Solutions to CSEC Maths P2 JUNE 2012
Transcript of Solutions to CSEC Maths P2 JUNE 2012
Solutions to CSEC Maths P2 JUNE 2012
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 1a
Calculate the exact value of
3
1
5β
2
3
2 4
5
Numerator: 3 1
5β
2
3
= 16
5β
2
3
=48β10
15
=38
15
Denominator: 2 4
5=
14
5
ππ’πππππ‘ππ
π·ππππππππ‘ππ =
38
15Γ·
14
5
=38
15Γ
5
14
=19
21
Question 1b
Data Given: The table below shows the cost price, selling price and profit or loss as a
Percentage of the cost price.
Cost Price Selling Price Percentage
Profit or Loss
$55.00 $44.00 ______
________ $100.00 25% profit
Required to Copy and Complete the table below inserting the missing values.
Cost Price Selling Price Percentage
Profit or Loss
$55.00 $44.00 20% loss
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
$80.00 $100.00 25% profit
For the 1st item:
Since the selling price is less than the cost price for the first item,
βΉ πΏππ π = πΆππ π‘ πππππ β πππππππ πππππ
= $55 β $44
= $11.00
β΄ ππππππ‘πππ πΏππ π =πΏππ π
πΆππ π‘ πππππ Γ 100
=11
55Γ 100
= 20%
For the 2nd item:
π΄ 25% ππππππ‘ πππππππ π‘βππ‘ π‘βπ π ππππππ πππππ ππ 125% ππ π‘βπ πππ π‘ πππππ.
β΄ πβπ πππ π‘ πππππ =100
125Γ 100
= $80.00
Question 1c part (i)
Data Given: The table below shows some rates of exchange
US $1.00 = EC $2.70
TT $1.00 = EC $0.40
Required to calculate the value of EC $1 in TT $
πΈπΆ $0.40 = ππ π1.00
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
πΈπΆ $1.00 =ππ $1.00
40Γ 100
πΈπΆ $1.00 = ππ $2.50
Question 1c part (ii)
Required to calculate the value of US $80 in EC $
ππ $1.00 = πΈπΆ $2.70
ππ $80.00 = πΈπΆ $2.60 Γ 80
= πΈπΆ $216.00
Question 1c part (iii)
Required to calculate the value of TT $648 in US $
ππ $1.00 = πΈπΆ $0.40
ππ $648 = πΈπΆ $0.40 Γ 648
πΈπΆ $259.20
Now,
πΈπΆ $2.70 = ππ $1.00
πΈπΆ $1.00 =ππ $1.00
πΈπΆ 2.70
πΈπΆ $259.20 =ππ $1.00
πΈπΆ 2.70Γ πΈπΆ $259.20
= $96.00
β΄ ππ $648.00 = ππ $96.00
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 2a part (i)
Required to Factorize
2π₯3π¦ + 6π₯2π¦2
Step 1: We factor out the common factor in each term
2π₯2π¦ is a common factor in each term
Thus, 2π₯2π¦(π₯) + 2π₯2π¦(3π¦)
Step 2: Factorize to get 2π₯2π¦(π₯ + 3π¦)
Question 2a part (ii)
Required to Factorize
9π₯2 β 4
Step 1: We re-write the expression as the difference of two squares:
(3π₯)2 β (2)2
Step 2: We factorize to get:
(3π₯ β 2)(3π₯ + 2)
Question 2a part (iii)
Required to Factorize
4π₯2 + 8π₯π¦ β π₯π¦ β 2π¦2
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Step 1: Group like terms
4π₯(π₯ + 2π¦) β π¦(π₯ + 2π¦)
Step 2: Factorize to get:
(π₯ + 2π¦)(4π₯ β π¦)
Question 2b
Required to Solve
2π₯β3
3+
5βπ₯
2= 3
Step 1: Find the LCM
2(2π₯β3)+3(5βπ₯)
6= 3
4π₯β6+15β3π₯
6= 3
Step 2: Simplify
π₯+9
6= 3
Step 3: Make π₯ the subject of the formula
π₯ + 9 = 3(6)
π₯ + 9 = 18
