University Maths Assignment: Matrices, Calculus and Equation Solving

This document presents a complete solution to a Maths assignment covering a range of topics. The assignment begins with matrix operations, including finding a matrix C given a matrix equation, calculating the inverse of a matrix, and evaluating determinants. It then moves on to solving a system of linear equations using Gaussian elimination. The assignment also involves a practical problem involving trigonometry to calculate the height of two buildings using angles of elevation and depression. Furthermore, the solution includes finding stationary points of a function and solving a related calculus problem, including integration. Finally, it covers a volume calculation problem involving a tank and the time required to fill it. This comprehensive solution provides detailed steps and explanations for each problem, offering a valuable resource for students studying mathematics.

Running head: MATHS
MATHS
Name of the Student:
Name of the University:
Author note

1
MATHS
Answer to question number
Part i:
Given, A = [−1 −1
−2 1 ]
Now let C = [ a b
c d ]
Therefore, CA = [ −a−c −b−d
−2 a+c −2 b+ d ]
Since, CA = B. Therefore, [ −a−c −b−d
−2 a+c −2 b+ d ]= [ 2 −1
−2 2 ]
Therefore,
- a – c = 2
- 2a + c = - 2
- b – d = - 1
- 2b + d = 2
Therefore, a = 0, b = -1/3, c = -2, d = 4/3
Hence, C = [ 0 −1/3
−2 4 /3 ]
Part ii:
Provided matrix: [ 1 2 −1
3 5 −1
−2 −1 −2 ]

2
MATHS
Augment with identity matrix: [ 1 2 −1 1 0 0
3 5 −1 0 1 0
−2 −1 −2 0 0 1 ]
=
[1 0 0 11
2
−5
3
−3
2
0 1 0 −4 2 1
0 0 1 −7
2
3
2
1
2 ]
= right of augmented matrix is the inverse matrix:
[ 11
2
−5
3
−3
2
−4 2 1
−7
2
3
2
1
2 ]
Part iii:
Determinant of
[ 0 1 2 0
1 0 3 9
−1 −3 1 2
0 1 −1 −2 ]
=
[ 1 0 3 9
1 −3 4 11
0 0 10
3
11
3
0 0 0 13
10 ]
= - 13
Answer to question number 2
The given equations:

3
MATHS
a + 2b – 3c = - 2
3a – b – 2c = 1
2a + 3b – 5c = - 3
The Gaussian form: [1 2 −3 −2
3 −1 −2 1
2 3 −5 −3 ]
= [ 1 2 −3 −2
3 −1 −2 1
2 3 −5 −3 ]
= [ 1 2 −3 −2
0 1 −1 −1
0 −1 1 1 ]
= [ 1 2 −1 −2
0 1 0 −1
0 −1 0 1 ]
Therefore, a + 2b – c = -2, and a = c.
Therefore, b = c and b = c – 1
Answer to question number 3
Given that, the distance in between the 2 buildings = 18 m.
The height of the building from base to eye level = 18.tan (53)
The height of the building from eye level to level of elevation = 18.tan (44)
Therefore, the total height of the building is = 41.26920 m.

4
MATHS
Answer to question number 4
Given that, y = (x - 1)2 (x - 2)
dy/dx = (x - 1)2 + (x – 2). 2. (x - 1)
= 3x2 - 8x + 5
Therefore, 3x2 - 8x + 5 = 0
3x2 - 3x – 5x + 5 = 0
3x (x – 1) – 5(x – 1) = 0
Or, x = 1, 5/3

5
MATHS
Therefore, the stationary points are 1, 5/3.
Answer to question number 5
Height of the tank = 20 feet
Width = 5 feet
Depth = 2 feet
Therefore, r = 2.5 feet.

6
MATHS
Volume of the triangle = pi. r2.h/3 = 2x (22/7) x2.5x2.5x2/3 = 26.19 cube feet
Volume of tank = pi. r2.h = (22/7) x2.5x18 = 45 cube feet
Total volume = 71.19 cube feet
Time required to fill the tank fully – 8.9 mins approximately.
Answer to question number 6
Part i
Lim x -> -1 (x3 + 1) /(x + 1) = Lim x -> -1 (x2 – x + 1)
Therefore, the value = (-1)2 – (-1) + 1 = 3
Part ii
Given, ∫ 3 xdx
( x2+ 1 )
7
= 3 ∫ xdx
( x2+ 1 )
7
Let u = x2+ 1
Or, 3∫ xdx
( x2+ 1 ) 7 = 3∫ du
( u )7
= - 3/2 (1/6u6)
= - 1/4(x2 + 1)6 + C
Part iii

7
MATHS
Area of the region: ∫
0
1
( x2 + √ x+1 ) dx
= 1/3 + (4 √2 ¿¿/3 – 2/3
= (4 √ 2 – 1)/3