Samacheer Kalvi Class 12 Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

 

Question 1. Find the rank of the matrix \( A = \begin{pmatrix} 1 & -3 & 4 & 7 \\ 9 & 1 & 2 & 0 \end{pmatrix} \)
Answer: The rank of a matrix is the maximum number of linearly independent row or column vectors. We use elementary row transformations to convert the given matrix into its echelon form to find its rank.
The given matrix is \( A = \begin{pmatrix} 1 & -3 & 4 & 7 \\ 9 & 1 & 2 & 0 \end{pmatrix} \).
The order of matrix A is \( 2 \times 4 \). Therefore, the rank of A, denoted as \( \rho(A) \), will be less than or equal to 2 (the minimum of the number of rows and columns).
Let us transform matrix A into an echelon form using row operations.

After the transformation, the matrix is in echelon form. We count the number of non-zero rows.
The number of non-zero rows in the echelon form is 2.
Therefore, the rank of matrix A is \( \rho(A) = 2 \).
In simple words: We changed the matrix using row operations until it looked like a triangle of numbers, which is called echelon form. Then we counted how many rows still had at least one number that wasn't zero. That count is the rank of the matrix.

🎯 Exam Tip: When finding the rank of a matrix, ensure all row operations are performed carefully to avoid calculation errors. The echelon form is key to correctly identifying non-zero rows.

 

Question 2. Find the rank of the matrix \( A = \begin{pmatrix} -2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7 \end{pmatrix} \)
Answer: The rank of a matrix helps understand its properties and is found by converting it to echelon form. The given matrix is \( A = \begin{pmatrix} -2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7 \end{pmatrix} \).
The order of matrix A is \( 3 \times 4 \). So, the rank of A, \( \rho(A) \), will be less than 3.
Let's transform the matrix A into an echelon form using elementary row operations.

The last equivalent matrix is in echelon form. We then count the number of rows that are not entirely zeros.
The number of non-zero rows is 3.
Therefore, the rank of matrix A is \( \rho(A) = 3 \).
In simple words: We used several row changes to make the matrix easier to look at, putting it into a special "echelon" shape. Once it was in this shape, we counted all the rows that had at least one number in them (not just zeros). This count tells us the matrix's rank.

🎯 Exam Tip: When performing multiple row operations, double-check each calculation to ensure accuracy. Mistakes in elementary transformations can lead to an incorrect rank.

 

Question 3. Find the rank of the matrix \( A = \begin{pmatrix} 4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0 \end{pmatrix} \)
Answer: The rank of a matrix is a fundamental property that helps characterize its linear transformation. The given matrix is \( A = \begin{pmatrix} 4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0 \end{pmatrix} \).
The order of matrix A is \( 3 \times 4 \). So, the rank of A, \( \rho(A) \), will be less than 3.
Let's transform the matrix A into an echelon form using elementary row operations.

The last equivalent matrix is in echelon form. By counting the non-zero rows, we can determine the rank.
The number of non-zero rows is 3.
Therefore, the rank of matrix A is \( \rho(A) = 3 \).
In simple words: We changed the matrix using specific math steps to put it into a staircase-like shape called echelon form. Then, we counted how many rows still had numbers in them (not just zeros). This count gives us the matrix's rank, which helps us understand its size in a special mathematical way.

🎯 Exam Tip: When dealing with larger numbers or multiple row operations, it's often helpful to keep a rough work column to check calculations, especially for multiplications and subtractions in row transformations.

 

Question 4. Examine the consistency of the system of equations; x + y + z = 7, x + 2y + 3z = 18, y + 2z = 6
Answer: A system of linear equations is consistent if it has at least one solution. We can check this by comparing the ranks of the coefficient matrix and the augmented matrix. The given equations are:
\( x + y + z = 7 \)
\( x + 2y + 3z = 18 \)
\( y + 2z = 6 \)
The matrix form of these equations is \( AX = B \), where:
\( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), \( B = \begin{pmatrix} 7 \\ 18 \\ 6 \end{pmatrix} \)
The augmented matrix is \( [A, B] = \begin{pmatrix} 1 & 1 & 1 & 7 \\ 1 & 2 & 3 & 18 \\ 0 & 1 & 2 & 6 \end{pmatrix} \).
Now, let's transform the augmented matrix to echelon form to find its rank.

From the echelon form, the rank of the coefficient matrix \( A \) (first three columns) is 2, because the third row has all zeros except the last element from the augmented part.
So, \( \rho(A) = 2 \).
The rank of the augmented matrix \( [A, B] \) (all four columns) is 3, because the third row is \( \begin{pmatrix} 0 & 0 & 0 & -5 \end{pmatrix} \), which is a non-zero row.
So, \( \rho(A, B) = 3 \).
Since \( \rho(A) \neq \rho(A, B) \), the given system of equations is inconsistent and has no solution. This means there are no values for x, y, and z that can satisfy all three equations at the same time.
In simple words: We put all the numbers from the equations into a big table called a matrix, then changed it until it was in a simple staircase shape. By comparing the number of "good" rows in the original part of the matrix versus the full table, we saw they weren't the same. This means there's no way to find x, y, and z that work for all the equations.

🎯 Exam Tip: A system of linear equations is consistent if and only if the rank of the coefficient matrix is equal to the rank of the augmented matrix. If the ranks are different, the system is inconsistent.

 

Question 5. Find k if the equations 2x + 3y - z = 5, 3x - y + 4z = 2, x + 7y - 6z = k are consistent.
Answer: For a system of equations to be consistent, the rank of the coefficient matrix (A) must be equal to the rank of the augmented matrix ([A, B]). We will use this condition to find the value of k. The given equations are:
\( 2x + 3y - z = 5 \)
\( 3x - y + 4z = 2 \)
\( x + 7y - 6z = k \)
The matrix form of these equations is \( AX = B \). The augmented matrix is:
\( [A, B] = \begin{pmatrix} 2 & 3 & -1 & 5 \\ 3 & -1 & 4 & 2 \\ 1 & 7 & -6 & k \end{pmatrix} \)
Let's transform the augmented matrix to echelon form.

