ALLAMA IQBAL OPEN UNIVERSITY
(Department of Mathematics)
WARNING
1. Plagiarism or hiring of ghost writer(s) for solving the assignment(s) will debar the student from award of degree/certificate if found at any stage.
2. Submitting assignment(s) borrowed or stolen from other(s) as one's own will be penalized as defined in the "Aiou Plagiarism Policy".
| Assignment Submission Schedule |
| 6 Credit Hours |
Due Date |
3 Credit Hours |
Due Date |
| Assignment 1 |
15-12-2025 |
Assignment 1 |
08-01-2026 |
| Assignment 2 |
08-01-2026 |
| Assignment 3 |
30-01-2026 |
Assignment 2 |
20-02-2026 |
| Assignment 4 |
20-02-2026 |
| Course: Introduction to Business Mathematics (1349) |
Semester: Autumn-2025 |
| Level: HSSC |
|
Please read the following instructions for writing your assignments. (SSC, HSSC & BA Programmes)
1. All questions are compulsory and carry equal marks but within a question the marks are distributed according to its requirements.
2. Read the question carefully and then answer it according to the requirements of the questions.
3. Late submission of assignments will not be accepted.
4. Your own analysis and synthesis will be appreciated.
5. Avoid irrelevant discussion/information and reproducing from books, study guide of allied material.
| Total Marks: 100 |
Pass Marks: 40 |
ASSIGNMENT No. 1
Q1(a). Define "ratio" and discuss how it is different from a rate? Convert the given ratio 30:360 into its simplest form.
▶
Definition of Ratio
A ratio is a comparison of two quantities that shows how many times one value contains or is contained within the other. It is usually written in the form \(a:b\) or \(\frac{a}{b}\). For example, if there are 2 apples and 3 oranges, the ratio of apples to oranges is \(2:3\).
Difference Between Ratio and Rate
- A ratio compares two quantities of the same kind, such as 2 apples to 3 oranges.
- A rate compares two quantities of different kinds, usually with units, such as 60 kilometers per hour (km/h) or 5 rupees per kilogram.
Converting the Ratio 30:360 into Simplest Form
Step 1: Find the greatest common divisor (GCD) of 30 and 360, which is 30.
Step 2: Divide both terms by 30:
\(
\frac{30}{30} : \frac{360}{30} = 1:12
\)
\(
30:360 = 1:12
\)
Q1(b). Calculate the continued ratio; (a:b:c) when (a:b = 5:6) and (b:c = 6:7).
▶
Calculate the Continued Ratio \((a:b:c)\)
We are given the ratios:
\(
a:b = 5:6 \quad \text{and} \quad b:c = 6:7
\)
Step 1: Equalize the middle term
The term \(b\) appears in both ratios. In this case, \(b = 6\) in the first ratio and \(b = 6\) in the second ratio. Since they are already equal, we can directly combine the ratios.
Step 2: Write the continued ratio
\(
a:b:c = 5:6:7
\)
\(
a:b:c = 5:6:7
\)
Q1(c). Convert 22.5% into a fraction and a decimal. If a student scores 84% in 150 exams, how many marks did they secure?
▶
Converting 22.5% into Fraction and Decimal
To convert a percentage into a decimal, divide by 100:
\(
22.5\% = \frac{22.5}{100} = 0.225
\)
To convert a percentage into a fraction:
\(
22.5\% = \frac{22.5}{100} = \frac{225}{1000} = \frac{9}{40}
\)
Answer:
Decimal: \(0.225\), Fraction: \(\frac{9}{40}\)
Calculating Marks Secured by the Student
Given that the student scored 84% in 150 marks, the marks secured can be calculated as:
\(
\text{Marks secured} = \frac{84}{100} \times 150 = 0.84 \times 150 = 126
\)
The student secured \(126\) marks out of 150.
Q1(d). If the cash price distributed between two students in the ratio of 4:5 is Rs. 900, calculate the amount that each student received.
▶
The cash price of Rs. 900 is to be divided between two students in the ratio \(4:5\). We need to find the amount each student received.
Step 1: Find the total parts of the ratio
\(
4 + 5 = 9 \text{ parts}
\)
Step 2: Value of one part
\(
\text{Value of one part} = \frac{900}{9} = 100
\)
Step 3: Calculate each student's share
First student: \(4 \times 100 = 400\) Rs.
Second student: \(5 \times 100 = 500\) Rs.
The first student received Rs. 400 and the second student received Rs. 500.
Q1(e). A manager earns Rs. 200,000 per month and spends Rs. 190,000. An assistant earns Rs. 40,000 and saves Rs. 8,000. Who saves more regarding their income?
▶
We are asked to compare the savings of a manager and an assistant in terms of their income.
Step 1: Calculate savings percentage for the manager
Manager's monthly income = Rs. 200,000
Manager's savings = Income - Expenditure = 200,000 - 190,000 = Rs. 10,000
Savings percentage = \(\frac{\text{Savings}}{\text{Income}} \times 100 = \frac{10,000}{200,000} \times 100 = 5\%\)
Step 2: Calculate savings percentage for the assistant
Assistant's monthly income = Rs. 40,000
Assistant's savings = Rs. 8,000
Savings percentage = \(\frac{8,000}{40,000} \times 100 = 20\%\)
Step 3: Comparison
Even though the manager saves more money in absolute terms (Rs. 10,000 vs Rs. 8,000), the assistant saves a higher proportion of their income (20% vs 5%).