π₯ = 18 β 9
π₯ = 9
Question 2c
Given Data: We are given two equations:
3π₯ β 2π¦ = 10
2π₯ + 5π¦ = 13
Required to solve both equations simultaneously
Using the Elimination Method
Step 1: Let 3π₯ β 2π¦ = 10 be Equation 1
Let 2π₯ + 5π¦ = 13 be Equation 2
Step 2: Multiply Equation 1 by 2 to get Equation 3
2(3π₯ β 2π¦) = 2(10)
6π₯ β 4π¦ = 20 [Equation 3]
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Step 3: Multiply Equation 2 by 3 to get Equation 4
3(2π₯ + 5π¦) = 3(13)
6π₯ + 15π¦ = 39 [Equation 4]
Step 4: Equation 4 β Equation 3
6π₯ + 15π¦ = 39 β
6π₯ β 4π¦ = 20
19π¦ = 19
Step 5: Make y the subject of the formula to find the value of π¦
π¦ =19
19
π¦ = 1
Step 6: Substitute π¦ = 1 into Equation 1 to find the value of π₯
3π₯ β 2π¦ = 10
3π₯ β 2(1) = 10
3π₯ β 2 = 10
3π₯ = 10 + 2
3π₯ = 12
π₯ =12
3
π₯ = 4
Thus π₯ = 4 and π¦ = 1
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 3a part (i)
Data Given: In a survey of 36 students,
30 play tennis π₯ play volleyball only
9π₯ play both tennis and volleyball
4 do not play either tennis or volleyball
π = { ππ‘π’ππππ‘π ππ π π’ππ£ππ¦}
π = {ππ‘π’ππππ‘π π€βπ ππππ¦ π£πππππ¦ππππ}
π = {ππ‘π’ππππ‘π π€βπ ππππ¦ π‘πππππ }
Required to Copy and Complete the Venn Diagram showing the number of students in subsets
marked y and z
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 3a part (ii)(a)
Required to write an expression in π₯ for the total number of students in the survey
The number of students who play volleyball only = π₯
The number of students who play tennis only = 30 β π₯
The number of students who play both tennis and volleyball = 9π₯
The number of students who do not play tennis nor volleyball = 4
Total number of students in the survey = π₯ + (30 β 9π₯) + 9π₯ + 4
= π₯ + 30 β 9π₯ + 9π₯ + 4
= π₯ + 30 + 4
= π₯ + 34
Question 3a part (ii)(b)
Required to write an equation in π₯ for the total number of students in the survey and to solve
for π₯
Total number of students = 36
Total number of students = π₯ + 34
U
V
T
4
30-9x
9x
x
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Thus π₯ + 34 = 36
π₯ = 36 β 34
π₯ = 2
Question 3b part (i)
Data Given: Diagram showing the journey of a ship which started at port π , sailed 15ππ due
South to port π and then further 20ππ due West to port π
Required to Copy and label the diagram to show points π and π and distances 20ππ and 15ππ
Question 3b part (ii)
Required to the shortest distance of the ship from the port to where the journey started
Using Pythagorasβ Theorem
ππ 2 = ππ2 + π π2
ππ = β(15)2 + (20)2
= 25ππ
Question 3b part (iii)
P
Q R
15km
20km
N
S
W E
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Required to Calculate the measure of angle πππ , giving answer to the nearest degree
π ππποΏ½ΜοΏ½π =πππ
βπ¦π
π ππποΏ½ΜοΏ½π =20
25
β΄ ποΏ½ΜοΏ½π = (20
25)
= 53.1Β°