The last equivalent matrix is in echelon form. For the system to be consistent, the rank of \( A \) must be equal to the rank of \( [A, B] \).
From the matrix, \( \rho(A) = 2 \) (because the third row of the coefficient part is all zeros).
For \( \rho(A, B) \) to also be 2, the last row \( \begin{pmatrix} 0 & 0 & 0 & k-8 \end{pmatrix} \) must be a zero row. This requires the last element to be zero.
So, \( k-8 = 0 \)
\( \implies k = 8 \)
When \( k=8 \), both ranks are 2, making the system consistent. This means there are many solutions, or a unique solution depending on other factors, but at least one solution exists.
In simple words: To make the equations work together (be consistent), the numbers in the matrices must follow a certain rule about their "rank". After changing the matrix into a simpler form, we found that a part of it should become all zeros for the ranks to match. This helped us figure out that 'k' must be 8 for the equations to have solutions.

🎯 Exam Tip: Remember that for a system to be consistent, \( \rho(A) = \rho(A, B) \). Use this condition to set the appropriate element in the echelon form to zero to solve for unknown variables like 'k'.

 

Question 6. Find k if the equations x + y + z = 1, 3x - y - z = 4, x + 5y + 5z = k are inconsistent.
Answer: A system of linear equations is inconsistent if it has no solution. This occurs when the rank of the coefficient matrix (A) is not equal to the rank of the augmented matrix ([A, B]). We will find k based on this condition. The given equations are:
\( x + y + z = 1 \)
\( 3x - y - z = 4 \)
\( x + 5y + 5z = k \)
The matrix form of these equations is \( AX = B \). The augmented matrix is:
\( [A, B] = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 3 & -1 & -1 & 4 \\ 1 & 5 & 5 & k \end{pmatrix} \)
Let's transform the augmented matrix to echelon form.

The last equivalent matrix is in echelon form. For the system to be inconsistent, we need \( \rho(A) \neq \rho(A, B) \).
From the matrix, \( \rho(A) = 2 \) (because the third row of the coefficient part, \( \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \), is all zeros).
For the system to be inconsistent, \( \rho(A, B) \) must be different from 2. This means the last row of the augmented matrix, \( \begin{pmatrix} 0 & 0 & 0 & k \end{pmatrix} \), must be a non-zero row.
This happens if \( k \neq 0 \).
Therefore, the equations are inconsistent when k takes any real value other than 0.
In simple words: For these equations to have no solution (be inconsistent), the "rank" of the main number table must be different from the rank of the bigger table that includes the answers. After simplifying the big table, we saw that this would only happen if the number 'k' was not zero. If 'k' is any number except zero, the equations will not have a solution.

🎯 Exam Tip: When determining inconsistency, ensure that the last row of the augmented matrix (after echelon form) is non-zero only in the constant column. This clearly indicates \( \rho(A) \neq \rho(A, B) \).

 

Question 7. Solve the equations x + 2y + z = 7, 2x - y + 2z = 4, x + y - 2z = -1 by using Cramer's rule.
Answer: Cramer's rule is a method for solving systems of linear equations using determinants. The given system of equations is:
\( x + 2y + z = 7 \)
\( 2x - y + 2z = 4 \)
\( x + y - 2z = -1 \)
First, we calculate the determinant of the coefficient matrix, \( \Delta \):
\( \Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{vmatrix} \)
\( = 1((-1)(-2) - (2)(1)) - 2((2)(-2) - (2)(1)) + 1((2)(1) - (-1)(1)) \)
\( = 1(2 - 2) - 2(-4 - 2) + 1(2 + 1) \)
\( = 1(0) - 2(-6) + 1(3) \)
\( = 0 + 12 + 3 \)
\( = 15 \)
Since \( \Delta = 15 \neq 0 \), we can apply Cramer's Rule, and the system is consistent with a unique solution.
Next, we calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms:
\( \Delta_x = \begin{vmatrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{vmatrix} \)
\( = 7((-1)(-2) - (2)(1)) - 2((4)(-2) - (2)(-1)) + 1((4)(1) - (-1)(-1)) \)
\( = 7(2 - 2) - 2(-8 + 2) + 1(4 - 1) \)
\( = 7(0) - 2(-6) + 1(3) \)
\( = 0 + 12 + 3 \)
\( = 15 \)
Then, we calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms:
\( \Delta_y = \begin{vmatrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{vmatrix} \)
\( = 1((4)(-2) - (2)(-1)) - 7((2)(-2) - (2)(1)) + 1((2)(-1) - (4)(1)) \)
\( = 1(-8 + 2) - 7(-4 - 2) + 1(-2 - 4) \)
\( = 1(-6) - 7(-6) + 1(-6) \)
\( = -6 + 42 - 6 \)
\( = 30 \)
Finally, we calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms:
\( \Delta_z = \begin{vmatrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{vmatrix} \)
\( = 1((-1)(-1) - (4)(1)) - 2((2)(-1) - (4)(1)) + 7((2)(1) - (-1)(1)) \)
\( = 1(1 - 4) - 2(-2 - 4) + 7(2 + 1) \)
\( = 1(-3) - 2(-6) + 7(3) \)
\( = -3 + 12 + 21 \)
\( = 30 \)
Now, by Cramer's rule, we find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{15}{15} = 1 \)
\( y = \frac{\Delta_y}{\Delta} = \frac{30}{15} = 2 \)
\( z = \frac{\Delta_z}{\Delta} = \frac{30}{15} = 2 \)
The solution to the system of equations is \( (x, y, z) = (1, 2, 2) \). Cramer's rule offers a direct way to solve such systems, particularly useful when unique solutions exist.
In simple words: We used a special math trick called Cramer's rule to solve these three equations. First, we found a main number called Delta. Since Delta wasn't zero, we knew there was a clear answer. Then, we found three other Deltas (Delta-x, Delta-y, Delta-z) by swapping out some numbers. Finally, we divided each of these new Deltas by the main Delta to find the values of x, y, and z.

