MATH 106
Final Examination, Fall Semester, 2004. Solutions
x + ky = 1
(1) (a) Determine for what values of k ∈ R the linear system
has
kx + y = 1
no solution, unique solution and infinitely many solutions. When the system has
solution(s), write down the (general) solution.
Solution: By performing row-operations in the augmented matrix of
..
1 k . 1 −kr1 +r2 →r2
the given linear system we have succesively
−→
.
k 1 .. 1
..
1
k
.
1
. If k = 1 the initial linear system is equivalent with
.
0 1 − k 2 .. 1 − k
x + y = 1 and it has infinitely solutions x = 1 − t, y = t, t ∈ R,
in this case. If
..
1
−1
.
1
k = −1 the last matrix in the above row-reduction process is
..
0
0 . 2
and it shows that, in this case, the linear system has no solutions. Finally, if
k ∈ R\{−1,
1}, then the row-reduction
process canbe continued in the following
.. 1
..
..
1
r
2
1
0
.
1
k
.
1
1
k
.
1
1−k2
−kr2 +r1 →r1
1+k
−→
way
−→
.
.
.
0 1 .. 1
0 1 .. 1
0 1 − k 2 .. 1 − k
1+k
1+k
the last matrix in the above row-reduction process shows that the initial linear
system has, in this case, the unique solution x = y =
1
(b) For which values of x ∈ R, is the matrix A =
1
1
.
1+k
1 x
x x
invertible ?
x x x
Solution:
The given
matrix A is invertible if and only if det(A) 6= 0. But det(A) =
1
0 ⇔ 1
x
1 x x x = 0 ⇔ x2 + x2 + x2 − x3 − x2 − x = 0 ⇔ 2x2 − x3 − x = 0 ⇔
x x
x(x − 2x + 1) = 0 ⇔ x(x − 1)2 = 0 ⇔ x ∈ {0, 1}. Therefore A is invertible if
2
and only if x ∈ R\{0,
1}.
2 −1
(c) Let A =
.
3
1
(i) Find elementary matrices E1 , E2 , E3 , E4 such that E1 E2 E3 E4 A = I2 ;
(ii) Write A as a product of elementary matrices.
Solution: By performing row-reduction operations in the given matrix we have succesievly:
1
1
1
1
2
1
1
2 −1 2 r1 1 − 2 −3r1 +r2 →r2 1 − 2 5 r2 1 − 2 2 r2 +r1 →r1 1 0
−→
−→
−→
−→
.
5
3
1
3
1
0
0
1
0
1
2
Thus,
the
elementary
can be obtained in the following way:
required
matrices
1
1 0 2 r1
−→
0 1
1
2
0
1 0 −3r1 +r2 →r2 1 0
−→
=: E4
=: E3
0 1
0 1
−3 1
0 1
1 0
1
−→
2
r
5 2
0
1 0
=: E2
0 25
0 1
1
r +r1 →r1
2 2
−→
1
1
2
0 1
=: E1
The inverses of E4 , E3 , E2 , E1 are respectively E4−1
1 0
1
−1
, E2 =
3 1
0
5
2
0
1
−1
, E1 =
0
b c
=
d e f
a 1
−1 −1 −1 −1
and obviously A = E4 E3 E2 E1 .
b
d e
, B =
commute if and only if
0 c
0 f
.
Solution: We have successively: AB
E3−1 =
a
(2) (a) Prove that the matrices A =
b
e
− 12
2 0
=
,
0 1
BA
⇔
=
a b d e
0 c
0 f
=
e a b
da db + ec
ad ae + bf
⇔ ae+bf = db+ec ⇔
=
⇔
0
fc
0
cf
0 c
0 f
d
b
bd − ae = bf − ce ⇔ e
a b c
=
d e f
.
(b) If u, v ∈ Rn are such that ||u + v|| = 1 and ||u − v|| = 5, find u · v.
Solution: ||u + v|| = 1 ⇒ ||u + v||2 = 1 ⇔ u2 + 2u · v + v 2 = 1 ||u − v|| = 5 ⇒
||u − v||2 = 25 ⇔ u2 − 2u · v + v 2 = 25. By subtracting the obtained relations one
can get that 4u · v = −24 or, equivalently u · v = −6.
(c) Decide whether the following mappings are linear and justify the answer in each
case
(i) T1 : R2 → R, T1 (x, y) = x + y;
(ii) T2 : R2 → R, T2 (x, y) = |x| + |y|.
