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Strictly use Geogebra to solve all the questions

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Module 0

• Plot the points $$(2,3)$$ and $$(6,9)$$ and find the distance between them without using the formula for distance.

• Plot two vectors $$u=⟨3,−2⟩$$ and $$v=⟨−1,5⟩$$.

• Find the angle between them.

• Construct a perpendicular bisector of a given segment whose end points are $$(0,6)$$ and $$(6,0)$$.

• Plot line $$y=3x$$.

• Graph the equation $$y=2x+3$$ over the interval $$[−5,5]$$ and find the points where the graph intersects the x-axis and y-axis.

• Explore the behavior of the quadratic function $$f(x)=x^2−4x+3$$. Determine its vertex, axis of symmetry, and roots.

• Find the angle between the lines with equations $$y=2x+3$$ and $$y=−3x+2$$.

• Determine the angle of depression from the top of a building to a point on the ground 50 meters away from the base of the building, given that the height of the building is 30 meters.

• Construct a right angled triangle with legs of lengths 3 units and 4 units. Find the length of the hypotenuse.

• Find the centriod of the above triangle.

• Enter a set of parametric equations, such as $$x(t)=cos(t)$$ and $$y(t)=sin(t)$$ to plot a circle.

• Create two planes: Plane 1: $$2x−3y+z=6$$, Plane 2: $$x+4y−2z=8$$. Plot the two planes in a 3D coordinate system.

• Determine the angle between the two planes. Are the planes intersecting? Is there any unique point in common? What do you infer from this?

Module A

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1.Ram and Lakshman were two brothers, Ram's pocket money was twice as Lakshman. The good boys that Ram and Lakshman were, they did not spend their pocket money on anything. They instead saved the same in their piggy bank. Every week, they would check their savings so far. Assume the first week's savings was $$(R_1,L_1)$$ and second week's $$(R_2,L_2)$$ and so on. They try plotting their weekly savings on a graph sheet. How will the points look like?

Sol.

In GeoGebra, if you plot the weekly savings of Ram and Lakshman on a graph with the x-axis representing Lakshman's savings and the y-axis representing Ram's savings the points will form a straight line. This is because Ram's savings are always proportional to Lakshman's savings.

The points will follow a straight line passing through the origin as:

$$y = 2x$$

2.Atul's house is centered at origin $$(0,0)$$ he walks straight (along the x-axis) for 2 units and then takes a left and walks 1 unit to reach Bala's house, after that he takes a right turn and walks for one unit and then a left turn and walks for one unit and reaches Chetan's house. He continues in a similar style, takes a right turn 1 unit and then left turn one unit and reaches Divya's house. Are the houses of Bala, Chetan and Divya on a straight line? What is the equation of this line? Plot this on Geogebra

Sol.

Atul starts his journey from the origin \(O = (0, 0).\) He first walks 2 units along the x-axis to reach \((2, 0)\), then takes a left turn and walks 1 unit to reach Bala's house at \((2, 1).\) Next, he continues by taking a right turn from Bala's house and walking 1 unit to \((3, 1),\) followed by another left turn and walking 1 unit to reach Chetan's house at \((3, 2).\) Continuing in the same pattern, Atul takes a right turn from Chetan's house and walks 1 unit to \((4, 2),\) then takes a left turn and walks 1 unit to reach Divya's house at \((4, 3).\)

Now, we will plot these points on GeoGebra and make a line out of them taking two at a time. We can clearly observe that whether the points make a straight line.

3.Plot the lines $$y=x$$, $$y=2x$$, $$y=10x$$.

Sol.

To plot these lines using GeoGebra, in the input bar at the bottom, type the equations and they will be plotted on the graph. Using GeoGebra to plot these lines provides a clear visual representation of how different linear equations with varying slopes behave. We can observe that the line $$y = 10x$$ is the steepest of them all. and the line $$y = x$$ is the shallowest.

4.Observe that they all pass through the origin. Why?

Sol.

All of the lines provided in the previous question are of the form: $$y = kx,$$ where k is 1, 2 and 10. Here, y is defined as k times x, i.e., when we put \(k = 0,\) y will definitely be 0 as \(0 * t = 0\) for all t belonging to R. As Origin is \((0,0)\) point, All the lines pass through it.

5.Plot $$y=2x+1$$. Observe, Why doesn't it pass through the origin?

Sol.

The plotting is done similarly to the previous questions in thew Module. The line does not pass through the Origin because when x is considered to be 0, y comes out to be 2 * 0 + 1, which gives 1. Hence, it does not pass through the Origin, i.e., \((0,0).\)

6.Plot $$y=ax+b$$, with $$a$$ and $$b$$ as parameters which you should be able to vary. What do you observe?

Sol.

To explore the behavior of the linear equation \(y=ax+b\) with varying parameters \(a\) (slope) and \(b\) (y-intercept) using GeoGebra, you can use sliders to dynamically adjust these values and observe the resulting changes on the graph. As you adjust the slider for \(a\), you will notice that increasing \(a\) makes the line steeper, while decreasing it makes the line gentler. Adjusting the slider for \(b\) shifts the entire line vertically. Increasing \(b\) moves the line upwards, while decreasing \(b\) moves it downwards. This interactive approach in GeoGebra helps visualize and understand how the parameters a and b influence the equation of a line, providing a clear and dynamic way to grasp the fundamental properties of linear equations.

6(a). Let a line be $$y=5x+6$$. For what values of $$\alpha$$ and $$\beta$$ will the line $$y= \alpha x + \beta$$ be parallel to the given line? When will it intersect the given line in the 3rd quadrant?

Sol.

Two lines are only parallel if there slopes are equal. So for the line to be parallel to $$y = 5x + 6,$$ Alpha must be equal to 5, whereas the value of Beta does not change the lines being parallel. This can be observed by plotting the lines and using sliders on GeoGebra.

For the lines to intersect in third quadrant, solve the two equations and when we will get the values of x and y in the terms of (\alpha) and (\beta), make both x and y less than 0. We will get the values of (\alpha) and (\beta) upon solving the inequalities.

7.Consider the following simultaneous equation:

$$2x+3y=7$$ $$3x+4y=10$$

Do you see a 2x2 matrix here? What is the importance of seeing a matrix in this problem? Why study matrices in general?
Do you observe that this problem can be retold as:
$$ \left( \begin{matrix} 2 & 3 \3 & 4 \\end{matrix}\right) $$ $$\left( \begin{matrix} x\ y\ \end{matrix} \right)$$ =$$\left( \begin{matrix} 7\ 10\ \end{matrix} \right)$$

Sol.

Yes, I see a 2x2 matrix here. Recognizing the system of equations in matrix form is crucial because it allows us to use matrix operations to solve the system more efficiently, especially when dealing with larger systems. By representing the coefficients of the variables and the constants in matrix form, we can apply many useful techniques to find the solutions. Studying matrices in general is important because they offer a powerful way to represent and manipulate data in various fields. Matrices provide a concise and structured way to represent complex systems and relationships, making it easier to analyze and solve problems.

When we rewrite the system of equations in matrix form, it looks like the matrix equation mentioned in the question.

This representation allows us to apply matrix operations to find the solution to the system of equations.

8.Consider a simple function $$f(x) = 3x+2$$. This function is invertible right? Can you tell us what is $$\alpha$$ such that $$f(\alpha)=17$$? Is such an $$\alpha$$ unique? How did you find such an $$\alpha?$$. Is this always possible?

Sol.

An invertible function is one that has a unique inverse function, meaning that for every output value of the function, there is exactly one input value that produced it. This property ensures that each y in the function's range corresponds to exactly one x in the domain. So, yes, this function indeed is invertible. This equation can be solved by putting the value of the function as 17 in the equation and then solving it. The value of (\alpha) comes out to be 5. Linear functions with non-zero slopes are one-to-one (bijective) and thus have a unique inverse. For any given output y, there is exactly one input x such that f(x)=y. For such functions, it is always possible to find such a value.

