Category: Physics

A concise Introduction to thin lenses
Table of Contents
Introduction to thin lenses involves general description about properties of thin lenses.

A lens can be defined as a piece of curved glass or plastic that makes things look larger, smaller or clearer when you look through it.

In human eye, one component is a lens and so we can also define lens as the transparent part of the eye, behind the pupil, that focuses light so that you can see clearly.

The idea behind lens operations is that when a light ray passes from air which is more optically denser than the lens material, it is refracted.

When many rays passes through the lens, they all refracted the same way and so they meet at a common point. Sometimes they don’t meet but instead they are scattered after refraction but they are seemed to be spreading from a common point.

Lenses are usually made of glass, transparent plastic or perspex.

common application of lenses includes cameras, spectacles,telescopes, microscopes, film projectors and the human eye.

A thin lens means a lens whose thickness is negligible compared to the radius of curvature of the lens surfaces.

Types of thin lenses

The basic two types of lenses are convex and concave lenses.

Convex lenses are also called converging lenses as they cause the rays that passes through it to meet at a point. Convex lenses are thickest at the middle and they thin in as you move towards their edge.

In this lesson we will be talking about biconvex lenses meaning that it is symmetrical if we cut it long it’s edges. Both sides of it’s services at the center are bulging outwards and the edges are curved inwards uniformly on both sides.see the figure below

Bi convex lens

showing symmetrical in bi-convex lens

Concave lenses are also called diverging lenses as they cause the rays passing through them to be spreading from a common pint. Concave lenses are thinnest at the middle and they they become thicker as you move towards the edges.

Illustrating concave lens

illustrating bi-concave lens

There are variations of convex and concave lenses as illustrated in figures below

plano convex lens

convex meniscus lens

plano concave lens

concave meniscus lens

Effects of lenses on Parallel rays of light

A cardboard with parallel slits is placed between the mirror and a bi-convex lens as in figure below

The mirror is set such that it reflects the sun rays so that the rays passes through the slits before they reach the lens.

After making observations, the bi-convex lens is replaced with a concave lens

Observation

when a convex lens is used, the rays are converged at a point on the paper and then diverge as they continue as shown.

illustrations of parallel rays as they pass through a bi-convex lens

When concave lens is used , the rays diverge as if they were from the focal point in front of the lens as shown.

illustrations of parallel rays as they pass through a biconcave lens

Investigating convergence and divergence of light by thin lenses using a ray box

A ray box acts as a source of parallel beam. A spot light can also be used.

A parallel beam is directed incident to the the lens as shown

Parallel rays of light incident to a convex lens

A white paper is placed on the other side of the lens and it’s position adjusted until a sharp point is observed.

observation

When a convex lens is used, the rays are converged at a point on the paper and then diverges as they continue as shown below

Parallel beam after passing through a converging lens

If convex lens was replaced with concave (diverging) lens, the rays will be observed diverging as if they are coming from a point on the other side of the lens. see the diagram below.

Parallel beam incident to diverging lens

Explanations

Light is usually refracted when it passes through a glass prism. A lens can be considered as an assembly of many tiny prisms where each prism refracts light as in figure below.

Illustrations of bi-convex lens as an assembly of prisms

Please note that, the middle part of the prism is like a rectangular glass prism and a ray that is incident to it at a perpendicular angle passes through without being refracted. As we may see in other lessons, a ray of light that passes normally through the geometrical center of the lens, passes through undeviated.

The figure below shows representation of concave lens as an assembly of prisms.

illustrations of concave lens as an assembly of prisms

conclusions

Rays of light that passes through a lens converges at a fixed point from the lens if the lens is a converging lens or diverge from a common imaginary point if the lens is a diverging lens.

The point at which the rays emerging from the lens converge or seems to diverge from is referred to as the principal focus.

A convex lens has a real principal focus while a concave lens has a virtual (imaginary) principal focus.

