon form of A has a pivot in each line.
(C) There exist vectors b in Rm for which the system Ax = b is
not consistent
(D) The echelon form of A has a pivot in each column
(b) (A) The transformation T : Rn → Rm , T (x) = Ax is surjective
(B) The transformation T : Rn → Rm , T (x) = Ax is not surjective
(C) The transformation T : Rn → Rm , T (x) = Ax is injective
(D) The transformation T : Rn → Rm , T (x) = Ax is not injective

6

12. The coordinate of the polynomial p(t) = 3 + t − 6t2 associated with
the first member of the base B = {1 − t2 ,t − t2 ,2 − 2t + t2 } of P2 is
(A)
(B)
(C)
(D)

1
2
7
6

13. Let A be an m × n matrix with linearly independant columns.
(a) When m = n:
(A) KerA = Rn
(B) RowA = Rm
(C) the nullspace of A is empty
(D) ColA is ortogonal to RowA
(b) The solution set of the equation Ax = 0
(A) is empty
(B) is a straight line
(C) has only one element
(D) has dimension m − n
(c) For the following values of m and n the equation Ax = 0 has
non-zero solutions
(A)
(B)
(C)
(D)

no possible values
m=n
n=m+1
n&lt;m

(d) The matrix AT A has the following properties:
(A) it has rank m
(B) it is singular
(C) ColAT A = Rm
(D) it has rank n

7

14. Let ·
¸
·
¸
3 −1
3 −1
A=
, and B =
. Then
2 0
4 −1
(A)
(B)
(C)
(D)

A and B are diagonalizable
A is diagonalizable but B is not
B is diagonalizable but A is not
Neither A nor B is diagonalizable

15. Let ·
¸
¸
·
3 4
3 −1
. Then
, and D =
C=
4 −1
2 0
(A)
(B)
(C)
(D)

C and D are orthogonally diagonalizable
C is orthogonally diagonalizable but D is not
D is orthogonally diagonalizable but C is not
Neither C nor D is orthogonally diagonalizable

16. Let ·
¸
−2 1
A=
1 −2
Then, for all x 6= 0 in R2 , the quadratic form Q(x) = xT Ax is such
that
(A)
(B)
(C)
(D)

Q(x) &gt; 0
Q(x) &gt; 0 (and there exists y 6= 0 such that Q(y) = 0)
Q(x) &lt; 0
Q(x) 6 0 (and there exists y 6= 0 such that Q(y) = 0)

8

17. Let 

0 1 1
A =  0 1 0 .
1 1 2
(a) The result of applying the Gram-Schmidt algorithm to the columns of the matrix

0 1 0
(A) P =  0 0 1  .
1 0 0

0 1 1
(B) P =  0 1 −1  .
1 0 0

0 1 2
(C) P =  0 1 −2  .
1 0 0

0 1 1/2
(D) P =  0 1 −1/2  .
1 0
0

1
(b) The orthogonal projection of the vector  2  . on the sub-space
3
generated by the first two columns of the matrix A is the vector:
 
1

(A) v = 2  .
3

3/2
(B) v =  3/2  .
3
 
1
(C) v =  1  .
3
 
2
(D) v =  2  .
3

9

18. The matrix

5
9
 1
7 

A=
 −3 −5 
1
5

can be factored in the form A = QR, where

5 −1
¸
·
1 1
5 
 ,R = r11 r12
Q= 
0 r22
6  −3 1 
1
3
with
(A)
(B)
(C)
(D)

r12
r12
r12
r12

=2
= −2
= 12
=3

19. Let

The
(A)
(B)
(C)
(D)

 
−1 2
4
A =  2 −3  ,b =  1  .
−1 −3
2
least squares solution x
ˆ = (ˆ
x1 ,ˆ
x2 ) of Ax = b has
x
ˆ1 = 1
x
ˆ1 = 2
x
ˆ1 = 3
x
ˆ1 = 4

20. The non-zero singular values of
·
¸
3 2 2
A=
,
2 3 −2
are
(A)
(B)
(C)
(D)

3, -1, 1
5, 3
25, 9
-5, 5, 3, -3

10