9.Consider the function $$f(x)=x^2-10$$, what is $$f(5)$$?

Sol.

f(5) signifies the value of the function at x = 5, which is given by x*x - 10, i.e., 25 - 10 = 15.

10.Consider the function $$f(x)=x^2-10$$, if $$f(\alpha)=54$$, what is $$\alpha$$?

Sol.

Here, we have to find x and we know the value of the function. So, 54 = x*x - 10, i.e., x = 8 or -8.

11.Consider the function $$g(x)=x^3-x^2-10x+2$$, if $$g(x)=-22$$ what is $$x$$?

Sol.

We can find this by plotting the curve of g(x) + 22 = 0 on GeoGebra and then checking the roots of the equation. On plotting the curve, we observe that it has only one real root and two roots are imaginary as the degree of the equation is 3.

12.Do you know what is $$\mathbb{R}, \mathbb{R}^2 and\ \mathbb{R}^3$$ ?

Sol.

\(R\) represents the set of all real numbers (one-dimensional), \(R^2\) represents the two-dimensional space of ordered pairs of real numbers, and \(R^3\) represents the three-dimensional space of ordered triples of real numbers.

13.Consider the function $$\phi : \mathbb{R}^2\rightarrow \mathbb{R}^2$$ defined by $$\phi (x,y)=(2x+3y,3x+4y)$$. Find x and y such that $$\phi (x,y)=(5,6)$$. Obseve that (5,6) as well as (x,y) lies in $$\mathbb{R}^2$$.

Sol.

To find the values of (x) and (y) such that the function (\varphi: \mathbb{R}^2 \to \mathbb{R}^2), defined by (\varphi(x, y) = (2x + 3y, 3x + 4y)), maps to the point ((5, 6)), we need to solve the corresponding system of linear equations. Specifically, we set (\varphi(x, y) = (5, 6)), which gives us the system of equations:

[ 2x + 3y = 5 ] [ 3x + 4y = 6 ]

To solve this system, we can use the elimination method. First, we multiply the first equation by 3 and the second equation by 2 to align the coefficients of (x). This yields

[ 6x + 9y = 15 ] [ 6x + 8y = 12 ]

Next, we subtract the second equation from the first to eliminate (x), resulting in

[ 6x + 9y - 6x - 8y = 15 - 12 ]

which simplifies to

[ y = 3 ]

With (y) determined, we substitute (y = 3) back into one of the original equations to solve for (x). Using the first equation, (2x + 3(3) = 5), we get

[ 2x + 9 = 5 ]

Subtracting 9 from both sides, we find

[ 2x = -4 ]

and dividing by 2, we obtain

[ x = -2 ]

Therefore, the values of (x) and (y) that satisfy (\varphi(x, y) = (5, 6)) are (x = -2) and (y = 3).

It is important to observe that both the point ((5, 6)) and the pair ((x, y)) lie in (\mathbb{R}^2). This means that ((5, 6)) and ((-2, 3)) are both elements of the two-dimensional real number space, (\mathbb{R}^2), which is consistent with the domain and codomain of the function (\varphi).

14.Is the function $$\phi$$ invertible? In the question above on matrices, we see that it is of the form $$A\vec{x}=b$$. Note that we can invert the matrix, using the method that was taught to us in our high school to find out the value for the variables $$x$$ and $$y$$. This is one of the many applications of matrices.

Sol.

To determine if the function (\varphi) is invertible, we need to analyze the function (\varphi: \mathbb{R}^2 \to \mathbb{R}^2) defined by (\varphi(x, y) = (2x + 3y, 3x + 4y)). This problem can be expressed in matrix form as (A \vec{x} = \vec{b}), where (A) is the matrix of coefficients, (\vec{x}) is the column vector of variables ((x, y)), and (\vec{b}) is the result vector ((5, 6)).

The matrix (A) corresponding to the linear transformation is:

[ A = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix} ]

For (\varphi) to be invertible, the matrix (A) must be invertible. A matrix is invertible if its determinant is non-zero. We calculate the determinant of (A):

[ \det(A) = \begin{vmatrix} 2 & 3 \ 3 & 4 \end{vmatrix} = (2 \cdot 4) - (3 \cdot 3) = 8 - 9 = -1 ]

Since the determinant of (A) is (-1), which is non-zero, the matrix (A) is invertible. Consequently, the function (\varphi) is also invertible.

To find the inverse function, we use the inverse of the matrix (A). The inverse of (A) is calculated using the formula for the inverse of a (2 \times 2) matrix:

[ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} = \frac{1}{-1} \begin{pmatrix} 4 & -3 \ -3 & 2 \end{pmatrix} = \begin{pmatrix} -4 & 3 \ 3 & -2 \end{pmatrix} ]

Therefore, the inverse function (\varphi^{-1}) can be written as:

[ \varphi^{-1}(x', y') = \begin{pmatrix} -4 & 3 \ 3 & -2 \end{pmatrix} \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} -4x' + 3y' \ 3x' - 2y' \end{pmatrix} ]

Applying this to find the values of (x) and (y) for (\varphi(x, y) = (5, 6)):

[ \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -4 & 3 \ 3 & -2 \end{pmatrix} \begin{pmatrix} 5 \ 6 \end{pmatrix} = \begin{pmatrix} -4(5) + 3(6) \ 3(5) - 2(6) \end{pmatrix} = \begin{pmatrix} -20 + 18 \ 15 - 12 \end{pmatrix} = \begin{pmatrix} -2 \ 3 \end{pmatrix} ]

Thus, (x = -2) and (y = 3).

14(a). Take a random looking 2*2 matrix. Is it invertible? How often is it invertible?

Sol.

To determine whether a random (2 \times 2) matrix is invertible, we need to consider the properties of the matrix. A (2 \times 2) matrix is invertible if and only if its determinant is non-zero.

Let's take a random (2 \times 2) matrix:

[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} ]

The determinant of this matrix (A) is calculated as follows:

[ \det(A) = ad - bc ]

For the matrix (A) to be invertible, (\det(A)) must not be equal to zero. Therefore, the condition for invertibility is:

[ ad - bc \neq 0 ]

15.We will now see matrices as functions. Instead of $$\phi$$ we will write the matrix itself:
$$\left( \begin{matrix} 2 & 3 \3 & 4 \\end{matrix}\right) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$$.

Sol.

We can view matrices as functions that map vectors from one space to another. For example, consider the matrix:

[ A = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix} ]

This matrix (A) can be seen as a function (A: \mathbb{R}^2 \to \mathbb{R}^2) that takes a vector from (\mathbb{R}^2) and maps it to another vector in (\mathbb{R}^2). Specifically, for a vector (\vec{x} = \begin{pmatrix} x \ y \end{pmatrix}), the matrix function (A) acts on (\vec{x}) as follows:

[ A \vec{x} = \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 2x + 3y \ 3x + 4y \end{pmatrix} ]

16.Consider the function $$\left( \begin{matrix} 1 & 2 \2 & 4 \\end{matrix}\right) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$$. This matrix takes a few elements to the origin. What are those elements? Plot this using Geogebra.

Sol.

Consider the function represented by the matrix (A):

[ A = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} ]

This matrix (A: \mathbb{R}^2 \to \mathbb{R}^2) maps vectors from (\mathbb{R}^2) to (\mathbb{R}^2). We are interested in finding which vectors (\vec{x} = \begin{pmatrix} x \ y \end{pmatrix}) are mapped to the origin by this matrix, i.e., we want to solve the equation:

[ A \vec{x} = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} ]

This leads to the system of linear equations:

[ \begin{cases} 1x + 2y = 0 \ 2x + 4y = 0 \end{cases} ]

We can simplify this system by noting that the second equation is just twice the first equation. Therefore, it suffices to solve the first equation:

[ x + 2y = 0 \implies x = -2y ]

This means that any vector of the form (\vec{x} = \begin{pmatrix} -2y \ y \end{pmatrix}) will be mapped to the origin by the matrix (A). In other words, the vectors that are mapped to the origin lie along the line (x = -2y). This line will be called as the Null Space of the Matrix.