The assistant saves more regarding their income, with a savings percentage of 20% compared to the manager's 5%.
Q2(a). Define discount and explain its types and calculate the discounted price of an item originally priced at Rs. 20,000 with a discount of Rs. 800.
▶
A discount is a reduction in the original price of a product or service, usually offered by sellers to encourage purchases or reward customers. It is the difference between the marked price and the selling price.
Types of Discount
1. Trade Discount: A reduction given by the manufacturer or wholesaler to the retailer, usually expressed as a percentage of the list price.
2. Cash Discount: A reduction offered to buyers for making prompt payment or paying in cash.
3. Seasonal Discount: A reduction offered during a particular season to boost sales.
4. Quantity Discount: A reduction given for purchasing goods in bulk or large quantities.
Calculation of Discounted Price
Given: Original price = Rs. 20,000, Discount = Rs. 800
Discounted Price = Original Price - Discount
\(
= 20,000 - 800 = 19,200
\)
The discounted price of the item is Rs. 19,200.
Q2(b). If a person buys a book rack for Rs. 18,000 with a discount of 12%, what was the original price?
▶
Step 1: Let the original price be \(P\)
Discount = 12% of \(P\) = \(0.12P\)
Selling price (after discount) = Original price - Discount = \(P - 0.12P = 0.88P\)
Step 2: Set up the equation
\(
0.88P = 18,000
\)
Step 3: Solve for \(P\)
\(
P = \frac{18,000}{0.88} = 20,454.55 \text{ (approximately)}
\)
The original price of the book rack was approximately Rs. 20,454.55.
Q2(c). Calculate the profit percentage if a person bought a T.V. for Rs. 5,000 and sold it for Rs. 6,250.
▶
Step 1: Calculate the profit
Profit = Selling Price - Cost Price
\(
= 6,250 - 5,000 = 1,250
\)
Step 2: Calculate the profit percentage
Profit Percentage = \(\frac{\text{Profit}}{\text{Cost Price}} \times 100\)
\(
= \frac{1,250}{5,000} \times 100 = 25\%
\)
The profit percentage is 25%.
Q2(d). Straight-Line Depreciation Question: A company purchases a computer for Rs. 30,000. The computer is expected to have a salvage value of Rs. 3,000 and a useful life of 5 years. Calculate the annual depreciation expense using the straight-line method.
▶
Step 1: Formula for straight-line depreciation
\(
\text{Annual Depreciation} = \frac{\text{Cost Price} - \text{Salvage Value}}{\text{Useful Life}}
\)
Step 2: Substitute the given values
\(
\text{Annual Depreciation} = \frac{30,000 - 3,000}{5} = \frac{27,000}{5} = 5,400
\)
The annual depreciation expense is Rs. 5,400.
Q2(e). Diminishing Balance Depreciation Question: A piece of machinery is bought for Rs. 80,000 with a depreciation rate of 15% per annum. Calculate the depreciation expense for the first year using the diminishing balance method.
▶
Step 1: Formula for diminishing balance depreciation
\(
\text{Depreciation Expense} = \text{Cost Price} \times \text{Depreciation Rate}
\)
Step 2: Substitute the given values
\(
\text{Depreciation Expense} = 80,000 \times 0.15 = 12,000
\)
The depreciation expense for the first year is Rs. 12,000.
Q3(a). Define simple interest and provide the formula for its calculation. If a principal amount of Rs. 60,000 is borrowed at an interest rate of 12% per annum for 3 years, calculate the total interest.
▶
Simple Interest (SI) is the interest calculated on the original principal amount of a loan or investment for a specific period of time at a fixed rate of interest. It does not include interest on previously earned interest.
Formula for Simple Interest
\(
SI = \frac{P \times R \times T}{100}
\)
Where:
P = Principal amount
R = Rate of interest per annum (%)
T = Time in years
Given:
Principal, \(P = 60,000\)
Rate of interest, \(R = 12\%\)
Time, \(T = 3\) years
Step 1: Apply the formula
\(
SI = \frac{60,000 \times 12 \times 3}{100} = \frac{2,160,000}{100} = 21,600
\)
The total simple interest for 3 years is Rs. 21,600.
Q3(b). Explain the difference between simple interest and compound interest. Calculate the compound amount for a principal of Rs. 9,000 at an interest rate of 9% compounded quarterly for one year.
▶
Difference Between Simple Interest and Compound Interest
- Simple Interest (SI): Interest is calculated only on the original principal. It does not include interest on previously earned interest.
- Compound Interest (CI): Interest is calculated on the principal as well as on any interest accumulated from previous periods. The interest is "compounded" over each period.