Solution: T1 is linear because T1 ((x, y) + (x0 , y 0 )) = T1 (x + x0 , y + y 0 ) = (x + x) +
(y + y 0 ) = (x + y) + (x0 + y 0 ) = T1 (x, y) + T1 (x0 , y 0 ) for all (x, y), (x0 , y 0 ) ∈ R2 and
also T1 (k(x, y)) = T1 (kx, ky) = kx + ky = k(x + y) = kT1 (x, y) for all k ∈ R and
all (x, y) ∈ R2 .
2
T2 is not linear because T2 (−1, −1) = | − 1| + | − 1| = 2 6= −2 = −T2 (1, 1).
x +
y +
z=1
(d) Solve the following linear system −x + 2y + 3z = 1 by Cramer’s rule.
Solution: The determinant
x + 4y + 9z = 1
of the coefficient
matrix A of the given linear system
1 1 1 1 1 1 is det(A) = −1 2 3 = 0 3 4
0 3 8
1 4 9 the system
can be
indeed
solved
1 1 1 1 1 1
x=
1
12
y=
1
12
z=
1
12
= 24 − 12 = 12 6= 0, which means that
by
the Cramer’s rule and
1 =
1 2 3 12 0 1 2 = 16 ;
0 3 8 1 4 9 1 1 1 1 1 1 1 =
−1 1 3 12 0 2 4 = 34
0 0 8 1 1 9 1 1 1 1 1 1 1 =
−1 2 1 12 0 3 2 = − 21 .
1 4 1
0 3 0 (3) (a) Decide whether the following subsets
(i) A = {(a, b, c, d) ∈ R4 | a − b = 2},
(ii) B = {(a, b, c, d) ∈ R4 | c = a + 2b and d = a − 3b},
are subspaces of R4 and justify the answer in each case.
Solution: A is not a subspace of R4 because 0 6∈ R4 , while B is a subspace of R4
being the solution set of a linear system.
(4) Find a basis of the subspace W = Span{v1 , v2 , v3 , v4 , v5 } of R3 , where v1 =
(1, 0, 1), v2 = (0, 1, 1), v3 = (1, 1, 2), v4 = (1, 2, 1), v5 = (−1, 1, −2).
Solution: We will find a solution of the given subspace of R3 by performing
row-reduction
operations
in the following matrix:
1 0 1 1 −1
0 1 1 2
1
1 1 2 1 −2
1 0 1
0 1 1
−r1 +r3 →r3
−→
1 −1
2
1
0 0 0 −2 −2
1 0 1 1 −1
0 1 1 2
1
0 1 1 0 −1
1 0 1 1 −1
0 1 1 2
1
−→
3
−→
− 12 r3
−r2 +r3 →r3
0 0 0 1
1
−2r3 +r2 →r2
−→
1 0 1 1 −1
+r1 →r1
0 1 1 0 −1 −r3−→
0 0 0 1
basis of W .
(5) Given A =
1
1 0 1 0 −2
0 1 1 0 −1 , which shows that {v1 , v2 , v4 } is a
0 0 0 1
1
1
1 4 1 2
1 2 1 1
0
, find:
0 0 1 2
1 −1 0 0 2
0
2
1 6 0 1
(a) a basis and the dimension of the row space of A;
(b) a basis and the dimension of the column space of A;
(c) a basis and the dimension of the null space of A.
Solution: By performing row-reduction operations in the given matrix we have successively:
1
0
1 4 1 2
1 2 1 1
−r + r → r
1
4
4
0
→
0
0
1
2
−2r1 + r5 → r5
1 −1 0 0 2
2
1 6 0 1
1 1 4
0 1 2
1
1
2r2 + r4 → r4
→
0 0 0
r 2 + r5 → r5
0 0 0
1
1
−r3 +r2 →r2
→
−r3 +r1 →r1
0 0 0 1
0 0 0 0
0 0 0 0
2
0
0
4
1
1
2
1
0
0
0
1
0 −2 −4 −1
2
1
2r + r → r
2
4
4
2
r2 + r 5 → r5
0
→
0 −1 −2 −2 −3
−r + r → r
3
4
4
1 1 4 1 2
0 1 2 1 1
−r + r → r
3
2
2
→
→
0
0
0
1
2
r3 + r5 → r5
−r3 + r1 → r1
0 0 0 0 0
2
2
0
1 1 4 0
0 1 2 0 −1
1
2
1
0 0 0 −1 −2
1
0
0 0 0 0 0
−r2 +r1 →r1
−→
1
1 0 2 0
0 1 2 0 −1
0 0 0 1
0 0 0 0
0 0 0 0
4
.
