Samacheer Kalvi Class 12 Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1. Solve the following equations by using Cramer's rule
(i) \( 2x + 3y = 7; 3x + 5y = 9 \)
(ii) \( 5x + 3y = 17; 3x + 7y = 31 \)
(iii) \( 2x + y - z = 3; x + y + z = 1; x - 2y - 3z = 4 \)
(iv) \( x + y + z = 6; 2x + 3y - z = 5; 6x - 2y - 3z = -7 \)
(v) \( x + 4y + 3z = 2; 2x - 6y + 6z = -3; 5x - 2y + 3z = -5 \)
Answer:
(i) Given equations are:
\( 2x + 3y = 7 \)
\( 3x + 5y = 9 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ll} 2 & 3 \\ 3 & 5 \end{array}\right| = (2 \times 5) - (3 \times 3) = 10 - 9 = 1 \)
Since \( \Delta = 1 \neq 0 \), we can use Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ll} 7 & 3 \\ 9 & 5 \end{array}\right| = (7 \times 5) - (3 \times 9) = 35 - 27 = 8 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ll} 2 & 7 \\ 3 & 9 \end{array}\right| = (2 \times 9) - (7 \times 3) = 18 - 21 = -3 \)

Now, apply Cramer's Rule to find x and y:
\( x = \frac{\Delta_x}{\Delta} = \frac{8}{1} = 8 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-3}{1} = -3 \)
So, the solution is \( x = 8 \) and \( y = -3 \). This method uses determinants to efficiently solve for variables in a system of linear equations.
In simple words: We find three special numbers using the equation numbers. One number is for the main problem, and two others are for 'x' and 'y'. We then divide to get the final answers for x and y.

🎯 Exam Tip: Always check that the main determinant \( \Delta \) is not zero. If \( \Delta = 0 \), Cramer's Rule cannot be used, and the system either has no solution or infinitely many solutions.

Answer:
(ii) Given equations are:
\( 5x + 3y = 17 \)
\( 3x + 7y = 31 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ll} 5 & 3 \\ 3 & 7 \end{array}\right| = (5 \times 7) - (3 \times 3) = 35 - 9 = 26 \)
Since \( \Delta = 26 \neq 0 \), we can use Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ll} 17 & 3 \\ 31 & 7 \end{array}\right| = (17 \times 7) - (3 \times 31) = 119 - 93 = 26 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ll} 5 & 17 \\ 3 & 31 \end{array}\right| = (5 \times 31) - (17 \times 3) = 155 - 51 = 104 \)

Now, apply Cramer's Rule to find x and y:
\( x = \frac{\Delta_x}{\Delta} = \frac{26}{26} = 1 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{104}{26} = 4 \)
So, the solution is \( x = 1 \) and \( y = 4 \). The steps for Cramer's rule are consistent regardless of the numbers involved, making it a reliable method.
In simple words: We find the main determinant and the determinants for x and y using the numbers from the equations. Then we divide to find the answers for x and y.

🎯 Exam Tip: Double-check all multiplications and subtractions when calculating determinants, as a small error can lead to incorrect final answers.

Answer:
(iii) Given equations are:
\( 2x + y - z = 3 \)
\( x + y + z = 1 \)
\( x - 2y - 3z = 4 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ccc} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & -3 \end{array}\right| \)
\( = 2((1 \times -3) - (1 \times -2)) - 1((1 \times -3) - (1 \times 1)) + (-1)((1 \times -2) - (1 \times 1)) \)
\( = 2(-3 + 2) - 1(-3 - 1) - 1(-2 - 1) \)
\( = 2(-1) - 1(-4) - 1(-3) \)
\( = -2 + 4 + 3 = 5 \)
Since \( \Delta = 5 \neq 0 \), we can use Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & -3 \end{array}\right| \)
\( = 3((1 \times -3) - (1 \times -2)) - 1((1 \times -3) - (1 \times 4)) + (-1)((1 \times -2) - (1 \times 4)) \)
\( = 3(-3 + 2) - 1(-3 - 4) - 1(-2 - 4) \)
\( = 3(-1) - 1(-7) - 1(-6) \)
\( = -3 + 7 + 6 = 10 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ccc} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{array}\right| \)
\( = 2((1 \times -3) - (1 \times 4)) - 3((1 \times -3) - (1 \times 1)) + (-1)((1 \times 4) - (1 \times 1)) \)
\( = 2(-3 - 4) - 3(-3 - 1) - 1(4 - 1) \)
\( = 2(-7) - 3(-4) - 1(3) \)
\( = -14 + 12 - 3 = -5 \)