Module B

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17.$$A$$ is assigned $$0$$, $$B:1$$, $$C:2$$, and so on up to $$Z:25$$
Assume you denoted every letter with a number, as given in the table above.
You need to encrypt the word $$SUDARSHANA$$ which stands for the numbers: $$18, 20, 3, 0, 17, 18, 7, 0, 13,0$$.
You encrypt this using a matrix given by : $$ \left( \begin{matrix} 2 & 3 \3 & 4 \\end{matrix}\right) $$.
So $$SUDARSHANA$$ will end up becoming: $$96, 134, 6, 9, 88, 123, 14, 21, 26, 39$$.
Given these numbers, how will you decrypt the message and get back $$SUDARSHANA$$?
. This is a well known cryptographic protocol called the Hill Cipher. You can read more online.

Sol.

To decrypt the message encrypted using the Hill Cipher, we need to apply the inverse of the encryption matrix. Given the encrypted numbers representing "SUDARSHANA" as 96, 134, 6, 9, 88, 123, 14, 21, 26, and 39, we first identify the corresponding encryption matrix, denoted as ( \mathbf{E} ), which in this case is ( \begin{pmatrix} 2 & 3 \ 3 & 4 \end{pmatrix} ). Utilizing this matrix, we compute its inverse, considering modulo 26 arithmetic. The decryption matrix, denoted as ( \mathbf{D} ), is then obtained. Proceeding with decryption, each pair of encrypted numbers undergoes matrix multiplication with ( \mathbf{D} ), followed by modulo 26 reduction. This process retrieves the original numbers corresponding to each letter. The decryption process is represented as ( \mathbf{p} = \mathbf{D} \cdot \mathbf{c} \pmod{26} ), where ( \mathbf{p} ) represents the decrypted numbers and ( \mathbf{c} ) represents the encrypted numbers. Finally, mapping these numbers back to their corresponding letters reveals the decrypted message "SUDARSHANA." This cryptographic protocol, known as the Hill Cipher, leverages linear algebra principles for both encryption and decryption, ensuring a secure communication channel.

18.We encounter equations very often in our lives. Consider for example, the following situation at Baker's Cafe. The manager has a very important estimate to make. Mostly, visitors at his cafe happen to be families and they are often comprised of Children and/or Adults. He observes that there are 3 adults and 1 child at a table and their bill turns out to be Rs.1200/-. There is yet another table with 2 children and 1 adult and their bill comes out to be Rs.1000/-. Can the manager estimate the consumption of a Child/Adult? This is popularly called the Simultaneous Equations and we all remember from our school days, multiple ways in which these can be solved.
$$ 3A + 1C = 1200 $$
$$ 1A + 2C = 1000 $$

Sol.

Let's denote the consumption of an adult as (A) and that of a child as (C).

The first observation, with 3 adults and 1 child resulting in a bill of Rs. 1200, can be expressed as the equation: [3A + 1C = 1200].

Similarly, the second observation, with 2 children and 1 adult totaling Rs. 1000, can be represented as: [1A + 2C = 1000].

To solve this system of equations, we can employ various methods such as substitution, elimination, or matrix methods. The values of (A) and (C) are determined as (280) and (360) respectively.

19.While we were taught the so called two variables and two unknowns, what if there were more equations than unknowns?
$$ 3A + 1C = 1200 $$
$$ 1A + 2C = 1000 $$
$$ 1A + 1C = 900 $$

Sol.

When faced with a system of equations where the number of equations exceeds the number of unknowns, we encounter what is known as an overdetermined system. In this scenario, there may not exist a single solution that satisfies all equations simultaneously. For instance, considering the equations provided. We have three equations but only two unknowns, A and C. Consequently, the system is overdetermined. In such cases, it is unlikely that a unique solution exists that satisfies all equations simultaneously. However, we can still attempt to find a solution that best fits the given equations. We aim to minimize the error between the observed values (the left-hand sides of the equations) and the predicted values (the right-hand sides) by adjusting the unknowns. Therefore, while the system may not have a single exact solution, we can seek an approximation that minimizes the discrepancy between the observed and predicted values.

20.Note that the previous question can be modelled as a matrix:

$$ 3A + 1C = 1200 $$ $$ 1A + 2C = 1000 $$ $$ 1A + 1C = 900 $$

Observe this is same as :

$$ \left( \begin{matrix} 3 & 1 \1 & 2 \1 & 1 \\end{matrix}\right) $$ $$\left( \begin{matrix} A\ C\ \end{matrix} \right)$$

$$ \left( \begin{matrix} 1200\\ 1000\\ 900\\ \end{matrix} \right) $$

Sol.

Indeed, the system of equations provided can be represented in matrix form as: [ \begin{pmatrix} 3 & 1 \ 1 & 2 \ 1 & 1 \ \end{pmatrix} \begin{pmatrix} A \ C \ \end{pmatrix}

\begin{pmatrix} 1200 \ 1000 \ 900 \ \end{pmatrix} ]

Here, the coefficient matrix on the left represents the coefficients of the unknowns A and C in each equation, while the vector on the right represents the constants on the right-hand side of each equation.

21.One obvious way to solve this, is to guess the values :-). Can you get closer to the solution by guessing? Note that there is no solution to this question. You can just reduce the error. Do you see why?

Sol.

In situations where a system of equations is overdetermined, as in the case provided, where there are more equations than unknowns, there may not exist a solution that satisfies all equations simultaneously. Consequently, attempting to solve the system by guessing values is unlikely to yield an exact solution. However, we can employ guesswork to minimize the error between the observed and predicted values, effectively reducing the discrepancy between the equations. By iteratively adjusting the guessed values and evaluating their impact on the overall error, we can approximate a solution that minimizes the discrepancy as much as possible. This process, though not guaranteed to yield a precise solution, allows us to iteratively refine the estimates and improve the fit between the observed and predicted values.

22.In the figure below: If 1000 people were to start in one state, what will be the distribution of people eventually? Write down a python script to find the convergence.

Sol.

Put your solution here.

23.In the figure below: If 1000 people were to start in one state, what will be the distribution of people eventually? Write down a python script to find the convergence.

Sol.

Put your solution here.

Module C

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24.Use Geogebra: Draw the vector $$\begin{bmatrix}1 \ 1 \ \end{bmatrix}$$. Find out all those vectors which are perpendicular to this vector.

Sol.

The given vector can be drawn with the help of the vector tool in GeoGebra. All the vectors which have their dot product with the given vector as 0 will be orthogonal to it. They will be of the form: $$v = (t, -t),$$ where t is an arbitrary variable.

25.Do you observe that we are asking for vectors $$\begin{bmatrix}x \ y \ \end{bmatrix}$$ such that, $$\begin{bmatrix}1 & 1 \ \end{bmatrix} \begin{bmatrix} x\ y \ \end{bmatrix}=0 $$

Sol.

The givem set of matrices lead to the equation which we get while doing the dot of these vectors as 0, i.e., $$x + y = 0$$ So, This equation represents all the vectors perpendicular to the given one. These vectors lie along the line with slope -1 passing through the origin in the Cartesian coordinate system.

26.Use Geogebra and solve the above question with $$\begin{bmatrix} 1 \ 1 \ \end{bmatrix}$$ replaced by $$\begin{bmatrix}a \ b \ \end{bmatrix}$$.Use $$(a,b)$$ as parameters and check what happens to $$(x,y)$$.

Sol.

27.What is $$(x,y,z)$$ satisfying the following equation? (Use Geogebra) $$\begin{bmatrix}1 & 2 & 3\ \end{bmatrix} \begin{bmatrix} x\ y \ z\ \end{bmatrix}=0 $$

Sol.

The equation given is ( [123]\begin{bmatrix} x \ y \ z \end{bmatrix} = 0 ). To solve this, we can multiply the matrix ( [123] ) by the column vector ( \begin{bmatrix} x \ y \ z \end{bmatrix} ) using matrix multiplication.