Related Topics
References:
- Secondary Physics Student’s Book Four. 3rd ed., Kenya Literature Bureau, 2012. pp. 1-42.
- Abbot A. F. (1980), Ordinary Level Physics, 3rd Edition, Heinemann Books International,
  London.
- Nelkon M. and Parker P., (1987), Advanced Level Physics, Heinemann Educational
  Publishers, London.
- Tom D., and Heather K. Cambridge IGCSE Physics. 3rd ed., Hodder Education, 2018, https://doi.org/978 1 4441 76421. pp. 106-142.
March 10, 2026
Questions involving pressure in Physics
Questions involving pressure are physics questions that test your understanding of the concept of pressure, how it is calculated, and how it applies in solids, liquids, and gases.

They may involve:
- Calculating pressure using formulas
- Explaining real-life applications
- Understanding liquid pressure
- Understanding atmospheric pressure
- Applying Blaise Pascal’s Principle
Here are questions involving pressure
1. The figure below shows Hare’s apparatus used for comparing liquid densities.
Use the information given in the above diagram to calculate the density of liquid x given that density of water is 1000kgm^-3(2mks)

answer

$$ \delta_w g h_w = \delta_wgh_x $$ $$delta_x = \frac{1000 \times 30}{25} = 1200kgm^{-3}$$ $$1000 \times 30 = \delta_x \times 25 \ $$

2. Figure 3 below shows a student drinking a soda using a straw.

Figure 3

. Explain why the soda rises up a straw when the student sucked on it.                 (2mks)

answer

Sucking decreases pressure inside the straw; and the greater atmospheric pressure pushes the soda up.

3. Using the idea of particles explain why the pressure inside the tyre increases when the tyre is pumped up     (2mks)

answer

When more air is pumped into the tyre the number of particles colliding with the walls increases; They increase the rate of change in momentum, hence the force per unit area increases;

4. Explain why a hole in a ship near the bottom is more dangerous than one nearer the surface.(2mks)

answer

Pressure in liquids increases with depth ; hence water will enter the ship at a higher rate making it to sink

5. The atmospheric pressure at the foot of Mt. Longonot is 760mm of mercury while at the peak of the mountain, the atmospheric pressure is 580mm of mercury. Given that the density of mercury is 13600Kg/m3, Calculate the height of the mountain (density of air = 1.3 Kg/m3.) (3mks)

Answer

Pressure difference = 760 – 580 = 180mmHg;
Pressure due to liquid column = ρgh – 0.18 x 10 x 13600 = 1.3 x h x 10;
H = 1883.1m;

6. Use fig. 3 below to answer questions 3 and 4.

Three identical tubes containing mercury were inverted as shown.

Explain the effect on the level of mercury in tube A if region X is filled with some air. (2 mks)

answer

The vertical height of mercury would be less than 76 cm. The trapped air would exert pressure on the column of mercury.

Indicate on the diagram above the levels of mercury in tube B and C. (1 mk)

answer
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March 9, 2026
Perfect squares
A perfect square is a number that is obtained when a whole number is multiplied by itself. In other words, it is the result of squaring an integer. For example, $1 = 1 \times 1$ $1 = 1 \times 1$ 1=1×1, $4 = 2 \times 2$ $4 = 2 \times 2$ 4=2×2, $9 = 3 \times 3$ $9 = 3 \times 3$ 9=3×3, and $16 = 4 \times 4$ $16 = 4 \times 4$ 16=4×4. These numbers are called perfect squares because they can be arranged to form a perfect square shape.

Perfect squares are important in mathematics because they help us understand square roots, area, and patterns in numbers. The square root of a perfect square is always a whole number. For instance, the square root of 25 is 5, since $5 \times 5 = 25$ $5 \times 5 = 25$ 5×5=25.

in quadratic expressions, a perfect square occurs when the expression can be written as the square of a binomial. This means the quadratic can be expressed in the form: $(a + b)^2 \quad \text{or} \quad (a – b)^2$ $(a + b)^2 \quad \text{or} \quad (a - b)^2$

When expanded, these forms follow special patterns: $(a + b)^2 = a^2 + 2ab + b^2$ $(a – b)^2 = a^2 – 2ab + b^2$

