Solved problems on linear algebra and geometry

This article provides solutions to problems on linear algebra and geometry, including finding the rank of matrices, orthogonal projections, reflections, and rotations. It also discusses the product rule for matrices and the importance of matrix echelon form in determining rank.

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