[ \begin{bmatrix} 1 & 2 & 3 \ \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}

\begin{bmatrix} 1x + 2y + 3z \ \end{bmatrix}

0 ]

For the equation to hold true, the result of this multiplication must be zero. Therefore, the solution to the equation is any vector ( \begin{bmatrix} x \ y \ z \end{bmatrix} ) such that ( x + 2y + 3z = 0 ). We will now go plot the equation on GeoGebra. We observe that this represents an equation of a plane and it contains the collection of position vectors which satisfy the given equation.

28.Use Geogebra and plot all the points in the set below. $$T= { \alpha(1,2,1) | \alpha \in \mathbb{R}} $$

Sol.

To plot all the points, we will create the given vector in the graph. Then we will use the sequence command in desired range to show the possible vectors under the given expression as in the question. We observe that the points follow a straight line pattern passing through the Origin.

29.Use Geogebra and plot all the points in the set below. $$S= { \beta(2,7,3) | \beta \in \mathbb{R}} $$

Sol.

The question will be solved exactly similar to the previous one.

30.Use Geogebra and plot all the points in the set below. $$W= {\alpha(1,2,1) + \beta(2,7,3) | \alpha,\beta \in \mathbb{R}} $$

Sol.

We can simply add the two previous sets to create the third one as given in the question in GeoGebra.

31.In the above set $$W$$ find out all the points $$(x,y,z)$$ satisfying the following: (Use Geogebra) $$\begin{bmatrix}w_1 & w_2 & w_3\ \end{bmatrix} \begin{bmatrix} x\ y \ z\ \end{bmatrix}=0 $$ where $$(w_1,w_2,w_3) \in W$$. Note that $$w_i$$s are real numbers.

Sol.

Put your solution here.

32.Given the matrix $$A=\begin{bmatrix}1 & 2 & 3\ 4 & 5 &6\ 7 & 8 & 9\ \end{bmatrix}$$, find out all the possible $$(x,y,z)$$ such that: $$ \begin{bmatrix}1 & 2 & 3\ 4 & 5 &6\ 7 & 8 & 9\ \end{bmatrix} \begin{bmatrix} x\ y\ z\ \end{bmatrix}=0$$ Observe carefully, what has this question got to do with previous five questions in this module

Sol.

Put your solution here.

33.Given the matrix $$A=\begin{bmatrix}1 & 2 & 3\ 4 & 5 &6\ 7 & 8 & 9\ \end{bmatrix}$$ what does the following three sets represent?
(i) $$\mathscr{R}={\alpha(1,2,3) + \beta(4,5,6) + \gamma(7,8,9) |\alpha, \beta, \gamma\in \mathbb{R}} $$
(ii) $$C={\alpha(1,4,7) + \beta(2,5,8) + \gamma(3,6,9) | \alpha, \beta, \gamma \in \mathbb{R}}$$
(iii) $$N={(x,y,z)|x(1,4,7) + y(2,5,8) + z(3,6,9) = 0 }$$
Use only Geogebra :)

Sol.

Put your solution here.

34.Did you observe that every vector of $$\mathscr{R}$$ is perpendicular to every vector of $$N$$?

Sol.

Put your solution here.

35.Consider the matrix $$B=\begin{bmatrix} 1 & 2 \ 2 & 4\ \end{bmatrix}$$. Draw the line $$2y+x=4$$. Seeing the matrix $$B$$ as a function $$B:\mathbb{R^2}\mapsto \mathbb{R^2}$$,
where does $$B$$ takes the line $$2y+x=4$$?
Where does it take:
i)$$2y+x=10$$
ii)$$2y+x=62$$
iii)$$2y+x=1800$$

Sol.

Put your solution here.

35(a).In general $$B=\begin{bmatrix} 1 & 2 \ 2 & 4\ \end{bmatrix}:\mathbb{R^2}\mapsto \mathbb{R^2}$$, where does this function take $$2y+x=k$$?(where $$k$$ is a constant)

Sol.

Put your solution here.

36.Consider a matrix $$A=\begin{bmatrix} 1 & 4 \ 2 & 3\ \end{bmatrix}$$ and a vector $v$ = $$\begin{bmatrix} 1 \ 1\ \end{bmatrix}$$ to what is it transformed?
36(a).Is it rotated?
36(b).Is the magnitude preserved?
36(c).What is the ratio of magnitude of $$Av$$ to $$v$$?

Sol.

Put your solution here.

37.Given $$B=\begin{bmatrix} 1 & 2 \ 2 & 4\ \end{bmatrix}:\mathbb{R^2}\mapsto \mathbb{R^2}$$. What is the range of this function?

Sol.

Put your solution here.

38.You have achieved the required wisdom if you have realized that:
$$B=\begin{bmatrix} 1 & 2 \ 2 & 4\ \end{bmatrix}:\mathbb{R^2}\mapsto \mathbb{R^2}$$.
" $$B$$ collapses a dimension ".

Sol.

Put your solution here.

Module D

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39.Given the matrix $$M=\begin{bmatrix} 1 & 3 \ 2 & 6\ \end{bmatrix}$$. Use Geogebra to plot $$\mathscr{R}$$, $$\mathscr{C}$$ & $$\mathscr{N}$$. what do you observe?
(i) $$\mathscr{R}={\alpha(1,3) + \beta(2,6) | \alpha, \beta\in \mathbb{R}} $$
(ii) $$\mathscr{C}={\alpha(1,2) + \beta(3,6) | \alpha, \beta\in \mathbb{R}} $$
(iii) $$\mathscr{N}={(x,y)| x(1,3) + y(2,6) = 0, \forall x,y\in \mathbb{R} }$$

Sol.

\((A)\) Upon plotting these graphs on Geogebra, we observe that the set R, which shows the possible combinations of the rows of matrix M, leads to set of points which lie on a straight line, which can be plotted with the help of in built functions provided..

\((B)\) Similarly, the set C, that signifies the possible combinations of the columns of the matrix M, plots the points, lying on another straight line, on an angle to the previous one.

\((C)\) The set N represents the solutions to a specific equation involving the rows and columns of M. It also forms a straight line on the graph, starting from the origin and having a different orientation compared to R and C.

This observation clarifies the visual representation of the relations between rows, columns and their combinations.

40.Note that $$\mathscr{C}$$ and $$\mathscr{N}$$ are orthogonal.

Sol.

We observed in the previous question that the points on both C and N follow a straight line path.

When we plot the straight line on the graph with the function using origin as the point and our vector as defined in the question, we get the lines as $$y = 2x$$ and $$x = -2y.$$ Here, we can observe from both the equations and also by looking at the graph itself that the two lines come out to be perpendicular to each other.

41.What is the null-space of $$M=\begin{bmatrix} 1 & 3 \ 2 & 6\ \end{bmatrix}$$ & the null-space of $$M^T$$?

Sol.

Null Space of a Matrix is defined as the set of points such that their column matrix X when post multiplied by the matrix M itself produces the Null Matrix O of relevant order, i.e., $$MX = O.$$

When we multiply these matrices, we get two equations in two variables.

$$x + 3y = 0$$ $$2x + 6y = 0$$

We can observe that the two equations are same and the solution for the null space of the matrix comes out to be the straight line from the equation, i.e., $$x = -3y.$$

Similarly, we can do it for the transpose matrix of M.

Let the transpose matrix of M be N. Now, as we learnt earlier, Null space of N will be found by the following equation: $$NX = O$$

This leads us to the equations:

$$x + 2y = 0$$ $$3x + 6y = 0$$

Similar to the previous example, the solution for the null space for N, the transpose of the matrix M, comes out to be a straight line, $$x = -2y.$$

42.Do you observe that $$C(M)$$ ⊥ $$N(M^T)$$ ,   $$R(M)$$ ⊥ $$N(M)$$ ?

Sol.

This observation can easily be made by plotting the two graphs on Geogebra.