Given:
Principal, \(P = 9,000\)
Rate of interest, \(R = 9\%\) per annum
Compounding frequency = Quarterly
Time, \(T = 1\) year
Step 1: Formula for Compound Amount (A)
\(
A = P \left(1 + \frac{R}{n}\right)^{nT}
\)
Where:
P = Principal amount
R = Annual interest rate in decimal or %
n = Number of compounding periods per year
T = Time in years
Step 2: Substitute the given values
Quarterly compounding → \(n = 4\)
\(
A = 9,000 \left(1 + \frac{9}{4 \times 100}\right)^{4 \times 1}
= 9,000 \left(1 + 0.0225\right)^4
= 9,000 \left(1.0225^4\right)
\)
\(
1.0225^4 \approx 1.0927
\)
\(
A \approx 9,000 \times 1.0927 = 9,834.30
\)
Step 3: Calculate Compound Interest
\(
CI = A - P = 9,834.30 - 9,000 = 834.30
\)
The compound amount after one year is approximately Rs. 9,834.30, and the compound interest earned is Rs. 834.30.
Q3(c). If a person invests Rs. 5,000 at a simple interest rate of 10% per annum, how much will they have after 5 years?
▶
Step 1: Formula for Simple Interest
\(
SI = \frac{P \times R \times T}{100}
\)
Step 2: Calculate Simple Interest
\(
SI = \frac{5,000 \times 10 \times 5}{100} = \frac{250,000}{100} = 2,500
\)
Step 3: Total Amount
\(
A = P + SI = 5,000 + 2,500 = 7,500
\)
The total amount after 5 years will be Rs. 7,500.
Q3(d). If a principal amount of Rs. 10,000 is borrowed at 5% p.a. for 3 years, what is the total interest paid?
▶
Step 1: Formula for Simple Interest
\(
SI = \frac{P \times R \times T}{100}
\)
Step 2: Calculate Simple Interest
\(
SI = \frac{10,000 \times 5 \times 3}{100} = \frac{150,000}{100} = 1,500
\)
The total interest paid after 3 years is Rs. 1,500.
Q4(a). Define annuity and explain its types and calculate the future value of an ordinary annuity of Rs. 1,500 payable at the end of each year for 12 years at an interest rate of 6%.
▶
Definition of Annuity
An annuity is a series of equal payments made at regular intervals over a specified period of time. It can be used for loans, investments, or retirement planning.
Types of Annuity
1. Ordinary Annuity: Payments are made at the end of each period.
2. Annuity Due: Payments are made at the beginning of each period.
3. Perpetuity: An annuity that continues indefinitely without an end date.
Given:
Payment per period \(PMT = 1,500\)
Number of periods \(n = 12\) years
Interest rate \(i = 6\%\) per annum
Ordinary annuity (payments at the end of each year)
Step 1: Formula for Future Value of Ordinary Annuity
\(
FV = PMT \times \frac{(1 + i)^n - 1}{i}
\)
Where:
FV = Future Value
PMT = Payment per period
i = Interest rate per period (in decimal)
n = Number of periods
Step 2: Substitute the given values
\(
i = 6\% = 0.06
\)
\(
FV = 1,500 \times \frac{(1 + 0.06)^{12} - 1}{0.06}
\)
\(
(1.06)^{12} \approx 2.0122
\)
\(
FV = 1,500 \times \frac{2.0122 - 1}{0.06} = 1,500 \times \frac{1.0122}{0.06} \approx 1,500 \times 16.87 = 25,305
\)
The future value of the ordinary annuity is approximately Rs. 25,305.
Q4(b). What is the present value of an annuity due of Rs. 2,000 per year for 5 years at an interest rate of 5%?
▶
Given: Payment per period \(PMT = 2,000\)
Number of periods \(n = 5\) years
Interest rate \(i = 5\%\) per annum
Annuity type: Annuity Due (payments at the beginning of each period)
Step 1: Formula for Present Value of Annuity Due
\(
PV_{\text{due}} = PMT \times \frac{1 - (1 + i)^{-n}}{i} \times (1 + i)
\)
Where:
PV = Present Value
PMT = Payment per period
i = Interest rate per period (in decimal)
n = Number of periods
Step 2: Substitute the given values
\(
i = 5\% = 0.05
\)
\(
PV_{\text{due}} = 2,000 \times \frac{1 - (1 + 0.05)^{-5}}{0.05} \times 1.05
\)
\((1.05)^{-5} \approx 0.7835\)
\(
1 - 0.7835 = 0.2165
\)
\(
\frac{0.2165}{0.05} = 4.33
\)
\(
PV_{\text{due}} = 2,000 \times 4.33 \times 1.05 \approx 2,000 \times 4.5465 = 9,093
\)
The present value of the annuity due is approximately Rs. 9,093.
Q5(a). Solve the linear equation: \(5x - 10 = 3x + 6\).
▶
Given the equation:
\(5x - 10 = 3x + 6\)
Step 1: Move all x terms to one side
\(
5x - 3x - 10 = 6
\)
\(
2x - 10 = 6
\)
Step 2: Move constants to the other side
\(
2x = 6 + 10
\)
\(
2x = 16
\)
Step 3: Solve for x
\(
x = \frac{16}{2} = 8
\)
The solution of the equation is \(x = 8\).