🎯 Exam Tip: Before applying Cramer's rule, always check if the determinant of the coefficient matrix \( (\Delta) \) is non-zero. If \( \Delta = 0 \), Cramer's rule cannot be directly used, and the system either has no solution or infinitely many solutions.

 

Question 8. The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of wheat, 2kg. of sugar and 1kg of rice is Rs 220. Find the cost of each per kg. using Cramer's rule.
Answer: Let's assume x, y, and z represent the cost per kg of wheat, sugar, and rice, respectively. Based on the problem's solution using Cramer's rule, the system of equations is formed as follows:
1. The cost of 2kg of wheat and 1kg of sugar is Rs 100: \( 2x + y + 0z = 100 \)
2. The cost of 1kg of wheat and 1kg of rice is Rs 80: \( x + 0y + z = 80 \)
3. The cost of 3kg of wheat, 2kg of sugar, and 1kg of rice is Rs 220: \( 3x + 2y + z = 220 \)
We will use Cramer's rule to find x, y, and z. First, calculate the determinant of the coefficient matrix, \( \Delta \):
\( \Delta = \begin{vmatrix} 2 & 1 & 0 \\ 1 & 0 & 1 \\ 3 & 2 & 1 \end{vmatrix} \)
\( = 2((0)(1) - (1)(2)) - 1((1)(1) - (1)(3)) + 0 \)
\( = 2(0 - 2) - 1(1 - 3) \)
\( = 2(-2) - 1(-2) \)
\( = -4 + 2 \)
\( = -2 \)
Since \( \Delta = -2 \neq 0 \), we can apply Cramer's Rule, meaning there's a unique solution.
Next, we calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms:
\( \Delta_x = \begin{vmatrix} 100 & 1 & 0 \\ 80 & 0 & 1 \\ 220 & 2 & 1 \end{vmatrix} \)
\( = 100((0)(1) - (1)(2)) - 1((80)(1) - (1)(220)) + 0 \)
\( = 100(0 - 2) - 1(80 - 220) \)
\( = 100(-2) - 1(-140) \)
\( = -200 + 140 \)
\( = -60 \)
Then, we calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms:
\( \Delta_y = \begin{vmatrix} 2 & 100 & 0 \\ 1 & 80 & 1 \\ 3 & 220 & 1 \end{vmatrix} \)
\( = 2((80)(1) - (1)(220)) - 100((1)(1) - (1)(3)) + 0 \)
\( = 2(80 - 220) - 100(1 - 3) \)
\( = 2(-140) - 100(-2) \)
\( = -280 + 200 \)
\( = -80 \)
Finally, we calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms:
\( \Delta_z = \begin{vmatrix} 2 & 1 & 100 \\ 1 & 0 & 80 \\ 3 & 2 & 220 \end{vmatrix} \)
\( = 2((0)(220) - (80)(2)) - 1((1)(220) - (80)(3)) + 100((1)(2) - (0)(3)) \)
\( = 2(0 - 160) - 1(220 - 240) + 100(2 - 0) \)
\( = 2(-160) - 1(-20) + 100(2) \)
\( = -320 + 20 + 200 \)
\( = -100 \)
Now, using Cramer's rule to find the costs:
\( x = \frac{\Delta_x}{\Delta} = \frac{-60}{-2} = 30 \)
\( y = \frac{\Delta_y}{\Delta} = \frac{-80}{-2} = 40 \)
\( z = \frac{\Delta_z}{\Delta} = \frac{-100}{-2} = 50 \)
The solution is \( (x, y, z) = (30, 40, 50) \).
Therefore, the cost of 1 kg of wheat is Rs 30, the cost of 1 kg of sugar is Rs 40, and the cost of 1 kg of rice is Rs 50. Setting up and solving such systems is crucial for many real-world pricing and resource allocation problems.
In simple words: We turned the problem about buying wheat, sugar, and rice into three math equations. Then, we used Cramer's rule, which is a way to solve these equations using special numbers called determinants. After doing all the calculations, we found out how much each kilogram of wheat, sugar, and rice costs.

🎯 Exam Tip: When solving word problems using Cramer's rule, carefully define your variables and set up the system of linear equations correctly from the given information. A single mistake in forming an equation will lead to an incorrect solution.

 

Question 9. A salesman has the following record of sales during three items A, B, and C, rates of commission. Find out the rate of commission on items A, B, and C by using Cramer's rule.
Answer: Let x, y, and z be the commission rates (in Rs) for items A, B, and C, respectively. We can form a system of linear equations from the provided sales data and total commission drawn.