43.Consider $$A=\begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \end{bmatrix}$$. What is $$N(A)$$, $$C(A)$$, $$R(A)$$, $$N(A^T)$$.

Sol.

Let us first check the Null Space. As the square matrix used is of the order of 3 now, not of the order of 2, we will have three equations in 3 variables. But the basic approach for Null Space will be the same, i.e., $$AX = O.$$

We will get the following equations:

$$x + 2y + 3z = 0$$ $$4x + 5y + 6z = 0$$ $$7x + 8y + 9z = 0$$

We can use Geogebra to solve these equations.

We get the solution as:

$$x = - y/2 = z,$$ which is the null space of the provided matrix.

Similarly, for the transpose of the provided matrix, we will get the following equations:

$$x + 4y + 7z = 0$$ $$2x + 5y + 8z = 0$$ $$3x + 6y + 9z = 0$$

We get the solution as:

$$x = -y/2 = z,$$ which, we can observe, is exactly similar to the Null Space of A itself.

Now, we will check the Column Space of the given matrix. We have seen how $C(M)$ is defined previously in this module.

Going with the similar approach, we get the set C(A) as:

$$C(A) = {a(1,4,7) + b(2,5,8) + c(3,6,9) : a, b, c \text{ belong to } R}$$

Similarly,

$$R(A) = {a(1,2,3) + b(4,5,6) + c(7,8,9) : a, b, c \text{ belong to } R}$$

44.Consider a 4x4 matrix $$M$$: $$\mathbb{R^4}\mapsto \mathbb{R^4}$$ whose range is
a) $$4-Dimension$$
b) $$3-Dimension$$
c) $$2-Dimension$$
d) $$1-Dimension$$
e) $$0-Dimension$$
Give an example each for all the above 5 cases.

Sol.

(a) ( A=\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix} )

(b) ( A=\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 1 & 0 \end{bmatrix} )

(c) ( A=\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix} )

(d) ( A=\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix} )

(e) ( A=\begin{bmatrix} 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix} )

Note that any other example which satisfies the given condition is also a correct answer for the question.

45.Consider $$A : \mathbb{R^{3}} \to \mathbb{R^{3}} $$
a) Show that if the range contains a point $$(a,b,c)$$, then it should contain the entire set $$S$$, defined by: $$S= {\alpha(a, b,c)| \alpha \in \mathbb{R}}$$.
b) Show that if the range contains the points $$(a,b,c)$$ and $$(d,e,f)$$, then the range contains the entire set $$T$$ defined by: $$T={\alpha(a,b,c) + \beta(d,e,f) \thinspace|\thinspace \alpha,\beta\in \mathbb{R}}$$.
c) Note: $$S$$ is of the dimension $$1$$, but $$T$$ need'nt be of dimension $$2$$. Think!

Sol.

\((A)\) Given that the range contains a point \((a,b,c)\), it means that there exists a vector x such that \(A(x) = (a,b,c).\)

But, since A is a linear transformation; for any scalar a, $$A(ax) = a(a,b,c),$$ which implies that the entire set S is also in the range.

\((B)\) Similarly, as long as the linearity of the transformation is sustained, $$A(ax + by) = a.x + b.y$$ for 2 scalars a and b and existing vectors \(x = (a,b,c)\) and \(y = (d,e,f).\) This indicates that the entire set T is in the range.

\((C)\) Yes, the dimension of T needn't be 2 as it depends upon the linear dependency of the two vectors with each other.

Module E

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46.Give an example of two $$2-dim$$ subspaces in $$\mathbb{R^{3}}$$. Let us call it $$S_1, S_2$$.

Sol.

To provide an example of two 2 - dimensional subspaces in 3 - dimensional Real space, we need to define two different planes through the origin in the whole space. For example, they can be the xy and the zx - planes. Therefore,

$$S_1 = {(x,y,0) : x,y \text{ belong to } R}$$ $$S_2 = {(x,0,z) : x,z \text{ belong to } R}$$

47.Let $$S_3$$ be all those vectors perpendicular to $$S_1$$. $$S_4$$ be that of $$S_2$$.

Sol.

Let us define \(S_3\) and \(S_4\) on the basis of their orthogonality towards \(S_1\) and \(S_2\) respectively.

As \(S_3\) contains all the vectors perpendicular to \(S_1\), i.e., the xy - plane, they will lie along the z-axis. Therefore,

$$S_3 = \{x : x \text{ lies along the z - axis}\}$$

Similarly, $$S_4 = {x : x \text{ lies along the z - axis}}$$

48.Find a matrix $$M$$ whose Null-Space is $$S_3$$. column space is $$S_2$$.

Sol.

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

As we know that row space and null space of a matrix are orthogonal to each other. It can clearly be observed from this matrix that its row space is xy plane and hence, as z - axis is orthogonal to it, it is its null space.

49.What does $$S_1$$ and $$S_4$$ represent?

Sol.

\(S_1\) represents a 2 - dimensional subspace within \(R^3.\)

Specifically, it signifies any plane passing through the origin in 3 - dimensional space.

\(S_4\) represents the orthogonal complement of a 2 - dimensional subspace \((S_2)\).

In other words, it consists of all vectors that are perpendicular to every vector in \(S_2\). Geometrically, it could signify a line, a plane, or even all of \(R^3\) itself, depending upon the orientation ad the dimensionality of \(S_2.\)

50.Do you observe there is a bijection from $$S_1 \to S_2$$?

Sol.

There is indeed a bijection from \(S_1\) to \(S_2\) given by the map:

\(f(x,y,0) = (x,0,y)\)

This map pairs each vector in \(S_1\) with a unique vector in \(S_2\) and covers all vectors in \(S_2\) maintaining the conditions of both injectivity and surjectivity.

Module F

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51.Imagine a situation of war in 1800's.Country A wants to send a letter to Country B such that their enemy country can't understand the message.How can you help the country A in this situation?

Sol.

In the scenario of wartime communication in the 1800s, Country A could employ the Caesar cipher to encode their messages to Country B. The Caesar cipher is a substitution cipher where each letter in the plaintext is shifted a certain number of places down or up the alphabet. By agreeing on a specific shift value beforehand, known as the "key," Country A could encode their messages, making them unintelligible to their adversaries without the knowledge of the key. In the Caesar cipher, each letter in the plaintext is shifted by a fixed number of positions in the alphabet. Mathematically, this can be represented using modular arithmetic. Let's denote ( n ) as the shift value (the key) and ( P ) as the position of a letter in the alphabet. The Caesar cipher encryption function ( E ) can be expressed as:

[ E(P) = (P + n) \mod 26 ]

Where ( \mod 26 ) ensures that the result wraps around the alphabet. For example, if ( n = 3 ) and ( P = 1 ) (representing 'A'), the encrypted letter would be ( E(1) = (1 + 3) \mod 26 = 4 ), which corresponds to 'D'.

52.How about shifting the alphabets by 1 letter each?What is the problem here?

Sol.

Shifting each letter of the alphabet by one position, known as a Caesar cipher with a fixed key of 1, is a simple form of substitution cipher. While it provides a basic level of encryption, it suffers from a significant vulnerability: its lack of security due to its limited key space.

Since there are only 25 possible keys (each shift value from 1 to 25), an attacker can easily perform a brute-force attack by trying all possible keys to decrypt the message. This means that the encrypted message can be deciphered through only 25 trials, making it highly vulnerable to cryptanalysis.

53.Try encoding the word "VICHARANASHALA" using the above method (But shift 4 letters this time)

Sol.

To encode the word "VICHARANASHALA" using a Caesar cipher with a shift of 4 letters, we shift each letter in the word by four positions in the alphabet:

• V becomes Z
• I becomes M
• C becomes G
• H becomes L
• A becomes E
• R becomes V
• A becomes E
• N becomes R
• A becomes E
• S becomes W
• H becomes L
• A becomes E
• L becomes P
• A becomes E

So, "VICHARANASHALA" would be encoded as "ZMGLEREVREWEP".