A quadratic expression is called a perfect square trinomial if:
1. The first term is a perfect square.
2. The last term is a perfect square.
3. The middle term is twice the product of the square roots of the first and last terms.
Examples

x²+6x+9 is a perfect square

x² = (x) x (x) =(x)²

9 = 3 x 3 = (3)²

middle term : 6x = 2(x)(3)

Therefore:

x²+6x+9 = (x+3)²

Example 2

assume x²−10x+25 is a perfect square

x²= (x) (x) = (x)²

25 = (5)²

middle term:-10x = -2(x)(5)

therefore x²−10x+25 = (x-5)²

Importance in Solving Quadratics

Perfect square expressions are useful because:
- They make factorization easier.
- They help in completing the square, a method used to solve quadratic equations.
- They simplify graphing, since expressions like $(x – a)^2$ clearly show the vertex of a parabola.
Example problem

factorize x² – 8x +16

solution

The constant term is a perfect square

middle term is twice the constant term

hence 16 = (4)(4)

-8 = -4-4

hence x² – 8x +16 = (x-4)(x-4) = (x-4)²

example

factorize:

$$x^2+\frac{4}{3}x+\frac{4}{9}$$

solution

$$\frac{4}{9} = \frac{2}{3} \ or \ (\frac{-2}{3})$$

$$\frac{4}{3}= \frac{2}{3}+\frac{2}{3}$$

$$x^2 + \frac{4}{3}x + \frac{4}{9} = (x+\frac{2}{3})^2$$

Example

factorize 4x²+20x+25

solution

4x²= (2x)(2x)

25=(5)(5)

hence 4x²+20x+25 = (2x+5)(2x+5) = (2x+5)²

completing the square

Completing the square is a method used to solve quadratic equations and rewrite quadratic expressions in a special form. It helps us change a quadratic expression into a perfect square trinomial, which can then be factored easily.

consider the expression x² +bx

we can add a constant term such that the expression becomes a perfect square.

To complete the square:

Take half of ,square it and then add it to the expression.

that is: add (b/2)² to the expression

Example problem on completing the square

what must be added to x² -18x to make it a perfect square?

$$\text{we must add}: (\frac{1}{2} \times (-18)^2 =(-9)^2 =81$$

so x² -18x is transformed to x² -18x+81 to be a perfect square.

Related Topics
February 23, 2026
Upthrust in gases
Upthrust in gases is the upward force exerted on an object immersed in a gas, which opposes the object’s weight. Upthrust in gases acts as an upward force on an object immersed in a gas and opposes the object’s weight. It arises from the pressure difference between the top and bottom of the object due to the gas’s density. This concept is similar to upthrust in liquids but involves the behavior of gases, which are much less dense than liquids.

When an object is placed in a gas, the pressure at the bottom of the object is greater than the pressure at the top. This difference in pressure results in an upward force (upthrust). This is what that results to upthrust in gases. The size of this force depends on the volume of the displaced gas and the density of the gas.

Just like liquids, gases exerts upthrust on objects that are in them. Air is the most common gas of interest and so it will be used extensively to illustrate upthrust in gases. We float objects on air. Sometimes, as human beings, we use parachutes to float in air.

The upthrust in gases is small because air has lower density compared to most of substances.

The density of air is about 1.3kgm^-3 or 0.0013gcm^-3.

A balloon can float in air if it contains a gas with a lower density than air. For example hydrogen has a density of 0.09 kgm^-3 whereas helium has a density of 0.18kg^m-3. Therefore a balloon filled with helium or hydrogen will rise on air provided density of the balloon fabric and air will be less than density of air.

Problems involving upthrust in gases

Consider the figure below that illustrates a balloon filled with air .

If we consider the balloon filled with air to a certain volume, the weight of air in the balloon plus its fabric is greater than the weight of air displaced. This is because the volume of air in the balloon is nearly equal to the volume of air displaced.

The upthrust force on the balloon due to the air is thus less than the weight. The balloon thus stay grounded because the it’s weight is less than the upthrust force that could set it up to float on air.