From the table, we get the following system of equations:
For January: \( 90x + 100y + 20z = 800 \)
Dividing by 10, we get: \( 9x + 10y + 2z = 80 \) .....(1)
For February: \( 130x + 50y + 40z = 900 \)
Dividing by 10, we get: \( 13x + 5y + 4z = 90 \) .....(2)
For March: \( 60x + 100y + 30z = 850 \)
Dividing by 10, we get: \( 6x + 10y + 3z = 85 \) .....(3)
Now, we use Cramer's rule to solve this system.
First, calculate the determinant of the coefficient matrix, \( \Delta \):
\( \Delta = \begin{vmatrix} 9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3 \end{vmatrix} \)
\( = 9((5)(3) - (4)(10)) - 10((13)(3) - (4)(6)) + 2((13)(10) - (5)(6)) \)
\( = 9(15 - 40) - 10(39 - 24) + 2(130 - 30) \)
\( = 9(-25) - 10(15) + 2(100) \)
\( = -225 - 150 + 200 \)
\( = -175 \)
Since \( \Delta = -175 \neq 0 \), Cramer's Rule is applicable, and the system is consistent with a unique solution. Sales commission rates are typically unique for each product.
Next, calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms:
\( \Delta_x = \begin{vmatrix} 80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3 \end{vmatrix} \)
\( = 80((5)(3) - (4)(10)) - 10((90)(3) - (4)(85)) + 2((90)(10) - (5)(85)) \)
\( = 80(15 - 40) - 10(270 - 340) + 2(900 - 425) \)
\( = 80(-25) - 10(-70) + 2(475) \)
\( = -2000 + 700 + 950 \)
\( = -350 \)
Then, calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms:
\( \Delta_y = \begin{vmatrix} 9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3 \end{vmatrix} \)
\( = 9((90)(3) - (4)(85)) - 80((13)(3) - (4)(6)) + 2((13)(85) - (90)(6)) \)
\( = 9(270 - 340) - 80(39 - 24) + 2(1105 - 540) \)
\( = 9(-70) - 80(15) + 2(565) \)
\( = -630 - 1200 + 1130 \)
\( = -700 \)
Finally, calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms:
\( \Delta_z = \begin{vmatrix} 9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85 \end{vmatrix} \)
\( = 9((5)(85) - (90)(10)) - 10((13)(85) - (90)(6)) + 80((13)(10) - (5)(6)) \)
\( = 9(425 - 900) - 10(1105 - 540) + 80(130 - 30) \)
\( = 9(-475) - 10(565) + 80(100) \)
\( = -4275 - 5650 + 8000 \)
\( = -1925 \)
Now, using Cramer's rule to find the commission rates:
\( x = \frac{\Delta_x}{\Delta} = \frac{-350}{-175} = 2 \)
\( y = \frac{\Delta_y}{\Delta} = \frac{-700}{-175} = 4 \)
\( z = \frac{\Delta_z}{\Delta} = \frac{-1925}{-175} = 11 \)
The solution is \( (x, y, z) = (2, 4, 11) \).
Therefore, the rates of commission for items A, B, and C are Rs 2, Rs 4, and Rs 11 respectively.
In simple words: We used the salesman's sales records to create math equations for his commission rates on items A, B, and C. Then, we solved these equations using Cramer's rule, which involves finding special numbers called determinants. This helped us figure out exactly how much commission he earns for each item.

🎯 Exam Tip: When given data in a table, first extract the linear equations accurately. Simplifying the equations (like dividing by a common factor) before calculating determinants can make the numbers smaller and reduce calculation errors.

 

Question 10. The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. In the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Answer: This problem can be solved using a transition probability matrix, which models how states (subscribing or not subscribing) change over time. Let S represent "subscribes" and F represent "does not subscribe".
The transition probabilities are:
- From S to S (subscribes again): 60% or 0.60
- From S to F (subscribes then stops): \( 100\% - 60\% = 40\% \) or 0.40
- From F to S (does not subscribe then starts): 25% or 0.25
- From F to F (does not subscribe and continues not to): \( 100\% - 25\% = 75\% \) or 0.75
The transition probability matrix, T, is:
\[ T = \begin{pmatrix} 0.60 & 0.40 \\ 0.25 & 0.75 \end{pmatrix} \] Here, the rows represent the current state (S, F), and the columns represent the next state (S, F).
In the last letter, 40% ordered a subscription, meaning 60% did not. So, the initial state vector \( P_0 \) (current percentages of subscribers and non-subscribers) is:
\( P_0 = \begin{pmatrix} 0.40 & 0.60 \end{pmatrix} \) (40% subscribe, 60% do not).
To find the expected percentage of subscribers in the current letter (after one year/one mailing cycle), we multiply the initial state vector by the transition matrix:
\( P_1 = P_0 T \)
\[ P_1 = \begin{pmatrix} 0.40 & 0.60 \end{pmatrix} \begin{pmatrix} 0.60 & 0.40 \\ 0.25 & 0.75 \end{pmatrix} \] To calculate this, we do:
\( (0.40 \times 0.60) + (0.60 \times 0.25) \) for the new 'S' state
\( = 0.24 + 0.15 = 0.39 \)
And:
\( (0.40 \times 0.40) + (0.60 \times 0.75) \) for the new 'F' state
\( = 0.16 + 0.45 = 0.61 \)
So, the new state vector is \( P_1 = \begin{pmatrix} 0.39 & 0.61 \end{pmatrix} \).
This means 39% of those receiving the current letter can be expected to order a subscription, and 61% are not expected to subscribe. This method of using Markov chains is powerful for predicting future states in many fields like economics and biology.
In simple words: We used a special table called a "transition matrix" to show how people change from subscribing to not subscribing, or vice-versa. Since we knew how many people subscribed last time, we used this matrix to predict how many people will likely subscribe this time. It turned out that 39% are expected to subscribe.

🎯 Exam Tip: When setting up a transition matrix, ensure the rows (or columns, depending on convention) sum to 1, representing all possible outcomes from a given state. Also, correctly define your initial state vector (percentages for each state).

 

Question 1. Find the rank of the matrix \( A = \begin{pmatrix} 1 & -3 & 4 & 7 \\ 9 & 1 & 2 & 0 \end{pmatrix} \)
Answer: The given matrix is: \[ A = \begin{pmatrix} 1 & -3 & 4 & 7 \\ 9 & 1 & 2 & 0 \end{pmatrix} \] First, we apply an elementary transformation to simplify the matrix. We use the operation \( R_2 \rightarrow R_2 - 9R_1 \) to make the element in the first column of the second row zero. The transformed matrix in echelon form is: \[ \begin{pmatrix} 1 & -3 & 4 & 7 \\ 0 & 28 & -34 & -63 \end{pmatrix} \] Now, we count the number of non-zero rows in this echelon form. Both rows contain at least one non-zero element. Understanding echelon form helps simplify matrix analysis. Therefore, the number of non-zero rows is 2. So, the rank of the matrix \( P(A) \) is 2.
In simple words: We change the matrix step-by-step until it looks like a triangle of numbers, with zeros below the main diagonal. Then, we count how many rows still have numbers that are not zero. That count is the rank of the matrix.

