Engineering Mathematics Linear Algebra Solved Problems

This article contains solved problems on Engineering Mathematics Linear Algebra including topics like RREF, vector space, linear transformation, diagonalizability, and more. It also includes information on how to determine if a set of vectors is linearly independent and how to find an orthonormal basis for a vector space.

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 )

So, S is linearly independent
The set W is given for 3 elements in the vector space.
W = { x1 , x2 , x3 }
There are a set of simultaneous equations resulting from the vector space,
x1+ x2=0
x1+ x3 =0
Hence, the vector space, W, is linearly independent and is of 3-dimension subspace.
PROBLEM 4
Let A be the following 3x3 matrix:
A=
( 1 0 0
−3 4 −6
−1 1 −1 )
Test A for diagonalizability. If A is diagonalizable, find a 3x3 invertible matrix B and a 3x3
diagonal matrix D such that B−1 AB =D.
SOLUTION
A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is
equal to the multiplicity of the eigenvalue. As such, the matrices with distinct eigenvalues
(multiplicity=1) are considered diagonalizable. The basis of R3 consist of eigenvectors of A.
λI − A=
( λ 0 0
0 λ 0
0 0 λ )−
( 1 0 0
−3 4 −6
−1 1 −1 )= (−λ+1 0 0
−3 −λ+ 4 −6
−1 1 −λ−1 )
Finding the determinant of the resulting matrix using the cofactor expansion,
det ( λI −A )= ( (−λ+4 ) (−λ−1 ) +6 ) − ( 3 (−λ−1 ) +6 ) + (−3−λ+ 4 ) ¿
¿−6 λ+ (− λ−1 ) (−λ+ 1 ) (−λ +4 ) +6=0
Find the roots of the quadratic function resulting,
λ1=1 , λ2=1 , λ3=2 … . eigenvalues

Using the eigenvalues to find the multiplicity,
(− λ+1 0 0
−3 −λ +4 −6
−1 1 − λ−1 )… .replacing the value here .
Using λ1=1,
(−1+1 0 0
−3 −1+4 −6
−1 −1 −1−1 )= [ 0 0 0
−3 −3 −6
−1 1 −2 ]
Now perform the row operations to obtain the reduced row echelon form of the matrix,
[ 0 0 0
−3 −3 −6
−1 1 −2 ] [1 −1 2
0 0 0
0 0 0 ]
Solving further,
[ 1 −1 2
0 0 0
0 0 0 ] [ v1
v2
v3 ]= [ 0
0
0 ]
[−2 s+ t
t
s ]=
[1
1
0 ]t + [−2
0
1 ] s
Using λ2=2,
(−2+1 0 0
−3 −2+4 −6
−1 −1 −2−1 )=
[−1 0 0
−3 −2 −6
−1 1 −3 ]
Now perform the row operations to obtain the reduced row echelon form of the matrix,
[−1 0 0
−3 −2 −6
−1 1 −3 ] [1 0 0
0 1 −3
0 0 0 ]

[1 0 0
0 1 −3
0 0 0 ] [v1
v2
v3 ]=
[0
0
0 ]
[ 0
3 t
t ]=
[0
3
1 ]t
B−1 AB=D
Matrix A is diagonalizable as the matrix D is given as,
D= (2 0 0
0 1 0
0 0 1 )
B= (0 1 −2
3 1 0
1 0 1 )
B−1= (−1 1 −2
3 −2 6
1 −1 3 )
B−1 AB=
(−1 1 −2
3 −2 6
1 −1 3 )( 1 0 0
−3 4 −6
−1 1 −1 )(0 1 −2
3 1 0
1 0 1 )
B−1 AB=D=
(2 0 0
0 1 0
0 0 1 )
The matrix is diagonalizable
PROBLEM 5
Let L be the transformation from P3 to P3 defined by
L ( f ( x ) )=f ' ( x ) +f ( 1 )
Show that L is linear

Write the matrix of L with respect to the standard bases on P3
Write the matrix of L with respect to the basis {1+ x , 2+ x , 1+ x2 }

PROBLEM 6
Suppose
det ( a b c
d e f
g h i )=8
det ( A ) =a ( ei−fh ) −b ( di−fg ) + c ( dh−eg ) =8
Find the determinants of the following matrices:

(a) (2 a g d
2 b h e
2c i f ) det ( a )=2a ( hf −ei )−g ( 2 bf −2ce ) +d ( 2 bi−2 ch )
(b) ( −a −b −c
d e f
2 g+d 2h+ e 2i+ f ) det ( b )=−a ¿
PROBLEM 7
Let T:R3 → R2be the linear transformation
T
( x1
x2
x3
) =
( 2 x1 + x2−x3
x1 +3 x2 −4 x3 )
Find matrix of T
Does T have any eigenvectors? Why or why not?
Is T one-to-one? Onto?
SOLUTION
Let T be L,

PROBLEM 8
(i) Determine whether the set of vectors {1+x + x3 , 2+ x + x2 +2 x3 , 2−x +2 x2 } in P3 is line
independent. Justify your answer.
(ii) Does this set form a basis? Why or why not?
Each of the given sets has three polynomials. Since dimension of P3 is 3, it is sufficient
to test linear independency for the sets to be a basis for P2 (Note that dimension of a vector space
is the total number of basis vectors present in a basis set of that vector space). Test procedure is
as follows: Let the polynomials in each set be { p 1( x ), p 2( x), p 3( x )}. Solve
ap 1( x )+bp 2( x)+ cp 3(x)=0 for constants a, b and c. If these constants are all zero, then the set
is a linearly independent. Else, it is not. According to this, we obtain that the sets in the equation
listed are the basis for P3.
PROBLEM 9
Let {u1 , u2 ,u3 , u4 } be an orthogonal basis for a vector space V. If x=c1 u1+ C2 u2+ c3 u3 + c4 u4 is a
vector orthogonal to u1∧u2with the properties that ||x ||=10and x .u3 =6, then what are the
possible values of c1 , c2 , c3∧c4?
SOLUTION
Let u 1 ,u 2 , u3 , u 4 be the three vectors given, and U =span ({u 1 ,u 2 , u3 ,u 4 })⊆ R4 . To find
an orthonormal basis for U we would like to use the Gram-Schmidt Process, but the process has to start
from a basis for U. So, our first question is: is {u 1 , u2 , u 3 ,u 4 } a basis for U?
By definition the vectors span U, so we need only check that they are linearly independent: it turns out
they are,
x=c1 u1+ C2 u2+c3 u3 +c4 u4
10=c1 u1 +C2 u2 +c3 (6)+c4 u4