Q5(b). Find the two consecutive odd integers whose sum is 40.
▶
Finding Two Consecutive Odd Integers
Let the first odd integer be \(x\). Since the integers are consecutive odd numbers, the next odd integer will be \(x + 2\).
Step 1: Set up the equation
Sum of the two integers = 40
\(
x + (x + 2) = 40
\)
Step 2: Simplify the equation
\(
2x + 2 = 40
\)
\(
2x = 38
\)
\(
x = 19
\)
Step 3: Find the second integer
Second integer = \(x + 2 = 19 + 2 = 21\)
The two consecutive odd integers are 19 and 21.
Q5(c). Solve the quadratic equation using the factorization method: \(x^2 + 2x - 3 = 0\).
▶
Given equation:
\(x^2 + 2x - 3 = 0\)
Step 1: Factorize the quadratic
We look for two numbers whose product is \(-3\) (constant term) and sum is \(2\) (coefficient of x).
The numbers are 3 and -1.
\(
x^2 + 2x - 3 = x^2 + 3x - x - 3 = (x^2 + 3x) - (x + 3) = x(x + 3) - 1(x + 3) = (x - 1)(x + 3)
\)
Step 2: Solve for x
\(
x - 1 = 0 \quad \Rightarrow \quad x = 1
\)
\(
x + 3 = 0 \quad \Rightarrow \quad x = -3
\)
Answer:
The solutions of the equation are \(x = 1\) and \(x = -3\).
Q5(d). Use the quadratic formula to solve: \(2x^2 - 4x + 1 = 0\).
▶
Given equation:
\(2x^2 - 4x + 1 = 0\)
Step 1: Quadratic Formula
The quadratic formula is:
\(
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\)
Here, \(a = 2\), \(b = -4\), \(c = 1\)
Step 2: Substitute values
\(
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}
\)
\(
x = \frac{4 \pm \sqrt{16 - 8}}{4}
\)
\(
x = \frac{4 \pm \sqrt{8}}{4}
\)
\(
\sqrt{8} = 2\sqrt{2}
\)
\(
x = \frac{4 \pm 2\sqrt{2}}{4}
\)
Step 3: Simplify
\(
x = \frac{4}{4} \pm \frac{2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}
\)
The solutions of the equation are:
\(
x = 1 + \frac{\sqrt{2}}{2} \quad \text{and} \quad x = 1 - \frac{\sqrt{2}}{2}
\)
Q5(e). Solve the simultaneous equations: \(3x + 4y = 24\) and \(2x - 3y = 15\).
▶
Solving Simultaneous Equations
Given equations:
1) \(3x + 4y = 24\)
2) \(2x - 3y = 15\)
Step 1: Solve one equation for one variable
From equation (1): \(3x + 4y = 24\)
\(
3x = 24 - 4y \quad \Rightarrow \quad x = \frac{24 - 4y}{3}
\)
Step 2: Substitute into equation (2)
\(
2\left(\frac{24 - 4y}{3}\right) - 3y = 15
\)
\(
\frac{48 - 8y}{3} - 3y = 15
\)
Multiply through by 3:
\(
48 - 8y - 9y = 45
\)
\(
48 - 17y = 45
\)
\(
-17y = -3 \quad \Rightarrow \quad y = \frac{3}{17}
\)
Step 3: Find x
\(
x = \frac{24 - 4y}{3} = \frac{24 - 4 \cdot \frac{3}{17}}{3} = \frac{24 - \frac{12}{17}}{3} = \frac{\frac{408 - 12}{17}}{3} = \frac{396/17}{3} = \frac{396}{51} = \frac{132}{17}
\)
The solution of the system is:
\(
x = \frac{132}{17}, \quad y = \frac{3}{17}
\)
ASSIGNMENT No. 2
Q1(a). Define a function and provide an example; Explain the vertical line test and why \(y^2 = x\) is not a function.
▶
Definition of a Function
A function is a relation between two sets in which each input (usually x) is associated with exactly one output (usually y). In other words, for every value of x, there is only one corresponding value of y.
Example:
\(
f(x) = 2x + 3
\)
For each x, this equation gives a unique y.
Vertical Line Test
The vertical line test is a visual way to determine whether a graph represents a function. If a vertical line intersects the graph at more than one point at the same time, the graph does not represent a function. If it intersects at exactly one point for every vertical line, it is a function.
Why \(y^2 = x\) is Not a Function
The equation \(y^2 = x\) can be rewritten as \(
y = \pm \sqrt{x}
\). For some values of x (for example, x = 4), there are two corresponding y-values (y = 2 and y = -2). Therefore, it fails the vertical line test because a vertical line at x = 4 would intersect the graph at two points. Hence, \(y^2 = x\) is not a function.
Q1(b). What is the difference between even and odd functions?
▶
Difference Between Even and Odd Functions
Functions can be classified as even, odd, or neither based on their symmetry.
Even Function:
A function \(f(x)\) is called even if for every x in its domain:
\(
f(-x) = f(x)
\)
Even functions are symmetric about the y-axis.