🎯 Exam Tip: When finding the rank of a matrix, always ensure the final matrix is in a true echelon form to correctly count the non-zero rows.

 

Question 2. Find the rank of the matrix \( A = \begin{pmatrix} -2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7 \end{pmatrix} \)
Answer: The given matrix is: \[ A = \begin{pmatrix} -2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7 \end{pmatrix} \] The order of matrix A is \( 3 \times 4 \), which means it has 3 rows and 4 columns. Therefore, the rank of A, \( P(A) \), must be less than or equal to 3. We transform the matrix into echelon form using elementary row operations:

The final matrix is in echelon form. Now, we count the number of non-zero rows. All three rows have at least one non-zero element. Performing row operations is a key technique in linear algebra. Therefore, the number of non-zero rows is 3. So, the rank of the matrix \( P(A) \) is 3.
In simple words: We use row operations to make the matrix simpler, creating zeros in certain positions. After these steps, if all three rows still have some numbers, then the rank of the matrix is 3. This means it has full rank for its number of rows.

🎯 Exam Tip: When performing row operations, aim to create zeros below the main diagonal in a systematic way to reach the echelon form correctly. Always check your calculations carefully.

 

Question 3. Find the rank of the matrix \( A = \begin{pmatrix} 4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0 \end{pmatrix} \)
Answer: The given matrix is: \[ A = \begin{pmatrix} 4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0 \end{pmatrix} \] The order of matrix A is \( 3 \times 4 \). So, the rank \( P(A) \) must be less than or equal to 3. We transform the matrix into echelon form using elementary row operations:

The final matrix is in echelon form. Gaussian elimination helps achieve this form systematically. Now, we count the number of non-zero rows. All three rows have at least one non-zero element. Therefore, the number of non-zero rows is 3. So, the rank of the matrix \( P(A) \) is 3.
In simple words: We perform a series of row operations to transform the matrix. This means swapping rows, multiplying rows by numbers, or adding multiples of one row to another. The goal is to get zero in specific positions. Once done, we count the rows that still have at least one non-zero number; this count is the matrix's rank.

🎯 Exam Tip: When faced with a complex matrix, try to make the first element of the first row '1' (or '–1') by swapping rows or dividing, as this simplifies subsequent row operations.

 

Question 4. Examine the consistency of the system of equations; \( x + y + z = 7 \), \( x + 2y + 3z = 18 \), \( y + 2z = 6 \)
Answer: The given system of equations is: 1. \( x + y + z = 7 \) 2. \( x + 2y + 3z = 18 \) 3. \( y + 2z = 6 \) We write these equations in matrix form, \( AX = B \), and then form the augmented matrix \( [A, B] \). \[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 2 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, B = \begin{pmatrix} 7 \\ 18 \\ 6 \end{pmatrix} \] The augmented matrix \( [A, B] \) is: \[ [A, B] = \begin{pmatrix} 1 & 1 & 1 & 7 \\ 1 & 2 & 3 & 18 \\ 0 & 1 & 2 & 6 \end{pmatrix} \] Now, we transform this augmented matrix into echelon form using row operations:

This is the echelon form. Now, we find the rank of the coefficient matrix A (first three columns) and the rank of the augmented matrix \( [A, B] \). For matrix A: \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \] The number of non-zero rows in A is 2, so \( P(A) = 2 \). For the augmented matrix \( [A, B] \): \[ \begin{pmatrix} 1 & 1 & 1 & 7 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 0 & -5 \end{pmatrix} \] The number of non-zero rows in \( [A, B] \) is 3, because the last row contains the non-zero element -5. So, \( P(A, B) = 3 \). Since \( P(A) \neq P(A, B) \) (because \( 2 \neq 3 \)), the system of equations is inconsistent and has no solution. A system with different ranks for its coefficient and augmented matrices signifies that no single set of variable values can satisfy all equations simultaneously.
In simple words: To see if equations have a solution, we write them in a special matrix form. We then make the matrix simpler, like a staircase. If the "equation part" of the matrix has a different number of non-zero rows than the "full matrix," then there is no solution, and the equations are called inconsistent.

🎯 Exam Tip: A system of linear equations is consistent if and only if the rank of the coefficient matrix A is equal to the rank of the augmented matrix [A|B]. If the ranks are different, the system is inconsistent.

 

Question 5. Find k if the equations \( 2x + 3y - z = 5 \), \( 3x - y + 4z = 2 \), \( x + 7y - 6z = k \) are consistent.
Answer: The given system of equations is: 1. \( 2x + 3y - z = 5 \) 2. \( 3x - y + 4z = 2 \) 3. \( x + 7y - 6z = k \) The matrix form is \( AX = B \). The augmented matrix \( [A, B] \) is: \[ [A, B] = \begin{pmatrix} 2 & 3 & -1 & 5 \\ 3 & -1 & 4 & 2 \\ 1 & 7 & -6 & k \end{pmatrix} \] We transform this augmented matrix into echelon form using row operations:

This is the echelon form of the augmented matrix. For the system to be consistent, the rank of the coefficient matrix A must be equal to the rank of the augmented matrix \( [A, B] \). From the echelon form, the coefficient matrix A (first three columns) is: \[ \begin{pmatrix} 1 & 7 & -6 \\ 0 & -11 & 11 \\ 0 & 0 & 0 \end{pmatrix} \] The number of non-zero rows in A is 2, so \( P(A) = 2 \). For the augmented matrix \( [A, B] \), if \( k-8 \neq 0 \), then the last row would be non-zero, making \( P(A, B) = 3 \). In this case, \( P(A) \neq P(A, B) \), making the system inconsistent. For the system to be consistent, we must have \( P(A) = P(A, B) \). This means the last row of \( [A, B] \) must also be all zeros. So, \( k-8 = 0 \).
\( \implies k = 8 \) When \( k = 8 \), the last row becomes \( (0, 0, 0, 0) \). Then \( P(A, B) = 2 \), which equals \( P(A) \). Thus, for the equations to be consistent, the value of \( k \) must be 8. This ensures that the system has at least one solution.
In simple words: We change the matrix of the equations into a simpler form. For the equations to have solutions, the number of non-zero rows in the main part of the matrix must be the same as in the full matrix. This only happens if a specific value, \( k-8 \), becomes zero, which tells us \( k \) must be 8.