That is W-U > resultant downward forces.

If the balloon is filled with a gas that has a lower density than air, the gas and balloon fabric weigh less than the displaced air. The upthrust force U exerted by the air on the balloon is greater than the weight W of the inflated balloon. This results in the upthrust force being larger. The resultant upward force is greater than W-U and hence the balloon sets to accelerate upward.see the illustrations below.

Example Question

A meteorological balloon has a volume of of 55m³ and is filled with a helium gas of density 0.18kgm^-3. If the weight of the balloon fabric is 170N, calculate the maximum load the balloon can lift given that density of air is 1.3kgm^-3

solution

Volume of air displaced by the balloon is 55m³ which is volume occupied by the balloon.

mass of air displaced by the balloon = 55m³ x 1.3kgm^-3 = 71.5kg

weight of air displaced =71.5kg x 10 Nkg^-1 = 715N

mass of helium in the balloon = 55m³ x 0.18kgm^-3 = 9.9Kg

weight of helium in the ballon = 9.9kg x 10 Nkg^-1 = 99N

Total weight of the inflated balloon = 170N + 99N = 269 N

From the law of floatation, upthrust is the weight of air displaced which should also be weight of the balloon plus the weight of load it will carry. hence,

upthrust = weight of the balloon + load in the ballon = weight of air displaced

269N + load in the ballon = 715N

load in the balloon = (715-269)N=446N

So the total mass of the goods to be included in the balloon should never exceeded 44.6kg.

Related topics
February 6, 2026
Alphabetical letters (A B C)
Alphabetical letters are the basic symbols used to form words and communicate ideas in written language. They help us read, write, and understand information in an organized and meaningful way.

capita Alphabetical letters

small alphabetical letters

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February 2, 2026
Historical sites in Kenya
Kenya has a rich tapestry of history and heritage — from ancient Swahili cities and prehistoric fossil sites to colonial forts and caves that tell the story of the fight for independence. Here are some notable historical sites in Kenya you might find fascinating:

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February 1, 2026
Sources of water
Sources of water are the natural places from which humans, animals, and plants obtain water for their daily needs. The main source of water is rain, which supplies rivers, lakes, and underground water. Surface water comes from rivers, lakes, streams, and dams, while groundwater is found below the earth’s surface in wells, boreholes, and springs. Oceans and seas are also sources of water, although their water is salty and must be treated before it can be used for drinking. These sources of water are very important because they support life, agriculture, and economic activities.

Related topics
January 31, 2026
UPTHRUST IN GASES
1. Introduction

Gases, just like liquids, are fluids and can exert forces on objects placed in them. Although air cannot be seen, it has mass, occupies space, and exerts pressure. When an object is placed in air, it experiences an upward force known as upthrust. This force explains why balloons rise, smoke moves upward, and parachutes slow down falling objects.

2. What Is Upthrust?

Upthrust (buoyant force) is the upward force exerted by a fluid (liquid or gas) on an object immersed in it.

In gases, upthrust occurs because air pressure increases with depth. The pressure acting on the bottom of an object is greater than the pressure acting on the top, resulting in a net upward force.

3. Archimedes’ Principle in Gases

Archimedes’ principle also applies to gases and states that:

An object wholly or partially immersed in a gas experiences an upthrust equal to the weight of the gas displaced.

This means the greater the volume of air displaced by an object, the greater the upthrust acting on it.

4. Mathematical Expression for Upthrust in Gases

The upthrust in gases can be calculated using the formula: $U = \rho g V$ U=ρgV

Where:
- $U$ U = upthrust (N)
- $\rho$ ρ = density of the gas (kg/m³)
- $g$ g = acceleration due to gravity (m/s²)
- $V$ V = volume of gas displaced (m³)
5. Conditions for Floating, Rising, and Falling in Air

Illustrating upthrust in gases

Comparison of Forces Result
Upthrust > Weight of object Object rises
Upthrust = Weight of object Object floats
Upthrust < Weight of object Object falls

6. Illustrations of Upthrust in Gases

(a) Balloons

A helium or hot-air balloon rises because it displaces air that is denser than itself. The upthrust acting on the balloon is greater than its weight, causing it to rise upward.