53Β° [to the nearest degree]
Question 4a part(i)
Data Given: Diagram showing the cross-section of a prism
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Required to calculate the length of the arc π΄π΅πΆ
Length of the arc π΄π΅πΆ =270Β°
360°à πΆππππ’ππππππππ ππ π πππππππ‘π ππππππ
=3
4Γ 2π Γ π
=3
4Γ 2 Γ
22
7Γ 3.5
= 16.5ππ
Question 4a part(ii)
Required to calculate the perimeter of the sector ππ΄π΅πΆ
Perimeter of ππ΄π΅πΆ = πΏππππ‘β ππ π΄π + π΄ππ πΏππππ‘β π΄π΅πΆ + πΏππππ‘β ππ π ππππ’π πΆπ
= 16.5 + 3.5 + 3.5
= 23.5ππ
Question 4a part (iii)
Required to calculate the area of the sector ππ΄π΅πΆ
π΄πππ ππ π πππ‘ππ ππ΄π΅πΆ =270Β°
360°à π΄πππ ππ π πππππππ‘π ππππππ
=3
4Γ
22
7Γ (3.5)2
= 28.875
= 28.88ππ2 [to 2 decimal places]
Question 4b part (i)
A
B
C
O 3.5cm
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Data Given: A prism is 20cm long and made of tin
1ππ3 of tin has a mass of 7.3π
Required to calculate the volume of the prism
ππππ’ππ ππ π‘βπ ππππ π = πΆπππ π β π πππ‘πππππ π΄πππ Γ π»πππβπ‘
= 28.875 Γ 20
= 577.5ππ3
Question 4b part (ii)
Required to calculate the mass of the prism, to the nearest kg
1ππ3ππ π‘ππ π€πππβπ 7.3π
577.5ππ3ππ π‘ππ = 7.3 Γ 577.5π
= 4215.75π
=4215.75
1000
= 4.2ππ
=4 kg [to the nearest kg]
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 5a part(i)
Required to Construct triangle πππ with ππ = 8ππ, β πππ = 60Β° and β πππ = 45Β°
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 5a part (ii)
Required to measure and state the length of π π
Using a Ruler to measure the length of π π
The length of π π = 6.0ππ ]
Question 5b part (i)
Data Given: The line π passes through the points π(6,6) and π(0, β2)
Required to find the gradient of the line
Let (π₯1, π¦1) = (0, β2) and (π₯2, π¦2) = (6,6)
πΊπππππππ‘ =(π¦2βπ¦1)
(π₯2βπ₯1)
=6β(β2)
6β0
=8
6
=4
3
Question 5b part (ii)
Required to find the equation of the line π
The line π is of the form π¦ = ππ₯ + π
Since (0, β2) lies on the line π, then the y-intercept is β2 and the gradient m =4
3
Substituting π =4
3 and π = β2 into π¦ = ππ₯ + π to get π¦ =
4
3π₯ β 2
Thus the equation of the line π = π¦ =4
3π₯ β 2
Question 5b part (iii)
Required to find the midpoint of the line segment ππ
ππππππππ‘ = (π₯1+π₯2
2 ,
π¦1+π¦2
2)
=6+0
2,
6+(β2)
2
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
= (3,2)
Question 5b part (iv)
Required to find the length of the line segment ππ
πβπ πππππ‘β ππ ππ = β(π₯1 β π₯2)2 + (π¦2 β π¦1)2
= β(6 β 0)2 + (6 β (β2))2
= β62 + 82
= 10 π’πππ‘π
Question 6a
Data Given: Graph showing triangle πΏππ and its image πππ after an enlargement
Required to Locate the centre of enlargement
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
The centre of enlargement was found by finding the point of intersection between the end
points of triangle LNM triangle PRQ.