🎯 Exam Tip: Remember that for a system of linear equations to be consistent, the rank of the coefficient matrix must be equal to the rank of the augmented matrix [A|B]. Use this condition to solve for unknown variables like k.

 

Question 6. Find k if the equations \( x + y + z = 1 \), \( 3x - y - z = 4 \), \( x + 5y + 5z = k \) are inconsistent.
Answer: The given system of equations is: 1. \( x + y + z = 1 \) 2. \( 3x - y - z = 4 \) 3. \( x + 5y + 5z = k \) The matrix form is \( AX = B \). The augmented matrix \( [A, B] \) is: \[ [A, B] = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 3 & -1 & -1 & 4 \\ 1 & 5 & 5 & k \end{pmatrix} \] We transform this augmented matrix into echelon form using row operations:

This is the echelon form of the augmented matrix. For the system to be inconsistent, the rank of the coefficient matrix A must be different from the rank of the augmented matrix \( [A, B] \). From the echelon form, the coefficient matrix A (first three columns) is: \[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & -4 & -4 \\ 0 & 0 & 0 \end{pmatrix} \] The number of non-zero rows in A is 2, so \( P(A) = 2 \). For the augmented matrix \( [A, B] \): If \( k \neq 0 \), then the last row of \( [A, B] \) is \( (0, 0, 0, k) \) where k is non-zero. So \( P(A, B) = 3 \). In this case, \( P(A) \neq P(A, B) \) (because \( 2 \neq 3 \)), making the system inconsistent. If \( k = 0 \), then the last row of \( [A, B] \) would be \( (0, 0, 0, 0) \). Then \( P(A, B) = 2 \), which equals \( P(A) \). This would make the system consistent. Therefore, for the equations to be inconsistent, \( k \) must assume any real value other than 0. This ensures there's a contradiction in the system, preventing a solution.
In simple words: We simplify the matrix of the equations into its echelon form. For the equations to have no solution (be inconsistent), the rank of the main matrix must be different from the rank of the full augmented matrix. This happens when the value 'k' in the last position is not zero.

🎯 Exam Tip: For a system to be inconsistent, ensure that after reducing the augmented matrix to echelon form, you have a row where all elements of the coefficient matrix are zero, but the corresponding element in the augmented column is non-zero.

 

Question 7. Solve the equations \( x + 2y + z = 7 \), \( 2x - y + 2z = 4 \), \( x + y - 2z = -1 \) by using Cramer's rule.
Answer: The given system of equations is: 1. \( x + 2y + z = 7 \) 2. \( 2x - y + 2z = 4 \) 3. \( x + y - 2z = -1 \) We will use Cramer's rule to solve these equations. First, calculate the determinant of the coefficient matrix, \( \Delta \): \[ \Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{vmatrix} \]
\( \implies \Delta = 1((-1)(-2) - (2)(1)) - 2((2)(-2) - (2)(1)) + 1((2)(1) - (-1)(1)) \)
\( \implies \Delta = 1(2 - 2) - 2(-4 - 2) + 1(2 + 1) \)
\( \implies \Delta = 1(0) - 2(-6) + 1(3) \)
\( \implies \Delta = 0 + 12 + 3 \)
\( \implies \Delta = 15 \) Since \( \Delta = 15 \neq 0 \), we can apply Cramer's Rule, and the system is consistent with a unique solution. A non-zero determinant is a good indicator of solvability. Next, calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms: \[ \Delta_x = \begin{vmatrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{vmatrix} \]
\( \implies \Delta_x = 7((-1)(-2) - (2)(1)) - 2((4)(-2) - (2)(-1)) + 1((4)(1) - (-1)(-1)) \)
\( \implies \Delta_x = 7(2 - 2) - 2(-8 + 2) + 1(4 - 1) \)
\( \implies \Delta_x = 7(0) - 2(-6) + 1(3) \)
\( \implies \Delta_x = 0 + 12 + 3 \)
\( \implies \Delta_x = 15 \) Then, calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms: \[ \Delta_y = \begin{vmatrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{vmatrix} \]
\( \implies \Delta_y = 1((4)(-2) - (2)(-1)) - 7((2)(-2) - (2)(1)) + 1((2)(-1) - (4)(1)) \)
\( \implies \Delta_y = 1(-8 + 2) - 7(-4 - 2) + 1(-2 - 4) \)
\( \implies \Delta_y = 1(-6) - 7(-6) + 1(-6) \)
\( \implies \Delta_y = -6 + 42 - 6 \)
\( \implies \Delta_y = 30 \) Finally, calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms: \[ \Delta_z = \begin{vmatrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{vmatrix} \]
\( \implies \Delta_z = 1((-1)(-1) - (4)(1)) - 2((2)(-1) - (4)(1)) + 7((2)(1) - (-1)(1)) \)
\( \implies \Delta_z = 1(1 - 4) - 2(-2 - 4) + 7(2 + 1) \)
\( \implies \Delta_z = 1(-3) - 2(-6) + 7(3) \)
\( \implies \Delta_z = -3 + 12 + 21 \)
\( \implies \Delta_z = 30 \) Now we find x, y, and z using Cramer's rule: \( x = \frac{\Delta_x}{\Delta} = \frac{15}{15} = 1 \) \( y = \frac{\Delta_y}{\Delta} = \frac{30}{15} = 2 \) \( z = \frac{\Delta_z}{\Delta} = \frac{30}{15} = 2 \) The solution to the system of equations is \( (x, y, z) = (1, 2, 2) \).
In simple words: Cramer's rule helps solve equations by using special numbers called determinants. We calculate a main determinant and then three more determinants, one for each unknown (x, y, z). By dividing these numbers, we find the values of x, y, and z that solve the equations. This method works well when the main determinant is not zero.