Example: \(f(x) = x^2\)
Odd Function:
A function \(f(x)\) is called odd if for every x in its domain:
\(
f(-x) = -f(x)
\)
Odd functions are symmetric about the origin.
Example: \(f(x) = x^3\)
Key Differences:
- Even functions satisfy \(f(-x) = f(x)\); odd functions satisfy \(f(-x) = -f(x)\).
- Even functions are symmetric about the y-axis; odd functions are symmetric about the origin.
- Examples: Even → \(x^2, \cos x\); Odd → \(x^3, \sin x\)
Q1(c). Graph the linear function \(y = 2x + 3\) and identify its \(x - intercept\) and \(y - intercept\) Determine the domain and range of the function \(f(x) = \sqrt{x-4}\)
▶
Graph the linear function y = 2x + 3
Step 1: Identify slope and y-intercept
The equation is in slope-intercept form: \(y = mx + b\).
Slope: \(m = 2\)
Y-intercept: \(b = 3\), so the line crosses the y-axis at (0, 3)
Step 2: Find the x-intercept
Set \(y = 0\):
\(0 = 2x + 3\)
\(2x = -3\)
\(x = -\frac{3}{2}\)
X-intercept: \((-1.5, 0)\)
Step 3: Plot points and draw the line
Plot (0, 3) and (-1.5, 0) and draw a straight line through them.
Domain and range of \(f(x) =
Step 1: Find the domain
Square root requires non-negative input: \(x-4 \ge 0\)
Solve: \(x \ge 4\)
Domain: \([4, \infty)\)
Step 2: Find the range
Square root outputs non-negative numbers: \(y \ge 0\)
Range: \([0, \infty)\)
Step 3: Optional points for graph
If \(x = 4\), \(f(4) = 0\) → point (4, 0)
If \(x = 5\), \(f(5) = 1\) → point (5, 1)
If \(x = 8\), \(f(8) = 2\) → point (8, 2)
Graph starts at (4, 0) and increases to the right.
Q2(a). Matrix Types: Differentiate between a scalar matrix and an identity matrix. Provide examples.
▶
Matrix Types: Scalar Matrix vs Identity Matrix
1. Scalar Matrix
A scalar matrix is a square matrix in which all the elements on the main diagonal are the same, and all off-diagonal elements are zero.
General form:
\(
\begin{bmatrix}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{bmatrix}
\)
Here, \(k\) is any scalar (a constant number).
Example:
\(
\begin{bmatrix}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{bmatrix}
\)
2. Identity Matrix
An identity matrix is a special scalar matrix in which all the diagonal elements are 1 and all off-diagonal elements are 0.
General form:
\(
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\)
Example:
\(
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\)
Key Difference:
- Every identity matrix is a scalar matrix, but not every scalar matrix is an identity matrix.
- In a scalar matrix, diagonal elements can be any number; in an identity matrix, diagonal elements must be 1.
Q2(b). What is the determinant of a 2x2 matrix? Find the determinant of \(\begin{bmatrix} 2 & 3 \\ 1 & -5 \end{bmatrix}\) Is this matrix singular.
▶
Determinant of a 2x2 Matrix
Step 1: Formula for a 2x2 matrix
For a 2x2 matrix
\(
A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},
\)
the determinant is given by:
\(
\det(A) = ad - bc
\)
Step 2: Apply the formula
Given matrix:
\(\begin{bmatrix} 2 & 3 \\ 1 & -5 \end{bmatrix}\)
Here, \(a=2, b=3, c=1, d=-5\)
\(
\det(A) = (2)(-5) - (3)(1) = -10 - 3 = -13
\)
Step 3: Check if the matrix is singular
A matrix is singular if its determinant is 0.
Here, \(\det(A) = -13 \neq 0\), so the matrix is not singular.
Q2(c). Explain the concept of a singular matrix.
▶
Concept of a Singular Matrix
Definition:
A square matrix is called singular if its determinant is equal to zero.
In mathematical terms, for a square matrix \(A\):
\(
\text{If } \det(A) = 0, \text{ then } A \text{ is singular.}
\)
Properties of a Singular Matrix:
1. A singular matrix does not have an inverse.
2. Its rows or columns are linearly dependent, meaning at least one row or column can be expressed as a combination of others.
3. The system of linear equations represented by a singular matrix either has no solution or infinitely many solutions.
Example:
Consider the matrix:
\(
A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}
\)
Determinant:
\(
\det(A) = (2)(2) - (4)(1) = 4 - 4 = 0
\)
Since the determinant is 0, \(A\) is a singular matrix.
Q2(d). Find the transpose of the matrix \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\).
▶
Consider the matrix:
\(
A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
\)
The transpose of a matrix is obtained by swapping its rows and columns. So, the transpose of \(A\) is:
\(
A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}
\)
Here, the first row (1, 2) becomes the first column, and the second row (3, 4) becomes the second column.