54.What if you have only the encoded message? How will you get to the original message?

Sol.

If we only have the encoded message and no knowledge of the key (the shift value used in the Caesar cipher), we would need to employ cryptanalysis techniques to decrypt the message.

One common approach is frequency analysis, which relies on the fact that certain letters appear more frequently than others in natural language text. For example, in English, the most common letters are 'E', 'T', 'A', 'O', and 'I'. By analyzing the frequency of letters in the encoded message and comparing it to the expected frequency distribution of letters in English text, we can make educated guesses about the shift value.

Another method involves trying all possible shift values (from 1 to 25) and examining the decrypted text for meaningful words or patterns. This brute-force approach would involve decoding the message 25 times with different shift values until the original message is revealed.

55.What if we substitute each letter by some other letter using a pre- defined mapping (eg.a->t,b->f,c->y,...)?How many trails do we have to do so that we can reach the secret message if we only have the encoded text and not the mapping ?

Sol.

If we have an encoded message using a substitution cipher with a predefined mapping, and we don't know the mapping, we essentially face a cryptanalysis problem.

The number of possible mappings in a substitution cipher depends on the size of the alphabet used in the encoding. For example, if we're using the English alphabet, which consists of 26 letters, there are ( 26! ) possible permutations of the alphabet.

Therefore, without knowing the mapping, we would need to try each possible permutation to decipher the message. This brute-force approach would require checking all ( 26! ) mappings, which is clearly impractical due to the vast number of trials involved.

In summary, if we only have the encoded text and not the mapping used in a substitution cipher, it is practically infeasible to decipher the secret message by trying all possible mappings.

56.Is there any efficient approach for the second part of the 55th question?

Sol.

Yes, there are more efficient approaches for decrypting a message encoded with a substitution cipher when the mapping is unknown. One common technique is frequency analysis.

In most languages, including English, certain letters occur more frequently than others.

Here's how frequency analysis works:

1. Count the frequency of each letter in the encoded message.

2. Compare the frequency distribution to the expected frequency distribution of letters in the language being used (e.g., English).

3. Identify common patterns, such as single-letter words or repeated sequences, which may correspond to common letters or words in the language.

4. Use these patterns to make educated guesses about the mapping, such as which encoded letter corresponds to 'E' or 'T'.

5. Once a few letters are deciphered, use context and word patterns to further decrypt the message.

57.What do you think is the frequency of occurence of various letters in a sample English text? Which letter do you expect to be the most frequent ?

Sol.

In a typical English text, the frequency of occurrence of various letters follows a well-known distribution. The most frequent letter in English text is 'E', followed by 'T', 'A', 'O', 'I', 'N', 'S', 'H', 'R', 'D', 'L', 'C', 'U', 'M', 'W', 'F', 'G', 'Y', 'P', 'B', 'V', 'K', 'J', 'X', 'Q', and 'Z', in descending order of frequency. These frequencies can vary slightly depending on the specific text and context, but they provide a general guideline for the relative occurrence of letters in English text.

58.Assuming that an English text follows a particular order of frequency,can you solve the 56th question?

Sol.

Assuming that the English text follows the typical order of letter frequency, we can use this knowledge to help decrypt a message encoded with a substitution cipher.

Given the encoded message, we can analyze the frequency of letters in the text. By identifying the most frequently occurring letter in the encoded message, we can make an educated guess that it corresponds to the most frequent letter in English text, which is 'E'.

Once we determine the mapping for this letter, we can continue deciphering the rest of the message based on context and patterns. This process can be iterated, gradually revealing more letters and improving our understanding of the mapping until the entire message is decrypted.

While frequency analysis provides a powerful tool for decrypting substitution ciphers, it may still require some manual effort and linguistic knowledge, especially for longer messages or messages with less predictable patterns. However, by leveraging the knowledge of letter frequency in English text, we can significantly reduce the number of trials needed to decrypt the message compared to a brute-force approach.

59.Suppose we take a subset from a huge text i.e $$k^{th}$$, $$2k^{th}$$, $$3k^{th}$$... elements.Will they also follow the same pattern observed in the previous question?

Sol.

Yes, if we take a subset of characters from a large enough English text, such as every (k^{th}) character, (2k^{th}) character, (3k^{th}) character, and so on, they are likely to follow a similar pattern of letter frequency as observed in the previous question.

This is because the frequency distribution of letters in English text is relatively stable across different texts, assuming the text samples are large enough and representative of typical English language usage. Therefore, even when considering a subset of characters from a large text, we would still expect the most frequent letters to be 'E', 'T', 'A', 'O', 'I', and so on, in roughly the same order of frequency.

Of course, the specific frequencies may vary slightly depending on the particular text and context, but the overall pattern of letter frequency should remain consistent. This consistency is what allows frequency analysis to be an effective technique for decrypting substitution ciphers, even when working with subsets of text.

60.Assume you arrange two meaningful english text strings in front of each other.what is the expected number of collisions in the letters? Call it "collision frequency".

Sol.

Put your solution here.

61.Assume that in the previous question ,we apply the ceaser cypher(the one discussed inthe first few questions), on both the strings, and alphabet by 5 letters then will the collision frequency remain the same?What if we shift first string by 3 letters and second by 5?

Sol.

Put your solution here.

62.Suggest any such method using which we can be confident that the encoded text can't be decoded by the enemy. (We may discuss it in further classes)

Sol.

Put your solution here.

Project 1- Read and Solve this

Sol.

Put your solution here.

Module G

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63.A dart is thrown at random onto a board that has the shape of a circle as shown below. Calculate the probability that the dart will hit the shaded region.

Sol.

The shaded region is the difference in area between two concentric circles (a larger circle and a smaller circle).

• The radius of the larger circle ( R ) is 14 units.
• The radius of the smaller circle ( r ) is 7 units.

The area of the shaded region ( A_{\text{shaded}} ) is given by the difference in the areas of these two circles: [ A_{\text{shaded}} = \pi R^2 - \pi r^2 ]

Substituting the values: [ A_{\text{shaded}} = \pi (14^2 - 7^2) ] [ A_{\text{shaded}} = \pi (196 - 49) ] [ A_{\text{shaded}} = \pi \cdot 147 ]

The total area ( A_{\text{total}} ) is the area of the larger circle: [ A_{\text{total}} = \pi R^2 ]

Substituting the value: [ A_{\text{total}} = \pi \cdot 14^2 ] [ A_{\text{total}} = \pi \cdot 196 ]

The probability ( P ) of a dart hitting the shaded region is the ratio of the area of the shaded region to the total area of the circle: [ P = \frac{A_{\text{shaded}}}{A_{\text{total}}} ]

Substituting the areas calculated: [ P = \frac{\pi \cdot 147}{\pi \cdot 196} ] [ P = \frac{147}{196} ]

Simplifying the fraction: [ P = \frac{147 \div 49}{196 \div 49} ] [ P = \frac{3}{4} ]

Thus, the probability of a dart hitting the shaded region is ( \frac{3}{4} ) or 0.75.

64.Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean or expectation of X.

Sol.

Put your solution here.

65.A factory produces items, and each item is independently defective with probability 0.02. If 100 items are produced in a day, what is the expected number of defective items?

Sol.

To find the expected number of defective items produced in a day by the factory, we use the concept of expectation in probability theory.

Given:

• Each item is defective with probability ( p = 0.02 ).
• The number of items produced in a day ( n = 100 ).

The expected number of defective items ( E(X) ) can be calculated using the formula for the expectation of a binomial distribution: [ E(X) = n \cdot p ]

Substituting the given values: [ E(X) = 100 \cdot 0.02 ] [ E(X) = 2 ]

Thus, the expected number of defective items produced in a day is 2.

66.A point is chosen at random inside a sphere of radius R. What is the probability that this point is closer to the center of the sphere than to its surface?

Sol.

To find the probability that a randomly chosen point inside a sphere is closer to the center than to its surface, we analyze the problem geometrically.