(b) Parachutes

When a parachute opens, it increases the surface area and displaces more air. The upthrust (and air resistance) acting upward increases, reducing the downward motion of the parachutist and allowing a safe landing.

(c) Smoke and Hot Air

Hot air and smoke rise because they are less dense than the surrounding cooler air. The upthrust acting on them is greater than their weight.

7. Worked Examples

Example 1:

An object displaces 0.1 m³ of air. Given that its’ density of air = 1.2 kg/m³ and the acceleration due to gravity = 10 m/s². calculate upthrust in air.

Calculate the upthrust acting on the object.

Solution: $U = \rho g V$ U=ρgV $U = 1.2 \times 10 \times 0.1$ U=1.2×10×0.1 $U = 1.2 \, \text{N}$ U=1.2N

Upthrust = 1.2 N

Example 2: Determining Motion of an Object in Air

A balloon has a weight of 0.9 N.
Upthrust acting on it is 1.2 N.

Since: $\text{Upthrust} > \text{Weight}$ Upthrust>Weight

The balloon will rise.

Example 3: Volume of Air Required for Floating

An object has a weight of 2.4 N.
Density of air = 1.2 kg/m³, $g = 10 \, \text{m/s}^2$ g=10m/s2

Calculate the minimum volume of air the object must displace to float. $2.4 = 1.2 \times 10 \times V$ 2.4=1.2×10×V $V = \frac{2.4}{12}$ V=122.4 $V = 0.2 \, \text{m}^3$ V=0.2m3

Required volume = 0.2 m³

8. Problems Involving Upthrust in Gases
1. A balloon displaces 0.25 m³ of air.
  Density of air = 1.2 kg/m³, $g = 10 \, \text{m/s}^2$ g=10m/s2.
  Calculate the upthrust acting on the balloon.
2. An object weighs 3 N and experiences an upthrust of 2 N in air.
  State whether the object will rise, float, or fall, giving a reason.
3. Explain why a parachutist falls slowly after opening a parachute.
4. A hot-air balloon displaces 1.5 m³ of air.
  Calculate the upthrust acting on the balloon.
5. State two applications of upthrust in gases in everyday life.
9. Conclusion

Upthrust in gases is an important concept that explains the behavior of objects in air. It depends on the density of the gas, the volume displaced, and gravity. Understanding upthrust helps explain real-life phenomena such as balloons rising, parachutes slowing descent, and hot air moving upward.

Related topics
January 29, 2026

Comparison of Forces	Result
Upthrust > Weight of object	Object rises
Upthrust = Weight of object	Object floats
Upthrust < Weight of object	Object falls

The Key Concepts in Matrices

Key matrices concepts include identity, determinant, inverse, and singular matrices, which help describe matrix properties and transformations.

Matrices are important tools in mathematics and science. They are used to solve systems of equations, describe transformations, and model real-life problems such as economics, physics, and computer graphics.
In this lesson, we will learn four important concepts related to square matrices:

Identity Matrix
Determinant of a Matrix
Inverse of a Matrix
Singular Matrix

2. Identity Matrices

Definition

An identity matrix is a square matrix that has:

1’s on the main diagonal
0’s everywhere else

It is similar to the number 1 in multiplication, because multiplying any matrix by the identity matrix gives the same matrix.

Notation of matrices

The identity matrix is denoted by I.