The centre of enlargement was found to be (1,5)
Question 6b
Required to State the scale factor and the center of enlargement
ππππππ πΉπππ‘ππ =πΏππππ‘β ππ ππ
πΏππππ‘β ππ πΏπ
=6
3
= 2
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 6c
Required to Determine the value of π΄πππ ππ πππ
π΄πππ ππ πΏππ
Since π π
ππ=
2
1
Then π΄πππ ππ πππ
π΄πππ ππ πΏππ=
(2)2
(1)2
=4
1
= 4
Question 6d
Required to Draw and Label Triangle π΄π΅πΆ
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 6e
Describe fully the transformation which maps triangle πΏππ onto triangle π΄π΅πΆ
Triangle πΏππ undergo a rotation of 90β in an anti-clockwise direction about the origin
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
(0.0)
Question 7a
Data Given: Table showing the ages of persons who visited the clinic during the week
Required to Copy and Complete the table to show cumulative frequency
Age, π₯ LCL β UCL
LCB β€ π₯ β€ UCB Number of
Persons
Cumulative Frequency
Points to be
Plotted
(39.5,0)
40 β 49 39.5 β€ π₯ β€ 49.5
4 4 (49.5,4)
50 β 59 49.5 β€ π₯ β€ 59.5
11 15 (59.5,15)
60 β 69
59.5 β€ π₯ β€ 69.5
20 35 (69.5,35)
70 β 79
69.5 β€ π₯ β€ 79.5
12 47 (79.5,47)
80 β 89
79.5 β€ π₯ β€ 89.5
3 50 (89.5,50)
Question 7b
Required to Draw the Cumulative Frequency curve to represent the data
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 7c part (i)
Required to Estimate the median age for the data
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
The Median Age is calculated by finding the x value (years) when the y value is 1
2 ππ 50
on the graph
Thus, the Median Age is found to be 64.75 when y = 25
Question 7c part (i)
Required to Estimate the probability that a person who visited the clinic was 75 years or
younger
π(π΄ ππππ πππ ππ 75 π¦ππππ ππ π¦ππ’ππππ)
= ππ’ππππ ππ ππππ πππ 75 π¦ππππ ππ π¦ππ’ππππ Γ· πππ‘ππ ππ’ππππ ππ ππππ πππ
= 42.5 Γ· 50
=17
20
Question 8a
Given Data: Diagram showing three figures in a sequence of figures
Required to draw the fourth figure in the sequence of figures
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 8b
Required to copy and complete the table given
Figure Area of Triangle Number of Pins on Base
1 1 (2 Γ 1) + 1 = 3
2 4 (2 Γ 2) + 1 = 5
3 9 (2 Γ 3) + 1 = 7
4 42 = 16 (2 Γ 4) + 1 = 9
β100 = 10 100 (2 Γ 10) + 1 = 21
20 202 = 400 (2 Γ 20) + 1 = 41
π π2 2π + 1
Question 9a part (i)
Required to Solve π¦ = 8 β π₯ and 2π₯2 + π₯π¦ = β16
Let π¦ = 8 β π₯ [Equation 1]
2π₯2 + π₯π¦ = β16 [Equation 2]
Using the Method of Substitution
Step 1: Substitute Equation 1 into Equation 2
2π₯2 + π₯(8 β π₯) = β16
2π₯2 + 8π₯ β π₯2 = β16
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
π₯2 + 8π₯ = β16
π₯2 + 8π₯ + 16 = 0
(π₯ + 4)(π₯ + 4) = 0
π₯ = β4
Step 2: Substitute π₯ = 4 into Equation 1
π¦ = 8 β (β4)
= 8 + 4
= 12
Thus π₯ = β4 and π¦ = 12
Question 9a part (ii)
Required to State: With reason or not, π¦ = 8 β π₯ is a tangent to the curve with equation
2π₯2 + π₯π¦ = β16
Solve the Equation 2π₯2 + π₯π¦ = β16 to obtain the root π₯ = β4
Since the roots are real and equal, the straight line does not intersect the curve but rather
it touches the curve at the point (β4, 12)
Question 9b part (i)
Data Given: π₯ roses and π¦ orchids in each bouquet with the following constraints
The number of orchids must be at least half the number of roses
There must be at least 2 roses
There must be no more than 12 flowers in the bouquet
Required to write the three inequalities for the constraints given
π¦ β₯1
2π₯
π₯ β₯ 2
π₯ + π¦ β€ 12
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 9b part (ii)
Required to shade the region that satisfies all three inequalities
Question 9b part (iii)
Required to State the co-ordinates of the points which represent the vertices of the region
showing the solution set
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
β³ π΄π΅πΆ is the region which is shaded.