Next, calculate \( \Delta_z \) by replacing the z-coefficients with the constant terms:
\( \Delta_z = \left|\begin{array}{ccc} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{array}\right| \)
\( = 2((1 \times 4) - (1 \times -2)) - 1((1 \times 4) - (1 \times 1)) + 3((1 \times -2) - (1 \times 1)) \)
\( = 2(4 + 2) - 1(4 - 1) + 3(-2 - 1) \)
\( = 2(6) - 1(3) + 3(-3) \)
\( = 12 - 3 - 9 = 0 \)

Now, apply Cramer's Rule to find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{10}{5} = 2 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-5}{5} = -1 \)
\( \implies \) \( z = \frac{\Delta_z}{\Delta} = \frac{0}{5} = 0 \)
So, the solution is \( (x, y, z) = (2, -1, 0) \). This method is good for solving systems with three equations and three unknown variables.
In simple words: For three equations, we find four special numbers called determinants. One is for the main equation, and one for each variable (x, y, z). We then divide each variable's determinant by the main determinant to find its value.

🎯 Exam Tip: When working with 3x3 determinants, pay close attention to the alternating positive and negative signs for each term's expansion (e.g., \( + - + \)).

Answer:
(iv) Given equations are:
\( x + y + z = 6 \)
\( 2x + 3y - z = 5 \)
\( 6x - 2y - 3z = -7 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 6 & -2 & -3 \end{array}\right| \)
\( = 1((3 \times -3) - (-1 \times -2)) - 1((2 \times -3) - (-1 \times 6)) + 1((2 \times -2) - (3 \times 6)) \)
\( = 1(-9 - 2) - 1(-6 + 6) + 1(-4 - 18) \)
\( = 1(-11) - 1(0) + 1(-22) \)
\( = -11 - 22 = -33 \)
Since \( \Delta = -33 \neq 0 \), we can use Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ccc} 6 & 1 & 1 \\ 5 & 3 & -1 \\ -7 & -2 & -3 \end{array}\right| \)
\( = 6((3 \times -3) - (-1 \times -2)) - 1((5 \times -3) - (-1 \times -7)) + 1((5 \times -2) - (3 \times -7)) \)
\( = 6(-9 - 2) - 1(-15 - 7) + 1(-10 + 21) \)
\( = 6(-11) - 1(-22) + 1(11) \)
\( = -66 + 22 + 11 = -33 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ccc} 1 & 6 & 1 \\ 2 & 5 & -1 \\ 6 & -7 & -3 \end{array}\right| \)
\( = 1((5 \times -3) - (-1 \times -7)) - 6((2 \times -3) - (-1 \times 6)) + 1((2 \times -7) - (5 \times 6)) \)
\( = 1(-15 - 7) - 6(-6 + 6) + 1(-14 - 30) \)
\( = 1(-22) - 6(0) + 1(-44) \)
\( = -22 - 44 = -66 \)

Next, calculate \( \Delta_z \) by replacing the z-coefficients with the constant terms:
\( \Delta_z = \left|\begin{array}{ccc} 1 & 1 & 6 \\ 2 & 3 & 5 \\ 6 & -2 & -7 \end{array}\right| \)
\( = 1((3 \times -7) - (5 \times -2)) - 1((2 \times -7) - (5 \times 6)) + 6((2 \times -2) - (3 \times 6)) \)
\( = 1(-21 + 10) - 1(-14 - 30) + 6(-4 - 18) \)
\( = 1(-11) - 1(-44) + 6(-22) \)
\( = -11 + 44 - 132 = -99 \)

Now, apply Cramer's Rule to find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{-33}{-33} = 1 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-66}{-33} = 2 \)
\( \implies \) \( z = \frac{\Delta_z}{\Delta} = \frac{-99}{-33} = 3 \)
So, the solution is \( (x, y, z) = (1, 2, 3) \). Systems of three linear equations are often used to model real-world problems with three unknown quantities.
In simple words: We find four special numbers (determinants) from the given equations. We divide the determinants for x, y, and z by the main determinant to find the values of x, y, and z.

🎯 Exam Tip: Remember to simplify fractions to their lowest terms if the determinant calculations result in non-integer values for x, y, or z.