Q2(e). Calculate the cofactor of the element in the first row and first column of the matrix \(B = \begin{bmatrix} 5 & 7 \\ 8 & 6 \end{bmatrix}\)
▶
Consider the matrix:
\(B = \begin{bmatrix} 5 & 7 \\ 8 & 6 \end{bmatrix} \)
We want to calculate the cofactor of the element in the first row and first column, which is \(b_{11} = 5\).
Step 1: Remove the row and column containing the element. Removing the first row and first column leaves the submatrix:
\(\begin{bmatrix} 6 \end{bmatrix}\)
Step 2: Calculate the determinant of this submatrix. The determinant of a 1x1 matrix is the element itself:
\(\det \begin{bmatrix} 6 \end{bmatrix} = 6\)
Step 3: Apply the cofactor formula. The cofactor \(C_{ij}\) is given by:
\(C_{ij} = (-1)^{i+j} \cdot \text{determinant of the minor}\)
Here, \(i = 1\), \(j = 1\):
\(C_{11} = (-1)^{1+1} \cdot 6 = (+1) \cdot 6 = 6\)
So the cofactor of the element in the first row and first column is:
\(6\)
Q3(a). Matrix Operations: Explain the addition of matrices and add \(\begin{bmatrix} 2 & 6 \\ 5 & 8 \end{bmatrix}\) and \(\begin{bmatrix} 7 & 0 \\ 4 & 3 \end{bmatrix}\).
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Matrix Addition
Matrix addition is done by adding corresponding elements of two matrices. Both matrices must have the same size (same number of rows and columns).
Consider the matrices:
\(A = \begin{bmatrix} 2 & 6 \\ 5 & 8 \end{bmatrix}\) and \(B = \begin{bmatrix} 7 & 0 \\ 4 & 3 \end{bmatrix} \)
To add \(A + B\), add each element in the same position:
- First row, first column: \(2 + 7 = 9\)
- First row, second column: \(6 + 0 = 6\)
- Second row, first column: \(5 + 4 = 9\)
- Second row, second column: \(8 + 3 = 11\)
So, the resulting matrix is:
\(A + B = \begin{bmatrix} 9 & 6 \\ 9 & 11 \end{bmatrix}\)
Q3(b). What is scalar multiplication of a matrix?
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Scalar Multiplication of a Matrix
Scalar multiplication is an operation where we multiply every element of a matrix by a single number, called a scalar. The resulting matrix has the same size as the original, but each element is scaled by the scalar.
Definition:
If \(A\) is an \(m \times n\) matrix and \(k\) is a scalar, then:
\(k \cdot A = \begin{bmatrix} k \cdot a_{11} & k \cdot a_{12} & \dots & k \cdot a_{1n} \\ k \cdot a_{21} & k \cdot a_{22} & \dots & k \cdot a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ k \cdot a_{m1} & k \cdot a_{m2} & \dots & k \cdot a_{mn} \end{bmatrix} \)
Example:
Consider the matrix:
\(A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \)
and let the scalar \(k = 5\). Then:
\(5 \cdot A = \begin{bmatrix} 5 \cdot 1 & 5 \cdot 3 \\ 5 \cdot 2 & 5 \cdot 4 \end{bmatrix} = \begin{bmatrix} 5 & 15 \\ 10 & 20 \end{bmatrix} \)
Properties of Scalar Multiplication:
- Distributive over matrix addition: \(k \cdot (A + B) = k \cdot A + k \cdot B\)
- Associative with scalars: \(k \cdot (l \cdot A) = (k \cdot l) \cdot A\)
- Multiplying by 1: \(1 \cdot A = A\)
- Multiplying by 0: \(0 \cdot A = \text{zero matrix of same size as } A\)
Scalar multiplication is widely used in linear algebra, computer graphics, physics, and engineering to scale matrices while preserving their structure.
Q3(c). Describe the process of solving simultaneous linear equations using matrices.
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Solving Simultaneous Linear Equations Using Matrices
Simultaneous linear equations can be solved efficiently using matrices. The main idea is to represent the system in matrix form and then use matrix operations to find the solution.
Step 1: Write the system in matrix form
Consider the system of equations:
\(\begin{cases} 2x + 3y = 8 \\ 4x - y = 2 \end{cases}\)
This can be written in matrix form as:
\(AX = B\)
where
\(A = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}, \quad
X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad
B = \begin{bmatrix} 8 \\ 2 \end{bmatrix}\)
Step 2: Find the inverse of the coefficient matrix (if it exists)
If \(\det(A) \neq 0\), the inverse \(A^{-1}\) exists and the solution is:
\(X = A^{-1} B\)
Step 3: Calculate the determinant of \(A\)
\(\det(A) = (2 \cdot -1) - (3 \cdot 4) = -2 - 12 = -14\)
Since the determinant is not zero, the inverse exists.
