Engineering Mathematics Linear Algebra Solved Problems

Engineering Mathematics
Linear Algebra
Student Name
Student ID Number
Hashemite Kingdom of Jordan
Institutional Affiliation
Date of Submission

PROBLEM 1
(i) A matrix in RREF has the same solution set as the original matrix.
a. False
(ii) If V is a vector space, then every basis of V has the same size
a. True
(iii) If v , w are in R3, then the span (v, w) is a subspace of R3
(iv) The dimension of a vector space is the number of elements in the vector space
a. True
(v) If the matrix product AB is defined then the number of rows of A is equal to the
number of columns of B.
a. True
(vi) An eigenvalue can have more than one eigenvector
a. True
(vii) The multiplicity of an eigenvalue cannot be greater than the dimension of the
corresponding eigenspace.
a. False
(viii) The eigenspaces of two distinct eigenvalues are orthogonal to each other
a. True
(ix) If T : R5 → R4 is a linear transformation, then T is not 1-1.
a. False
(x) There exists v1 , v2∧v3 in R4 which are linearly independent
a. True
PROBLEM 2
Consider the matrices below,
A=(2 0 −1
4 2 1 ), B=
( 1 1
−1 5
0 2 ),C= ( 1 −3 2
−2 1 2
5 0 3 )∧D=(2 −1
3 4 )
Compute the following:
(i) BAD

BA= ( 1 1
−1 5
0 2 ) (2 0 −1
4 2 1 )=
( 6 2 0
18 10 6
8 4 2 )
BAD=
( 6 2 0
18 10 6
8 4 2 ) ( 2 −1
3 4 )
BAD=not possible
Matrix dimension mismatch
(ii) DB
DB= ( 2 −1
3 4 ) ( 1 1
−1 5
0 2 )
Matrix dimension mismatch
(iii) CT C+ BT
C=
( 1 −3 2
−2 1 2
5 0 3 )... .Transposing
CT= ( 1 −2 5
−3 1 0
2 2 3 )
CT C=
( 1 −2 5
−3 1 0
2 2 3 )( 1 −3 2
−2 1 2
5 0 3 )= ( 30 −5 13
−5 10 −4
13 −4 17 )
B= ( 1 1
−1 5
0 2 ) , BT = ( 1 −1 0
1 5 2 )
CT C+ BT =
( 30 −5 13
−5 10 −4
13 −4 17 ) + ( 1 −1 0
1 5 2 )
Dimension Error
There is a matrix dimension mismatch
(iv) CBA
CBA= ( 1 −3 2
−2 1 2
5 0 3 )( 1 1
−1 5
0 2 ) (2 0 −1
4 2 1 )

CBA=(3 x 3)(3 x 2)(2 x 3)
First work out BA,
BA= ( 1 1
−1 5
0 2 ) (2 0 −1
4 2 1 )=
( 6 2 0
18 10 6
8 4 2 )
CBA= ( 1 −3 2
−2 1 2
5 0 3 )( 6 2 0
18 10 6
8 4 2 )
CBA= (−32 −20 −14
22 14 10
54 22 6 )
PROBLEM 3
(i) Let W be a subspace of R3. Show that W ⊥= { v ∈ R3
| v . w=0 for every w∈ W } is also
a subspace.
(ii) Find the dimension of the subspace, W =
{(x1
x2
x3
)|x1 + x2=0∧x1 + x3=0 }of R3
SOLUTION
W ⊥= { v ∈ R3
| v . w=0 for every w∈ W }
Let v1 , v2 ∈W ⊥. Then {v1 , w }=0 for all w ∈ W
So, for any α , β ∈ R3
{ α v1 + β v2 , w }=a { v1 , w } + β { v2 , w }=0
For all, w ∈ W . Hence, α v1 + β v2 ∈ W ⊥
¿
{( x1
x2
x3
)| x1 + x2=0∧x1+ x3=0}of R3
R3=Span ( W )