2 × 2 Identity Matrix

$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

3 × 3 Identity Matrix

$I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ I3=100010001

Illustration

Main diagonal →   1   1   1
Other entries →   0   0   0

Property

$A \times I = I \times A = A$

3. Determinant of a Matrix

Definition

The determinant of a square matrix is a single number that tells us important information about the matrix, such as:

Whether the matrix has an inverse
Whether the system of equations has a unique solution
The scaling factor of a geometric transformation

Determinant of a 2 × 2 Matrix

For a matrix: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The determinant is: $|A| = ad – bc$

Example

$A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ A=[3124] $|A| = (3 \times 4) – (2 \times 1) = 12 – 2 = 10$ ∣A∣=(3×4)−(2×1)=12−2=10

The determinant represents area scaling:

If |A| = 2 → area doubles
If |A| = 0 → area collapses to a line

4. Inverse of a Matrices

Definition

The inverse of a matrix A is another matrix $A^{-1}$ A−1 such that: $A \times A^{-1} = I$

Inverse of a 2 × 2 Matrix

For: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $A^{-1} = \frac{1}{ad – bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Note: The inverse exists only if the determinant is not zero.

Example

$A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$

Step 1: Find determinant: $|A| = (2 \times 3) – (1 \times 5) = 6 – 5 = 1$

Step 2: Find inverse: $A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$

Illustration

Matrix A → changes a vector  
Inverse A⁻¹ → returns it back to original

5. Singular Matrix

Definition

A singular matrix is a matrix that has no inverse.

$$\matrix{(a & b &c)}$$

Condition

A matrix is singular if: ∣A∣=0

Example

$A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$ A=[2142] $|A| = (2 \times 2) – (4 \times 1) = 4 – 4 = 0$ ∣A∣=(2×2)−(4×1)=4−4=0

Therefore, A is singular and cannot be inverted.

Illustration

Transformation squashes shape into a line → no way to reverse it

$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

6. Summary Table

Concept	Meaning	Key Condition
Identity Matrix	Matrix with 1’s on diagonal	Acts like number 1
Determinant	Single number describing matrix properties	The matrix is not a zero matrix
Inverse Matrix	Matrix that reverses A	Exists if A exists
Singular Matrix	Matrix without inverse

7. Practice Questions

Question 1

Find the determinant of: $A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}$

Answer:
$|A| = (4×6) − (7×2) = 24 − 14 = 10$

Question 2

Write the 3 × 3 identity matrix.

Answer: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Question 3

Determine whether the matrix is singular: $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$

Answer:
$|A| = (1×6) − (3×2) = 6 − 6 = 0$

Matrix is singular.

Question 4

Find the inverse of: $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Answer: $|A| = (1×4) − (2×3) = 4 − 6 = -2$ $A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

8. Conclusion

Understanding identity, determinant, inverse, and singular matrices is important in solving systems of equations, physics, engineering, and computer science. These concepts help us understand when a matrix can be reversed and how it transforms.

Category: Physics

Table of Contents

Types of thin lenses

Effects of lenses on Parallel rays of light

Observation

Investigating convergence and divergence of light by thin lenses using a ray box

observation

Explanations

conclusions

Related Topics

References:

answer

answer

answer

answer

Answer

Related topics

Examples

Importance in Solving Quadratics

Example problem

solution

solution

completing the square

Example problem on completing the square

Problems involving upthrust in gases

Example Question

solution

Related topics

capita Alphabetical letters

small alphabetical letters

Related posts

Related topics

Related topics

1. Introduction

2. What Is Upthrust?

3. Archimedes’ Principle in Gases

4. Mathematical Expression for Upthrust in Gases

5. Conditions for Floating, Rising, and Falling in Air

6. Illustrations of Upthrust in Gases

(a) Balloons

(b) Parachutes

(c) Smoke and Hot Air

7. Worked Examples

Example 1:

Example 2: Determining Motion of an Object in Air

Example 3: Volume of Air Required for Floating

8. Problems Involving Upthrust in Gases

9. Conclusion

Related topics

2. Identity Matrices

Definition

Notation of matrices

2 × 2 Identity Matrix

3 × 3 Identity Matrix

Illustration

Property

3. Determinant of a Matrix

Definition

Determinant of a 2 × 2 Matrix

Example

4. Inverse of a Matrices

Definition

Inverse of a 2 × 2 Matrix

Example

Illustration

5. Singular Matrix

Definition

Condition

Example

Illustration

6. Summary Table

7. Practice Questions

Question 1

Question 2

Question 3

Question 4

8. Conclusion

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