Step 4: Find the inverse of \(A\)
For a 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the inverse is:
\(A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
Applying this to our matrix:
\(A^{-1} = \frac{1}{-14} \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 1/14 & 3/14 \\ 2/7 & -1/7 \end{bmatrix}\)
Step 5: Multiply \(A^{-1}\) by \(B\) to get \(X\)
\(X = A^{-1} B = \begin{bmatrix} 1/14 & 3/14 \\ 2/7 & -1/7 \end{bmatrix} \begin{bmatrix} 8 \\ 2 \end{bmatrix}\)
Performing the multiplication:
\(x = (1/14 \cdot 8) + (3/14 \cdot 2) = 8/14 + 6/14 = 14/14 = 1\)
\(y = (2/7 \cdot 8) + (-1/7 \cdot 2) = 16/7 - 2/7 = 14/7 = 2\)
Step 6: Write the solution
\(X = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\)
So the solution to the system is:
\(x = 1, \quad y = 2\)
Summary:
- Represent the system as \(AX = B\)
- Check that \(\det(A) \neq 0\) to ensure the inverse exists
- Find \(A^{-1}\)
- Multiply \(A^{-1} B\) to get the solution matrix \(X\)
This method can be extended to larger systems using the same principle with bigger matrices.
Q3(d). Use Cramer’s rule to solve the system of equations \(2x + 3y = 5\) and \(4x - y = 11\).
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Solving a System of Equations Using Cramer's Rule
Consider the system of equations:
\( 2x + 3y = 5\)
\( 4x - y = 11 \)
Step 1: Write the coefficient matrix
The coefficient matrix \(A\) is:
\(A = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}\)
Step 2: Find the determinant of \(A\)
For a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the determinant is \(ad - bc\):
\(\det(A) = (2 \cdot -1) - (3 \cdot 4) = -2 - 12 = -14\)
Since \(\det(A) \neq 0\), the system has a unique solution.
Step 3: Form matrices for \(x\) and \(y\)
Replace the columns of \(A\) with the constants from the right-hand side:
\(A_x = \begin{bmatrix} 5 & 3 \\ 11 & -1 \end{bmatrix}, \quad
A_y = \begin{bmatrix} 2 & 5 \\ 4 & 11 \end{bmatrix}\)
Step 4: Find determinants of \(A_x\) and \(A_y\)
\(\det(A_x) = (5 \cdot -1) - (3 \cdot 11) = -5 - 33 = -38\)
\(\det(A_y) = (2 \cdot 11) - (5 \cdot 4) = 22 - 20 = 2\)
Step 5: Apply Cramer's Rule
Cramer's Rule states:
\(x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}\)
\(x = \frac{-38}{-14} = \frac{19}{7}, \quad y = \frac{2}{-14} = -\frac{1}{7}\)
Step 6: Solution
\(\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 19/7 \\ -1/7 \end{bmatrix}\)
The system has a unique solution: \(x = \frac{19}{7}\), \(y = -\frac{1}{7}\).
Q4(a). Inverse Matrix: Find the inverse of \(\begin{bmatrix} 5 & 9 \\ 4 & 1 \end{bmatrix}\).
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Finding the Inverse of a 2x2 Matrix
Consider the matrix:
\(A = \begin{bmatrix} 5 & 9 \\ 4 & 1 \end{bmatrix}\)
Step 1: Calculate the determinant
For a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the determinant is:
\(\det(A) = ad - bc\)
Substitute the values:
\(\det(A) = (5 \cdot 1) - (9 \cdot 4) = 5 - 36 = -31\)
Since the determinant is not zero, the inverse exists.
Step 2: Use the formula for the inverse
For a 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the inverse is:
\(A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
Substitute the values from \(A\):
\(A^{-1} = \frac{1}{-31} \begin{bmatrix} 1 & -9 \\ -4 & 5 \end{bmatrix}\)
Step 3: Simplify using fractions
\(A^{-1} = \begin{bmatrix} \frac{-1}{31} & \frac{9}{31} \\[2mm] \frac{4}{31} & \frac{-5}{31} \end{bmatrix}\)
\(A^{-1} = \begin{bmatrix} \frac{-1}{31} & \frac{9}{31} \\[1mm] \frac{4}{31} & \frac{-5}{31} \end{bmatrix}\)
Q4(b). Conversion: Convert \((101101)_2\) to decimal.
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Converting Binary to Decimal
Let's convert the binary number \((101101)_2\) to decimal step by step.
Step 1: Write down the place values
Binary is base 2, so each digit represents a power of 2. From right to left, the powers are \(2^0, 2^1, 2^2, 2^3, 2^4, 2^5\).
\((101101)_2 = 1 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0\)
Step 2: Calculate each term
\(1 \cdot 2^5 = 32, \quad 0 \cdot 2^4 = 0, \quad 1 \cdot 2^3 = 8, \quad 1 \cdot 2^2 = 4, \quad 0 \cdot 2^1 = 0, \quad 1 \cdot 2^0 = 1\)
Step 3: Add the results
\(32 + 0 + 8 + 4 + 0 + 1 = 45\)
\((101101)_2 = (45)_{10}\)
Q4(c). Binary Arithmetic: Add \((1101)_2\) and \((1011)_2\).