The co-ordinates for β³ π΄π΅πΆ is π΄ = (2,10), π΅ = (2,1) and πΆ = (8,4)
Question 9b part (iv)
Data Given: A profit of $3 is made on each rose and $4 on each orchid
Required to Determine the maximum possible profit on the sale of a bouquet
Step 1: Find an expression for the Total Profit made
Total profit made on π₯ roses at $3 each and on π¦ orchids at $4 each = (π₯ Γ 3) + (π¦ Γ 4)
Step 2: Find an equation for the Total Profit
π = 3π₯ + 4π¦
Step 3: Find the Total Profit made at each of the three points.
At point π΄ = (2,10) π = 3(2) + 4(10)
π = $46.00
A point π΅ = (2,1) π = 3(2) + 4(1)
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
π = $10.00
A point πΆ = (8,4) π = 3(8) + 4(4)
π = $40.00
The Maximum Profit occurs at point π΄ since the Total Profit made at point π΄ is $46
Question 10a part(i)
Data Given: Diagram showing quadrilateral ππ ππ
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Required to Calculate the length of π π
Using the Sine Rule
ππ
π πππ =
π π
π πππ
7
π ππ60Β°=
π π
π ππ48Β°
7π ππ48Β° = π ππ ππ60Β°
π π =7π ππ48Β°
π ππ60Β°
= 6.006
= 6.01 [to 2 decimal places]
Question 10a part(ii)
Required to Calculate the measure of β πππ
Using the Cosine Rule
ππ2 = ππ2 + ππ2 β 2(ππ)(ππ)πππ οΏ½ΜοΏ½
72 = 82 + 102 β 2(8)(10)πππ οΏ½ΜοΏ½
πππ οΏ½ΜοΏ½ =(72β82β102)
(β2(8)(10)
=β115
β160
οΏ½ΜοΏ½ = (0.7187)
= 44.02
= 44.0Β° [to the nearest 0.1Β°]
Question 10b part(i)(a)
8cm
10cm 7cm
R
Q
T
S
48
60
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Required to Calculate the measure of angle πππ
ποΏ½ΜοΏ½π = 180Β° β 70Β°
= 110Β°
In triangle πππ, ππ = ππ [since they are radii of the same circle]
β triangle πππ is isosceles
πβπ’π , ποΏ½ΜοΏ½π =180 Β°β110 Β°
2
ποΏ½ΜοΏ½π = 35Β°
Question 10b part(i)(b)
Required to Calculate the measure of angle πππ
ποΏ½ΜοΏ½π =70 Β°
2
= 35 Β°
[The angle that is subtended by a chord at the circumference of a circle is half of that angle that
the chord subtends at the center of the circle, and standing on the same arc]
ποΏ½ΜοΏ½π ππ ποΏ½ΜοΏ½π = 90Β°
[The angle made by the tangent ππ to a circle and a radius ππ , at the point of contact, π is a
right angle.
ποΏ½ΜοΏ½π = 180Β° β (90Β° + 35Β°)
= 55Β°
[The sum of the interior angles of a triangle = 180Β°]
Question 10b part(i)(c)
Required to Calculate the measure of angle πππ
ποΏ½ΜοΏ½π = 70Β°
ποΏ½ΜοΏ½π = 90Β°
[The sum of the three interior angles of a triangle = 180Β°]
ποΏ½ΜοΏ½π = 180Β° β (90Β° + 70Β°)
= 20Β°
Question 10b part(ii)(a)
Required to Name the triangle which is congruent to triangle πππ
ππ = ππ and ππ = ππ
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
ποΏ½ΜοΏ½π = ποΏ½ΜοΏ½π
[Vertically opposite angles are equal when two straight lines intersect]
Therefore, β³ πππ is congruent to β³ πππ
Question 10b part(ii)(b)
Required to Name the triangle which is congruent to triangle πππ
Consider β³ πππ and β³ πππ
ππ = ππ
ποΏ½ΜοΏ½π = ποΏ½ΜοΏ½π = 90Β°
ππ is a common side to both triangles and therefore triangle πππ and triangle πππ have the
same hypotenuse and share a common side.