Answer:
(v) Given equations are:
\( x + 4y + 3z = 2 \)
\( 2x - 6y + 6z = -3 \)
\( 5x - 2y + 3z = -5 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ccc} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{array}\right| \)
\( = 1((-6 \times 3) - (6 \times -2)) - 4((2 \times 3) - (6 \times 5)) + 3((2 \times -2) - (-6 \times 5)) \)
\( = 1(-18 + 12) - 4(6 - 30) + 3(-4 + 30) \)
\( = 1(-6) - 4(-24) + 3(26) \)
\( = -6 + 96 + 78 = 168 \)
Since \( \Delta = 168 \neq 0 \), we can use Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ccc} 2 & 4 & 3 \\ -3 & -6 & 6 \\ -5 & -2 & 3 \end{array}\right| \)
\( = 2((-6 \times 3) - (6 \times -2)) - 4((-3 \times 3) - (6 \times -5)) + 3((-3 \times -2) - (-6 \times -5)) \)
\( = 2(-18 + 12) - 4(-9 + 30) + 3(6 - 30) \)
\( = 2(-6) - 4(21) + 3(-24) \)
\( = -12 - 84 - 72 = -168 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & -3 & 6 \\ 5 & -5 & 3 \end{array}\right| \)
\( = 1((-3 \times 3) - (6 \times -5)) - 2((2 \times 3) - (6 \times 5)) + 3((2 \times -5) - (-3 \times 5)) \)
\( = 1(-9 + 30) - 2(6 - 30) + 3(-10 + 15) \)
\( = 1(21) - 2(-24) + 3(5) \)
\( = 21 + 48 + 15 = 84 \)

Next, calculate \( \Delta_z \) by replacing the z-coefficients with the constant terms:
\( \Delta_z = \left|\begin{array}{ccc} 1 & 4 & 2 \\ 2 & -6 & -3 \\ 5 & -2 & -5 \end{array}\right| \)
\( = 1((-6 \times -5) - (-3 \times -2)) - 4((2 \times -5) - (-3 \times 5)) + 2((2 \times -2) - (-6 \times 5)) \)
\( = 1(30 - 6) - 4(-10 + 15) + 2(-4 + 30) \)
\( = 1(24) - 4(5) + 2(26) \)
\( = 24 - 20 + 52 = 56 \)

Now, apply Cramer's Rule to find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{-168}{168} = -1 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{84}{168} = \frac{1}{2} \)
\( \implies \) \( z = \frac{\Delta_z}{\Delta} = \frac{56}{168} = \frac{1}{3} \)
So, the solution is \( (x, y, z) = (-1, \frac{1}{2}, \frac{1}{3}) \). Fractions in the final answer are common, especially when working with coefficients that don't divide neatly. Always reduce fractions to their simplest form.
In simple words: We find four determinants from the three given equations. Then we divide the determinant for each letter (x, y, z) by the main determinant to find the values for x, y, and z.

🎯 Exam Tip: When the solution involves fractions, ensure all calculations are precise and reduce the fractions to their simplest form for full marks.

Question 2. A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is Rs 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is Rs 56. What is the cost per unit of labour and capital? (Use determinant method).
Answer: Let 'x' be the cost per unit of labour and 'y' be the cost per unit of capital.
From the problem, we can form two linear equations:
1. For 3 units of labour and 2 units of capital, total cost is Rs 62:
\( 3x + 2y = 62 \)
2. For 4 units of labour and 1 unit of capital, total cost is Rs 56:
\( 4x + y = 56 \)

Now, we use Cramer's Rule to solve this system.
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right| = (3 \times 1) - (2 \times 4) = 3 - 8 = -5 \)
Since \( \Delta = -5 \neq 0 \), we can apply Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ll} 62 & 2 \\ 56 & 1 \end{array}\right| = (62 \times 1) - (2 \times 56) = 62 - 112 = -50 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ll} 3 & 62 \\ 4 & 56 \end{array}\right| = (3 \times 56) - (62 \times 4) = 168 - 248 = -80 \)

Finally, apply Cramer's Rule to find x and y:
\( x = \frac{\Delta_x}{\Delta} = \frac{-50}{-5} = 10 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-80}{-5} = 16 \)
Therefore, the cost per unit of labour is Rs 10 and the cost per unit of capital is Rs 16. Real-world problems often simplify into systems of linear equations, making methods like Cramer's rule very practical.
In simple words: We first turn the problem into two math equations for labour and capital costs. Then, we use Cramer's rule by finding determinants to solve for the cost of each unit of labour and capital.

🎯 Exam Tip: Always define your variables clearly (e.g., 'x' for labour cost, 'y' for capital cost) before setting up equations in word problems.