Given:

• The sphere has a radius ( R ).
• We need to find the probability that a point is closer to the center than to the surface of the sphere.
• A point inside the sphere is closer to the center than to the surface if its distance from the center is less than half the radius of the sphere, ( \frac{R}{2} ). The volume ( V_{\text{inner}} ) of the sphere with radius ( \frac{R}{2} ) is given by: [ V_{\text{inner}} = \frac{4}{3} \pi \left( \frac{R}{2} \right)^3 = \frac{4}{3} \pi \cdot \frac{R^3}{8} = \frac{1}{6} \pi R^3 ] The volume ( V_{\text{total}} ) of the sphere with radius ( R ) is given by: [ V_{\text{total}} = \frac{4}{3} \pi R^3 ] The probability ( P ) that a randomly chosen point inside the sphere is closer to the center than to the surface is the ratio of the volume of the inner sphere to the volume of the entire sphere: [ P = \frac{V_{\text{inner}}}{V_{\text{total}}} = \frac{\frac{1}{6} \pi R^3}{\frac{4}{3} \pi R^3} = \frac{\frac{1}{6}}{\frac{4}{3}} = \frac{1}{6} \cdot \frac{3}{4} = \frac{1}{8} ]

Thus, the probability that a randomly chosen point inside the sphere is closer to the center than to its surface is ( \frac{1}{8} ).

67.A point is randomly chosen inside a cube with side length 𝑎. What is the probability that the point is closer to one of the vertices than to the center of the cube?

Sol.

Put your solution here.

68.Imagine you have a number line that ranges from 0 to 1. You randomly pick k points on this line. What is the expected distance of the closest point to the midpoint of the line?

Sol.

Put your solution here.

**Project 5-** [Read and Solve this](https://docs.google.com/document/d/1hjT8teTRsQiW-NDj37Y6JmzCMTEE2xNHOv6iC9NKkDU/edit)

Module H

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69.Plot and find the distance between points using Geogebra: $$A(1,2)$$, $$B(2,3)$$, and $$C(5,9)$$.

Sol.

Put your solution here.

70.Using Geogebra, which two points are the closest ones? What's the distance between them?

Sol.

Put your solution here.

71.Add four more points: $$D(0,-1)$$, $$E(4,11)$$, $$F(8,12)$$, $$G(-3,6)$$ to your graph. What’s the closest pair now? How many computations do you have to do?

Sol.

Put your solution here.

72.Suppose we have 10 points. How many pairs of points do you have to consider for finding the closest pair?

Sol.

Put your solution here.

73.What is the Y sorted order (by default assume ascending) of points $$A$$, $$B$$, $$C$$, $$D$$, $$E$$?

Sol.

Put your solution here.

74.Plot a line $$L$$ parallel to the Y-axis passing through the middle point in the X sorted order of the above points. Divide the set of points into left and right regions around the line.

Sol.

Put your solution here.

75.Find the closest pair of points in the left region and right region. What’s the minimum distance (say $$d$$) between them?

Sol.

Put your solution here.

76.Consider a band of width $$2d$$ around the Line $$L$$. Find the closest pair in this band. Compare this distance with $$d$$, the minimum value of the corresponding closest pair of our graph. Is the answer the same as the brute-force method you applied in question 71? (This divide and conquer method is known as the closest pair algorithm).

Sol.

Put your solution here.

77.Astronomers have recorded the positions of stars in a 3D coordinate system where each star is represented as a point. Given the coordinates of stars $$(1,2,3)$$, $$(4,5,6)$$, $$(6,7,8)$$, $$(10,11,12)$$, find the closest pair of stars. (Use Geogebra)

Sol.

Put your solution here.

78.If $$F(0) = 0$$, $$F(1) = 1$$, $$F(n) = F(n-1) + F(n-2)$$ for $$n \geq 2$$, find the value of $$F(5)$$.

Sol.

Put your solution here.

79.Dry run and find the output of the following python code: ``` def f(n): if n == 0: return 1 return n * f(n-1) print(f(5)) ```

Sol.

Put your solution here.

80.Does the closest pair algorithm assume that the $$x$$ coordinates (and $$y$$ coordinates) of the points are distinct? Is there a problem with the $$O( nlog(n))$$ performance if they are not distinct?

Sol.

Put your solution here.

81.Given a set of points where most points are far apart, but a few points are very close to each other, develop an algorithm to efficiently find the closest pair. For example, use the set: $$(100,200)$$, $$(300,400)$$, $$(5000,6000)$$, $$(1,2)$$, $$(1.001,2.001)$$.

Sol.

Put your solution here.

# Module I ---

82.Do you know the idea of equally likely events? What are these? Can you think of any event which is not equally likely?

Sol.

Equally likely events are events that have the same probability of occurring. When we say that events are equally likely, it means that each event has the same chance or likelihood of happening. For instance, coin toss, rollong a fair die, choosing a card from a deck of well-shuffled cards. An example of events not being equally likely may include result of sport games, which depends on various factors like relative strangths of the teams, the condition of the field, etc.

83.You are given two coins .What is the probability that one head and one tail shows up on tossing?

Sol.

To determine the probability of getting one head and one tail when tossing two coins, we need to consider all possible outcomes and then identify the outcomes that match the desired event (one head and one tail). When tossing two coins, each coin can land on heads (H) or tails (T). Therefore, the possible outcomes are:

• (H, H)
• (H, T)
• (T, H)
• (T, T) We are interested in the outcomes where there is one head and one tail. These outcomes are:
• (H, T)
• (T, H) The probability ( P ) of an event is given by the ratio of the number of favorable outcomes to the total number of outcomes. [ P(\text{one head and one tail}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2} ]

84.In a class in which all students practise at least one sport, 60% of students play soccer or basketball and 10% practice both sports. If there is also 60% that do not play soccer, calculate the probability that a student chosen at random from the class:

1. Plays soccer only.
2. Play basketball only.
3. Plays only one of the sports.
4. Plays neither soccer nor basketball.

Sol.

To solve this problem, we can use the principle of inclusion-exclusion and basic probability rules.

Given Data:

• Let ( S ) represent the set of students who play soccer.
• Let ( B ) represent the set of students who play basketball.
• ( P(S \cup B) = 0.60 ) (60% of students play soccer or basketball)
• ( P(S \cap B) = 0.10 ) (10% of students play both sports)
• ( P(S') = 0.60 ) (60% of students do not play soccer)
  
  

1. Probability that a student plays soccer only:
   ( P(S \setminus B) ). [ P(S \setminus B) = P(S) - P(S \cap B) ] Since ( P(S') = 0.60 ), the probability of playing soccer, ( P(S) ), is: [ P(S) = 1 - P(S') = 1 - 0.60 = 0.40 ] Therefore: [ P(S \setminus B) = P(S) - P(S \cap B) = 0.40 - 0.10 = 0.30 ]

2. Probability that a student plays basketball only:
   ( P(B \setminus S) ). [ P(B \setminus S) = P(B) - P(S \cap B) ] We need to find ( P(B) ). Using the principle of inclusion-exclusion for ( P(S \cup B) ): [ P(S \cup B) = P(S) + P(B) - P(S \cap B) ] Given ( P(S \cup B) = 0.60 ): [ 0.60 = 0.40 + P(B) - 0.10 ] Solving for ( P(B) ): [ 0.60 = 0.30 + P(B) ] [ P(B) = 0.30 ] Therefore: [ P(B \setminus S) = P(B) - P(S \cap B) = 0.30 - 0.10 = 0.20 ]

3. Probability that a student plays only one of the sports:
   This is the probability of playing either soccer only or basketball only, ( P((S \setminus B) \cup (B \setminus S)) ). [ P((S \setminus B) \cup (B \setminus S)) = P(S \setminus B) + P(B \setminus S) ] Using the results from above: [ P((S \setminus B) \cup (B \setminus S)) = 0.30 + 0.20 = 0.50 ]

4. Probability that a student plays neither soccer nor basketball:
   Given ( P(S \cup B) = 0.60 ), the probability of playing neither sport, ( P((S \cup B)') ), is: [ P((S \cup B)') = 1 - P(S \cup B) = 1 - 0.60 = 0.40 ]

Summary of Probabilities:

• Plays soccer only: ( 0.30 )
  
• Plays basketball only: ( 0.20 )

- Plays only one of the sports: \( 0.50 \)
- Plays neither soccer nor basketball: \( 0.40 \)

85.Imagine you are writing your semester exams . To write an exam , there are 70% chances that an alarm clock will wake you up successfully. If you hear the alarm clock then there are 95% chances you will write the exam and if you don't hear the alarm the chances are 50%. a)If you have written the exam what are the chances that you heard the alarm clock? b) What are the chances that you didn’t hear the alarm if you have not written the exam?