Thus, Triangle πππ is congruent to Triangle πππ
Question 11a part(i)(a)
Given Data: ππ΄ = (6 2 ), ππ΅ = (3 4 ) πππ ππΆ = (12 β 2 )
Required to Express the vector π΅π΄ in the form (π₯ π¦ )
Using the Vector Triangle Law
π΅π΄ = π΅π + ππ΄
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
= β(3 4 ) + (6 2 )
= (3 β 2 )
Question 11a part(i)(b)
Required to Express the vector π΅πΆ in the form (π₯ π¦ )
Using the Vector Triangle Law
π΅πΆ = π΅π + ππΆ
= β(3 4 ) + (12 β 2 )
= (9 β 6 )
Question 11a part(ii)
Required to State one geometrical relationship between BA and BC
|π΅πΆ| = 3|π΅π΄|
The vector π΅π΄ is a scalar multiple of the vector π΅πΆ
Question 11a part(iii)
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 11b part(i)
Data Given: (π β 4 1 π )(2 β 4 1 β 3 ) = (2 0 0 2 )
Required to Calculate the value of π and π
Step 1: Ensure that the dimensions of the matrices allows matrix multiplication
In this case, it does, as all three matrices are 2 Γ 2 matrices
Thus, the result will be of the form (π π π π )
Step 2: Multiply the matrices
((2 Γ π) + (β4 Γ 1) (β4 Γ π) + (β4 Γ β3) (1 Γ 2) + (π Γ 1) (1 Γ β4) + (π Γ β3) ) =
(2 0 0 2 )
(2π β 4 12 β 4π π + 2 β 4 β 3π ) = (2 0 0 2 )
Step 3: Equating entries to find the value of π and π
2π β 4 = 2 12 β 4π = 0
2π = 6 12 = 4π
π =6
2
12
4= π
π = 3 3 = π
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
π + 2 = 0 β4 β 3π = 2
π = β2 β4 β 2 = 3π
β6 = 3π
β6
3= π
β2 = π
Thus, π = 3 and π = β2
Question 11b part(ii)
Required to Find the inverse of (2 β 4 1 β 3 )
Let π΄ = (2 β 4 1 β 3 )
(3 β 4 1 β 2 )(2 β 4 1 β 3 ) = (2 0 0 2 )
(3 β 4 1 β 2 )(2 β 4 1 β 3 ) = 2(1 0 0 1 )
This is of the form π΄π΄β1 = πΌ
1
2(3 β 4 1 β 2 )(2 β 4 1 β 3 ) = (1 0 0 1 )
(3
2 β
4
2 1
2 β
2
2 ) (2 β 4 1 β 3 ) = (1 0 0 1 )
Therefore, π΄β1 = (3
2 β 2
1
2 β 1 )
Question 11b part(iii)
Required to Solve for π₯ and π¦ in (2 β 4 1 β 3 ) = (π₯ π¦ ) = (12 7 )
Step 1: Multiply the matrix equation by π΄β1
(3
2 β 2
1
2 β 1 ) (2 β 4 1 β 3 )(π₯ π¦ ) = (
3
2 β 2
1
2 β 1 ) (12 7 )
(π₯ π¦ ) = (3
2 β 2
1
2 β 1 ) (12 7 )
(π₯ π¦ ) = ((3
2Γ 12) + (β2 Γ 7) (
1
2Γ 12) + (β1 Γ 7) )
(π₯ π¦ ) = (4 β 1 )
π₯ = 4 and π¦ = β1
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub