Math 214 Homework 2 - Linear Algebra: Matrix Rank, Transformations

This document presents the solutions to Homework 2 of the Math 214 course from Fall 2018, focusing on linear algebra concepts. The solutions address three main problems. The first problem explores the rank of a 3x3 matrix, determining the values of a scalar 'a' for which the matrix has a rank less than 3 and finding a vector b for which the equation A*x = b has no solutions. The second problem delves into linear transformations in R2, identifying orthogonal projections, reflections, and rotations based on their transformation properties. The third problem examines matrix multiplication rules and rank, determining the dimensions of matrices to satisfy matrix product definitions and calculating the rank of the resulting matrices. Each solution includes detailed steps and justifications to provide a comprehensive understanding of the concepts.

Q1)
A = [ 1 a 2
2 1−a 1
−1 1 3 ]
a) For a rank less than 3 then |A| = 0
|A| = 1(3(1-a)-1)-a(6+1)+2(2+1-a)
= 3-2a-1 -6a-a + 6 -2a
|A| = 8 -12a
Therefore, 8 -12a = 0
a = 8/12
= a = 2/3
b) The system will have no solution
A⃗ x=⃗ b
Therefore ⃗
b ≠ 0
For no solution, Rank (A) ≠ Rank( AB)
a = 2/3, ⃗ b= [1
4
3 ]
any value for b but its condition is fulfilled that Rank (A) ≠ Rank ( AB )
since 2 ≠ 5
Q2)
a) Line passing through (0,0) and )100, 75) is given by 3y = 4x 0r y = 4/3(x)
if any of L2 or L3 were projected then they has to be projection onto this line, because image lie
only on this line passing through origin
we find the orthogonal projection of (75, 100) on this line
= [ ( 75,100 )∗ (100,75 )]
( 100,75 )∗(100,75) [ 100,75 ]
= [ ( 75100 ) + ( 7500 ) ]
( 10000 ) +(5625) [100,75 ] ≠[100,75 ]
Which is not (100,75), therefore none of L2 or L3 is a projection

b) L2(75,100) = (100,75) and L2(100, 75) = (75, 100) this gives us that L2 must be a reflection about
the line passing through the origin and the mid point of the line segment joining (75, 100) with
(100, 75)
c) L1 is projection , which means it must be projection onto the line containing (0,0) and (96, 72),
because the image of projection in |R2 is one dimensional if L3 is rotation and L3(75, 100) = (100,
75), then it must be rotation by tan−1
( 100
75 ) - tan−1 ( 75
100 ) clockwise.
tan−1
( 4
3 )−tan−1 3
4 =tan−1
( 4
3 − 3
4
1+ 4
3 + 3
4 )
= tan−1
( 16− 9
12
2 )=tan−1 ( 7
24 ) clockwise
So L3( 1
0 ) =( a
b )
L3( 1
0 ) =
[ cosθ −sinθ
sinθ cosθ ][ 1
0 ] =
[ cosθ
sinθ ]
Where θ=3600−tan−1
( 7
24 )
L3( 1
0 ) =¿
L1L3(1
0 )= L1¿
= ¿ ¿
= ¿
= 72
4032 ( 96 ,72 )
Q3)

(a) We have
A = [1
1
1 ]3*1 =
[1
2
3
4 ]4*1
For product to be defined according to matris product rule , A must be a 4 * 3 matrix, so m = 4 and n = 3
Lets take matrix A4*3 to which the above product is true
[1 0 0
1 1 0
1 1 1
1 1 2 ]4*3
[1
1
1 ]3*1 =
[1
2
3
4 ]4*1
In order to justify the statement we have to find the rank of the matrix A4*3 , lets A in echelon form As, A
is 4*3 matrix so the rank of A can be either 3 or less than 3, Asper proportion of rank
R2 → R2 – R1, R3 → R3 – R1 and R4 → R4 – R1
[1 0 0
0 1 0
0 1 1
0 1 2 ]
R3 → R3 – R2
[1 0 0
0 1 0
0 0 1
0 1 2 ]
R4 → R4 – R2
[1 0 0
0 1 0
0 0 1
0 0 2 ]
R4 → R4 – R3
[1 0 0
0 1 0
0 0 1
0 1 1 ]

In the above echelon for we found a 3 *3 identity matrix, therefore the rank of the matrix A is 3,
therefore we can say that the statement of (a) is True
(b)
Am*n =
[1
2
3
4 ]4*1 = [1
1
1 ]3*1
For the above product to be classified according to the product rule, A must be a 3 * 4 matrix, so m = 3
and n = 4
Lets take matrix A3*4 for which the above product is true
[1 0
0 1
2
0 0
0 0
1 0 0 0 ]3*4
[1
2
3
4 ]4*1 = [1
1
1 ]3*1
In order to justify the statement , we have to find the rank of te matrix A3*4. Now, rank of matrix A can
be equal 3 or less than in this case, because r(A) = min(m, n).
Lets convert the matrix A in echelon form
R2 → 2R4 and R3 → R3 – R1
[1 0
0 1
0 0
0 0
0 0 0 0 ]
We know that rank of a matrix is equal to the number of non zerorows in its row echelon form
Therefore, rank of A is 2, hence the statement is true