Sol.

Put your solution here.

86.Lets say an investment company “Future Wealth“ analyses stocks and predicts whether their price will go up or down. So far, half of the stocks analysed by the company have gone up, 3/4 of the stocks that went up were correctly predicted to go up, and 2/5 of the stocks that went down were incorrectly predicted to go up. Suppose that the company tells you that it will go up. Compute the probability that the stock will indeed go up.

Sol.

To solve this problem, we can use Bayes' Theorem. Let's define the following events:

• (U): The stock price goes up.
  
• (D): The stock price goes down.
  
• (P): The company predicts that the stock price will go up.
  

Given data:

• (P(U) = 0.5) (half of the stocks have gone up)
  
• (P(P \mid U) = \frac{3}{4}) (3/4 of the stocks that went up were correctly predicted to go up)
  
• (P(P \mid D) = \frac{2}{5}) (2/5 of the stocks that went down were incorrectly predicted to go up)
  

We need to compute (P(U \mid P)), the probability that the stock price will go up given that the company predicts it will go up.

Using Bayes' Theorem:
[ P(U \mid P) = \frac{P(P \mid U) \cdot P(U)}{P(P)} ]

First, we need to find (P(P)), the total probability that the company predicts the stock will go up. This can be calculated using the law of total probability: [ P(P) = P(P \mid U) \cdot P(U) + P(P \mid D) \cdot P(D) ]

Given:

• (P(U) = 0.5)
  
• (P(D) = 1 - P(U) = 0.5)
  
• (P(P \mid U) = \frac{3}{4} = 0.75)
  
• (P(P \mid D) = \frac{2}{5} = 0.4)
  

Substitute these values into the equation for (P(P)): [ P(P) = 0.75 \cdot 0.5 + 0.4 \cdot 0.5 = 0.375 + 0.2 = 0.575 ]

Now, use Bayes' Theorem to find (P(U \mid P)): [ P(U \mid P) = \frac{0.75 \cdot 0.5}{0.575} = \frac{0.375}{0.575} \approx 0.6522 ]

So, the probability that the stock will indeed go up given that the company predicts it will go up is approximately (0.6522) or (65.22%).

87.Imagine you are a bettor. You are watching a race between two horses A and B. Let’s say five races are conducted. Construct any three hypotheses defining winning probabilities of A and B. What confidence do you have in each of your hypotheses to be true? Lets say , out of 5 races A wins 3 and B wins the remaining 2. Then after 5 races , in which of your hypotheses will you have maximum confidence. As per your new hypothesis which horse has more chances to win the 6th round .

Sol.

Put your solution here.

88.You're training a spam filter . You have data on the frequency of certain words in both spam and non-spam emails. How would you update your beliefs about an email being spam or not spam based on the presence of specific words?Let’s say initially chances of an email being spam is 40%. Data: Word "free" appears in 80% of spam emails and 5% of non-spam emails

Sol.

Put your solution here.

89.Suppose that we use a perceptron to detect spam messages. Let's say that each email message is represented by the frequency of occurrence of keywords, and the output is +1 if the message is considered spam. a) Can you think of some keywords that will end up with a large number of positive weight in the perceptron? b)How about keywords that will get a negative weight? c)What parameters in the perceptron directly affects how many border-line messages end up being classified as spam?

Sol.

Put your solution here.

90.Lets have some parameters 3.1, 4.2 and 4 and the corresponding weights 5,1,and 3 respt. Calculate the weighted sum.

Sol.

Put your solution here.

91.Suppose you have set of numbers ranging from -infinity to infinity. Will it be easy to plot them in a limited screen size and compare them?

Sol.

Put your solution here.

92.Will simply dividing them by some large number work?

Sol.

Put your solution here.

93.Can you think of a way to fit these numbers in some finite range? Think about some kind of functions?

Sol.

Put your solution here.

94.Question $$a+b=5$$, $$a-b=3$$. What is the value of a and b?

Sol.

Put your solution here.

Module J

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95.Write the number 25 in its binary form.

Sol.

Put your solution here.

96.Given a text data, how will you convert it into binary form?

Sol.

Put your solution here.

97.What if we use the binary code for each character according to the ASCII convention? How much space would each character take up?

Sol.

Put your solution here.

98.Suppose I take the following notation for the letters s, o, n, h and a:

s: 00

o: 001

a: 010

h: 011

n: 1

Decode the following string: '00011010110011'

Sol.

Put your solution here.

99.Do you observe that the above string can have 2 different interpretations?

Sol.

Put your solution here.

100.Can this issue occur if we take each code of the same length? Can you define one such coding for the above example, i.e., s, o, n, h and a? At least how many digits would you have to take for each character?

Sol.

Put your solution here.

101.Observe that for 5 unique letters, I cannot have unique binary representations if I take length of each notation exactly 2. Why?

Sol.

Put your solution here.

102.Suppose you go to buy apples. There are three varieties of apples available. Your mom has given you a task that you have to buy 2 apples of any one type, 3 of the any other type, and 5 of the third. How will you minimise the total money spent?

Sol.

Put your solution here.

103.Given the text 'this is a new experience'. Write the frequency distribution of these characters for this sentence.

Sol.

Put your solution here.

104.To which characters should I give a shorter notation as compared to the others?

Sol.

Put your solution here.

105.Do you observe that the issue occurred in fifth question was because the code of 's' is a prefix of the code for 'o'?

Sol.

Put your solution here.

106.What all do you think should be the properties of a proper encoding rule?

Sol.

Put your solution here.

107.Get some quite basic knowledge about trees as data structures.

Sol.

Put your solution here.

108.Observe that the lower is a node in a tree, the more is the time you would take to reach till it from the top node.

Sol.

Put your solution here.

Project 4- Read and Solve this

Module K

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109.In a huge network of webpages, how does the browser decides which webpage should be appeared on top?

Sol.

Put your solution here.

110.Read about the Algorithms Random Walk and Equal points Distribution and try to figure out how they are able to find the importance of the webpages?

Sol.

Put your solution here.

111.Are you able to visualize that the Equal points distribution method is nothing but a repetitive matrix operation on a vector?

Sol.

Put your solution here.

112.Will the points of the webpages calculated the Equal points distribution method converge? If yes how can you be sure about it?

Sol.

Put your solution here.

113.Is this convergence dependent on the initial vector?

Sol.

Put your solution here.

114.If you have three websites, A, B, and C, and all are linked to each other, what happens to the importance of each website if:
Website A is linked to by both B and C.
B is linked to by A.
C is linked to by A.
How does the number of links pointing to a website affect its perceived importance?

Sol.

Put your solution here.

115.Given 4 buckets (A, B, C, and D):
A passes its coins to B and C
B passes its coins to D
C passes its coins to A
D passes its coins to B and C
If each bucket starts with 1 coin, calculate the number of coins in each bucket after first round and second round.

Sol.

Put your solution here.

116.In both the algorithms (Random Walk and Equal Distributions) what problem would you face if we have some highly connected nodes? Would it affect the evaluation of other nodes?

Sol.

Put your solution here.

117.What modifications in the algorithm can you think to solve this problem?

Sol.

Put your solution here.

Project 2- Read and Solve this

Project 3- Read and Solve this