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Binary Arithmetic: Addition
We want to add the binary numbers:
\((1101)_2 + (1011)_2\)
Step 1: Align the numbers
1101
+ 1011
Step 2: Add column by column (right to left)
Binary addition rules:
- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = 10 (write 0, carry 1)
Add each column:
Rightmost column: \(1 + 1 = 10\) → write 0, carry 1
Next column: \(0 + 1 + 1 \text{(carry)} = 10\) → write 0, carry 1
Next column: \(1 + 0 + 1 \text{(carry)} = 10\) → write 0, carry 1
Leftmost column: \(1 + 1 + 1 \text{(carry)} = 11\) → write 1, carry 1
Step 3: Write the result including final carry
\(\text{Sum} = (11000)_2\)
Q5(a). Find the product of the matrices and add. \(A = \begin{bmatrix} 2 & 0 \\ -5 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 5 & 0 \\ 8 & -3 \end{bmatrix}\)
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Matrix Operations: Addition and Multiplication
Consider the matrices:
\(A = \begin{bmatrix} 2 & 0 \\ -5 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 5 & 0 \\ 8 & -3 \end{bmatrix}\)
Step 1: Matrix Addition
Matrix addition is performed element by element:
\(A + B = \begin{bmatrix} 2+5 & 0+0 \\ -5+8 & 5+(-3) \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 3 & 2 \end{bmatrix}\)
Step 2: Matrix Multiplication
Matrix multiplication is done by taking the dot product of rows of \(A\) with columns of \(B\):
\(A \cdot B = \begin{bmatrix}
(2\cdot5 + 0\cdot8) & (2\cdot0 + 0\cdot-3) \\
(-5\cdot5 + 5\cdot8) & (-5\cdot0 + 5\cdot-3)
\end{bmatrix}\)
Step 3: Simplify each element
Top-left: \(2\cdot5 + 0\cdot8 = 10\)
Top-right: \(2\cdot0 + 0\cdot-3 = 0\)
Bottom-left: \(-5\cdot5 + 5\cdot8 = -25 + 40 = 15\)
Bottom-right: \(-5\cdot0 + 5\cdot-3 = 0 - 15 = -15\)
Step 4: Write the final product
\(A \cdot B = \begin{bmatrix} 10 & 0 \\ 15 & -15 \end{bmatrix}\)
Matrix Addition: \(A + B = \begin{bmatrix} 7 & 0 \\ 3 & 2 \end{bmatrix}\)
Matrix Multiplication: \(A \cdot B = \begin{bmatrix} 10 & 0 \\ 15 & -15 \end{bmatrix}\)
Q5(b). Discuss how the letter A is 01000001 in ASCII (a binary-based encoding).
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How the Letter "A" is Represented in ASCII
ASCII (American Standard Code for Information Interchange) is a binary-based encoding system used to represent text characters in computers. Each character is assigned a unique 7-bit or 8-bit binary number. Computers use this binary code to store and process text.
Step 1: Find the decimal value of "A"
In the ASCII table, the uppercase letter "A" has a decimal value of 65.
Step 2: Convert the decimal value to binary
To convert 65 to binary, divide by 2 repeatedly and record the remainders:
- 65 ÷ 2 = 32, remainder 1 (least significant bit)
- 32 ÷ 2 = 16, remainder 0
- 16 ÷ 2 = 8, remainder 0
- 8 ÷ 2 = 4, remainder 0
- 4 ÷ 2 = 2, remainder 0
- 2 ÷ 2 = 1, remainder 0
- 1 ÷ 2 = 0, remainder 1 (most significant bit)
Now, write the remainders from most significant bit to least significant bit:
\((65)_{10} = (01000001)_2\)
Step 3: Representation in ASCII
This binary number \((01000001)_2\) is how the computer stores and recognizes the letter "A". Each time "A" is typed or processed, the system internally uses this 8-bit binary code.
Summary:
- ASCII assigns decimal numbers to characters
- The letter "A" has a decimal value of 65
- Converting 65 to binary gives \((01000001)_2\)
- This binary code is how computers represent "A"
Q5(c). Adding Two Numbers in Binary. Suppose a computer adds two decimal numbers Find the sum of these two numbers in Binary.
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Adding Two Numbers in Binary
Suppose we want to add two decimal numbers using a computer. Computers perform addition in binary (base 2) rather than decimal (base 10).
Example: Add 13 and 11 in binary.
Step 1: Convert decimal numbers to binary
\((13)_{10} = (1101)_2\)
\((11)_{10} = 1011_2\)
Step 2: Align the binary numbers
1101
+ 1011
Step 3: Add column by column (right to left)
Binary addition rules:
- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = 10 (write 0, carry 1)
Add each column:
Rightmost column: \(1 + 1 = 10\) → write 0, carry 1
Next column: \(0 + 1 + 1 \text{(carry)} = 10\) → write 0, carry 1
Next column: \(1 + 0 + 1 \text{(carry)} = 10\) → write 0, carry 1
Leftmost column: \(1 + 1 + 1 \text{(carry)} = 11\) → write 1, carry 1
Step 4: Write the result including final carry
\(\text{Sum} = (11000)_2\)
Step 5: Verification (optional)
Convert back to decimal:
\((11000)_2 = (24)_{10}\)
\(13 + 11 = 24\)
The sum of 13 and 11 in binary is \((11000)_2\)
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