2
0
0
Thus a basis for the row space is {(1, 0, 2, 0, 1), (0,
0,−1), (0,0, 0,1, 2)} so that its
1, 2,
1
1
0 1
,
dimension is 3. Also a basis for the column space is 0 , 0
1 −1
1
2
1
1
1 , its dimension
0
0
being consequently 3.
The null space of the given matrix is the solution set of the corresponding homgeneous linear system of the given matrix, which is in its turn equivalent with the linear
system that corresponds to the last matrix in the above row-reduction process. Thus the
x1 = −2t − s
x2 = −2t + s
, t, s ∈ R which menas that the required null space
required null space is x3 = t
x4 = −2s
x5 = s
x1
−2t − s
−2
−1
1
x2
) ( −2
( −2t + s
is x3 =
t , t, s ∈ R, t, s ∈ R = t 1 + s 0
−2
x4
0
−2s
1
0
x5
s
−2 −1
−1
−2
−2 1
( −2 1 )
Span 1 , 0 . Because the vectors 1 , 0 are
0 −2
0 −2
0
1
)
, t, s ∈ R, t, s ∈ R =
obviously liniarly inde-
1
0
pendent, it follows that they form a basis of the required null space, its dimension being
consequently 2.
(1) Show that the mapping M22 × M22 → R, (U, V ) 7→ hU, V i where
hU, V i = u1 v1 + u2 v3 + u3 v2 + u4 v4 ,
whenever U =
u1
u2
u3
u4
,V =
v1
v2
v3
v4
, is not an inner product on M22 .
The given
mapping is indeed not an inner product
Solution:
*
+
0 1 0 1
,
= 0 · 0 + 1 · (−1) + (−1) · 1 + 0 · 0 = −2 < 0.
−1 0
−1 0
5
since
(2)
Find the values of a, b, c ∈ R for which the matrix A is symmetric, where A =
2 a + b + c 4b − 3c
5
2
1
2a + c
.
4
6
a+b+c=5
Solution: The required values are solutions of the linear system
4b − 3c = 2
2a + c = 4
that can be solved by performing
row-reduction
operations
in the corresponding
au
..
..
1 1
1 . 5
1 1 1 . 5
41 r2
−2r1 +r3 →r3
.
0 4 −3 ... 2 −→
.
−→
gumented matrix, namely:
0 4 −3 . 2
..
..
0 −2 −1 . −6
2 0 1 . 4
..
..
.. 9
7
1
1
1
.
5
1
1
1
.
5
1
0
. 2
4
2
1
−
r
r
2
3
2r
+r
→r
.
.
3
3
4
5
3 .
1 2 −→
1 −→
0 1 − 3 ..
0 1 − 3 ... 1 .
−→
0 1 −4 .
4
2
2
4
2
..
..
5 ..
0 −2 −1 . −6
0 0 − 2 . −5
0 0 1 . 2
The corresponding linear system of the last matrix in the above row-reduction process,
which is equivalent with the initial one, has the unique solution a = 1, b = c = 2.
(3) Let u1 = (0, −1, −1), u2 = (−1, 0, 1), u3 = (1, 1, 1) ∈ R3 .
(a) Show that {u1 , u2 , u3 } is a basis of R3 ;
(b) Apply the Gram-Schmidt process to convert the basis {u1 , u2 , u3 } into an orthogonal basis of R3 and find its corresponding orthonormal basis.
Solution: v1 = u1 = (0, −1, −1);
hu2 ,v1 i
1 1
v = (−1, 0, 1) − −1
2 (0, −1, −1) = − 1, − 2 , 2
||v1 ||2 1
1 1
−1
3 ,v1 i
3 ,v2 i
v3 = u3 −projSpan{v1 ,v2 } u3 = u3 − hu
v − hu
v = (1, 1, 1)− −2
2 (0, −1, −1)− 3/2 −1, − 2 , 2 =
||v1 ||2 1
||v2 ||2 2
(1, 0, 0) + − 2/3, − 34 , 34 = 31 , − 13 , 31 .
Thus v1 = (0, −1, −1), v2 = − 1, − 12 , 12 , v3 = 13 , − 13 , 13 form an orthogonal basis of R3 . The
v2 = u2 − projSpan{v1 } u2 = u2 −
norms of these vectors are:
||v1 || =
√
r
2, ||v2 || =
√
3
6
1
=
, ||v3 || = √
2
2
3
such that the corresponding orthonormal basis is
q1 =
v1
1
1 v2
2
1 1 v3
1
1 1 = 0, − √ , − √ , q2 =
= − √ , − √ , √ , q3 =
= √ , −√ , √ .
||v1 ||
||v2 ||
||v3 ||
2
2
6
6 6
3
3 3
6
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