🎯 Exam Tip: When using Cramer's rule, be very careful with the signs in the determinant expansion and ensure you substitute the constant terms into the correct columns for \( \Delta_x, \Delta_y, \Delta_z \).

 

Question 8. The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is Rs 220. Find the cost of each per kg. using Cramer's rule.
Answer: Let \( x \) be the cost of 1 kg of wheat, \( y \) be the cost of 1 kg of sugar, and \( z \) be the cost of 1 kg of rice. From the problem description, we form the following system of linear equations: 1. Cost of 2kg wheat and 1kg sugar is Rs 100: \( 2x + y + 0z = 100 \) 2. Cost of 1kg wheat and 1kg rice is Rs 80 (implied from solution): \( x + 0y + z = 80 \) 3. Cost of 3kg wheat, 2kg sugar, and 1kg rice is Rs 220: \( 3x + 2y + z = 220 \) We use Cramer's rule to find \( x, y, z \). First, calculate the determinant of the coefficient matrix, \( \Delta \): \[ \Delta = \begin{vmatrix} 2 & 1 & 0 \\ 1 & 0 & 1 \\ 3 & 2 & 1 \end{vmatrix} \]
\( \implies \Delta = 2((0)(1) - (1)(2)) - 1((1)(1) - (1)(3)) + 0 \)
\( \implies \Delta = 2(0 - 2) - 1(1 - 3) \)
\( \implies \Delta = 2(-2) - 1(-2) \)
\( \implies \Delta = -4 + 2 \)
\( \implies \Delta = -2 \) Since \( \Delta = -2 \neq 0 \), Cramer's Rule can be applied, and a unique solution exists for the system. Next, calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms: \[ \Delta_x = \begin{vmatrix} 100 & 1 & 0 \\ 80 & 0 & 1 \\ 220 & 2 & 1 \end{vmatrix} \]
\( \implies \Delta_x = 100((0)(1) - (1)(2)) - 1((80)(1) - (1)(220)) + 0 \)
\( \implies \Delta_x = 100(0 - 2) - 1(80 - 220) \)
\( \implies \Delta_x = 100(-2) - 1(-140) \)
\( \implies \Delta_x = -200 + 140 \)
\( \implies \Delta_x = -60 \) Then, calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms: \[ \Delta_y = \begin{vmatrix} 2 & 100 & 0 \\ 1 & 80 & 1 \\ 3 & 220 & 1 \end{vmatrix} \]
\( \implies \Delta_y = 2((80)(1) - (1)(220)) - 100((1)(1) - (0)(3)) + 0 \)
\( \implies \Delta_y = 2(80 - 220) - 100(1 - 0) \)
\( \implies \Delta_y = 2(-140) - 100(1) \)
\( \implies \Delta_y = -280 - 100 \)
\( \implies \Delta_y = -80 \) Finally, calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms: \[ \Delta_z = \begin{vmatrix} 2 & 1 & 100 \\ 1 & 0 & 80 \\ 3 & 2 & 220 \end{vmatrix} \]
\( \implies \Delta_z = 2((0)(220) - (80)(2)) - 1((1)(220) - (80)(3)) + 100((1)(2) - (0)(3)) \)
\( \implies \Delta_z = 2(0 - 160) - 1(220 - 240) + 100(2 - 0) \)
\( \implies \Delta_z = 2(-160) - 1(-20) + 100(2) \)
\( \implies \Delta_z = -320 + 20 + 200 \)
\( \implies \Delta_z = -100 \) Now, we find x, y, and z using Cramer's rule: \( x = \frac{\Delta_x}{\Delta} = \frac{-60}{-2} = 30 \) \( y = \frac{\Delta_y}{\Delta} = \frac{-80}{-2} = 40 \) \( z = \frac{\Delta_z}{\Delta} = \frac{-100}{-2} = 50 \) Therefore, the cost of 1 kg of wheat is Rs 30, the cost of 1 kg of sugar is Rs 40, and the cost of 1 kg of rice is Rs 50. Setting up equations from real-world scenarios is a practical application of algebra.
In simple words: We set up equations based on the prices of wheat, sugar, and rice. Using Cramer's rule, which involves calculating special numbers called determinants, we solve these equations. This helps us find the individual cost for one kilogram of each item.

🎯 Exam Tip: When solving word problems with Cramer's rule, always define your variables clearly and correctly translate the word problem into a system of linear equations before calculating determinants.

 

Question 9. A salesman has the following record of sales during three items A, B, and C, rates of commission. Find out the rate of commission on items A, B, and C by using Cramer's rule.
Answer: Let \( x, y, \) and \( z \) be the rates of commission (in Rs) for items A, B, and C respectively. From the given table, we can form a system of linear equations:

The equations are: 1. \( 90x + 100y + 20z = 800 \)
\( \implies 9x + 10y + 2z = 80 \) (Dividing by 10) 2. \( 130x + 50y + 40z = 900 \)
\( \implies 13x + 5y + 4z = 90 \) (Dividing by 10) 3. \( 60x + 100y + 30z = 850 \)
\( \implies 6x + 10y + 3z = 85 \) (Dividing by 10) We use Cramer's rule to solve this system. First, calculate the determinant of the coefficient matrix, \( \Delta \): \[ \Delta = \begin{vmatrix} 9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3 \end{vmatrix} \]
\( \implies \Delta = 9((5)(3) - (4)(10)) - 10((13)(3) - (4)(6)) + 2((13)(10) - (5)(6)) \)
\( \implies \Delta = 9(15 - 40) - 10(39 - 24) + 2(130 - 30) \)
\( \implies \Delta = 9(-25) - 10(15) + 2(100) \)
\( \implies \Delta = -225 - 150 + 200 \)
\( \implies \Delta = -175 \) Since \( \Delta = -175 \neq 0 \), Cramer's Rule can be applied, and a unique solution exists. Determinants are fundamental in solving linear systems and are derived from matrix properties. Next, calculate \( \Delta_x \) by replacing the first column of \( \Delta \) with the constant terms: \[ \Delta_x = \begin{vmatrix} 80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3 \end{vmatrix} \]
\( \implies \Delta_x = 80((5)(3) - (4)(10)) - 10((90)(3) - (4)(85)) + 2((90)(10) - (5)(85)) \)
\( \implies \Delta_x = 80(15 - 40) - 10(270 - 340) + 2(900 - 425) \)
\( \implies \Delta_x = 80(-25) - 10(-70) + 2(475) \)
\( \implies \Delta_x = -2000 + 700 + 950 \)
\( \implies \Delta_x = -350 \) Then, calculate \( \Delta_y \) by replacing the second column of \( \Delta \) with the constant terms: \[ \Delta_y = \begin{vmatrix} 9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3 \end{vmatrix} \]
\( \implies \Delta_y = 9((90)(3) - (4)(85)) - 80((13)(3) - (4)(6)) + 2((13)(85) - (90)(6)) \)
\( \implies \Delta_y = 9(270 - 340) - 80(39 - 24) + 2(1105 - 540) \)
\( \implies \Delta_y = 9(-70) - 80(15) + 2(565) \)
\( \implies \Delta_y = -630 - 1200 + 1130 \)
\( \implies \Delta_y = -700 \) Finally, calculate \( \Delta_z \) by replacing the third column of \( \Delta \) with the constant terms: \[ \Delta_z = \begin{vmatrix} 9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85 \end{vmatrix} \]
\( \implies \Delta_z = 9((5)(85) - (90)(10)) - 10((13)(85) - (90)(6)) + 80((13)(10) - (5)(6)) \)
\( \implies \Delta_z = 9(425 - 900) - 10(1105 - 540) + 80(130 - 30) \)
\( \implies \Delta_z = 9(-475) - 10(565) + 80(100) \)
\( \implies \Delta_z = -4275 - 5650 + 8000 \)
\( \implies \Delta_z = -1925 \) Now we find x, y, and z using Cramer's rule: \( x = \frac{\Delta_x}{\Delta} = \frac{-350}{-175} = 2 \) \( y = \frac{\Delta_y}{\Delta} = \frac{-700}{-175} = 4 \) \( z = \frac{\Delta_z}{\Delta} = \frac{-1925}{-175} = 11 \) Therefore, the rates of commission for items A, B, and C are Rs 2, Rs 4, and Rs 11 respectively. This problem demonstrates how real-world data can be solved using matrix methods.
In simple words: We use the sales data to create mathematical equations for the commission rates. Then, we apply Cramer's rule by calculating various determinants. This method helps us solve the equations and find out exactly how much commission the salesman gets for each item.

🎯 Exam Tip: When dealing with data presented in tables for Cramer's rule problems, ensure you correctly set up the equations before performing determinant calculations. Any error in forming the equations will lead to an incorrect final answer.

 

Question 10. The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. In the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Answer: Let S represent those who subscribe and F represent those who do not subscribe. We are given the following probabilities:

• 60% of subscribers will subscribe again: \( P(S \rightarrow S) = 0.60 \)
• 40% of subscribers will not subscribe again: \( P(S \rightarrow F) = 1 - 0.60 = 0.40 \)
• 25% of non-subscribers will subscribe: \( P(F \rightarrow S) = 0.25 \)
• 75% of non-subscribers will not subscribe again: \( P(F \rightarrow F) = 1 - 0.25 = 0.75 \)

We can form the transition probability matrix \( T \): \[ T = \begin{pmatrix} P(S \rightarrow S) & P(S \rightarrow F) \\ P(F \rightarrow S) & P(F \rightarrow F) \end{pmatrix} = \begin{pmatrix} 0.60 & 0.40 \\ 0.25 & 0.75 \end{pmatrix} \] In the last letter, it was found that 40% ordered a subscription. This means that initially, 40% were subscribers (S) and 60% were non-subscribers (F). So, the initial state matrix (or distribution of subscribers/non-subscribers) is \( \begin{pmatrix} 0.40 & 0.60 \end{pmatrix} \). To find the expected percentage of subscribers after the current letter (representing one year or one cycle), we multiply the initial state matrix by the transition matrix \( T \): Let \( P_1 \) be the state after one letter.\( P_1 = \begin{pmatrix} 0.40 & 0.60 \end{pmatrix} \begin{pmatrix} 0.60 & 0.40 \\ 0.25 & 0.75 \end{pmatrix} \)
\( \implies P_1 = ( (0.40)(0.60) + (0.60)(0.25) \quad (0.40)(0.40) + (0.60)(0.75) ) \)
\( \implies P_1 = ( 0.24 + 0.15 \quad 0.16 + 0.45 ) \)
\( \implies P_1 = ( 0.39 \quad 0.61 ) \) This result means 39% can be expected to subscribe (S) and 61% can be expected not to subscribe (F). Transition matrices are a useful tool in modeling dynamic systems where states change over time. Therefore, 39% of those receiving the current letter can be expected to order a subscription.
In simple words: We track how people change their subscription habits using a special matrix called a transition matrix. We start with the percentage of people who subscribed last time. Then, by multiplying this with the transition matrix, we can predict what percentage will subscribe this time. This helps predict future trends in customer behavior.

🎯 Exam Tip: When working with transition matrices, ensure the matrix is correctly set up with the probabilities (e.g., sum of probabilities in each row is 1). The initial state vector should also represent the correct distribution before the transition.