Question 3. A total of Rs 8,600 was invested in two accounts. One account earned \( 4\frac{3}{4} \% \) annual interest and the other earned \( 6\frac{1}{2} \% \) annual interest. If the total interest for one year was Rs 431.25, how much was invested in each account? (Use determinant method)
Answer: Let 'x' be the amount invested in the first account (at \( 4\frac{3}{4} \% \) interest) and 'y' be the amount invested in the second account (at \( 6\frac{1}{2} \% \) interest).
From the problem, we can form two linear equations:
1. Total investment is Rs 8,600:
\( x + y = 8600 \) (Equation 1)
2. Total interest earned is Rs 431.25:
Convert mixed percentages to improper fractions and then to decimals or regular fractions:
\( 4\frac{3}{4} \% = \frac{19}{4} \% = \frac{19}{400} \)
\( 6\frac{1}{2} \% = \frac{13}{2} \% = \frac{13}{200} \)
So, \( \frac{19}{400}x + \frac{13}{200}y = 431.25 \)
Multiply the entire equation by 400 to clear the denominators:
\( 19x + 26y = 431.25 \times 400 \)
\( 19x + 26y = 172500 \) (Equation 2)

Now, we use Cramer's Rule to solve the system:
\( x + y = 8600 \)
\( 19x + 26y = 172500 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ll} 1 & 1 \\ 19 & 26 \end{array}\right| = (1 \times 26) - (1 \times 19) = 26 - 19 = 7 \)
Since \( \Delta = 7 \neq 0 \), we can apply Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ll} 8600 & 1 \\ 172500 & 26 \end{array}\right| = (8600 \times 26) - (1 \times 172500) = 223600 - 172500 = 51100 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ll} 1 & 8600 \\ 19 & 172500 \end{array}\right| = (1 \times 172500) - (8600 \times 19) = 172500 - 163400 = 9100 \)

Finally, apply Cramer's Rule to find x and y:
\( x = \frac{\Delta_x}{\Delta} = \frac{51100}{7} = 7300 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{9100}{7} = 1300 \)
Therefore, the amount invested at \( 4\frac{3}{4}\% \) is Rs 7,300 and the amount invested at \( 6\frac{1}{2}\% \) is Rs 1,300. Interest problems are a classic application of linear equations, helping people calculate earnings on investments.
In simple words: We set up two equations from the problem details, one for the total money and one for the total interest. Then, we use Cramer's rule to find out how much money was put into each account.

🎯 Exam Tip: When dealing with percentages in equations, convert them to their decimal or fractional form (by dividing by 100) before calculations to ensure accuracy.

Question 4. At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Bentia spent Rs 780 and Rs 560 during the month of May

Find the hourly charges for the two games (rides). (Use determinant method).
Answer: Let 'x' be the hourly charge for Horse Riding and 'y' be the hourly charge for Quad Bikes Riding.
From the table, we can form two linear equations:
1. For Keren:
\( 3x + 4y = 780 \)
2. For Benita:
\( 2x + 3y = 560 \)

Now, we use Cramer's Rule to solve this system.
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ll} 3 & 4 \\ 2 & 3 \end{array}\right| = (3 \times 3) - (4 \times 2) = 9 - 8 = 1 \)
Since \( \Delta = 1 \neq 0 \), we can apply Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ll} 780 & 4 \\ 560 & 3 \end{array}\right| = (780 \times 3) - (4 \times 560) = 2340 - 2240 = 100 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ll} 3 & 780 \\ 2 & 560 \end{array}\right| = (3 \times 560) - (780 \times 2) = 1680 - 1560 = 120 \)

Finally, apply Cramer's Rule to find x and y:
\( x = \frac{\Delta_x}{\Delta} = \frac{100}{1} = 100 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{120}{1} = 120 \)
Therefore, the hourly charge for Horse Riding is Rs 100 and the hourly charge for Quad Bikes Riding is Rs 120. This method allows us to find unknown prices or rates based on total costs and quantities used by different individuals.
In simple words: We create two equations from the table about how much Keren and Benita spent. Then, we use Cramer's rule to figure out the hourly cost for horse riding and quad bike riding.

🎯 Exam Tip: Always make sure the equations are correctly formed from the given data in the table before starting the determinant calculations.

Question 5. In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides the information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Find the weights assigned to the three varieties by using Cramer's Rule.
Answer: Let 'x', 'y', and 'z' be the weights assigned to varieties I, II, and III respectively.
From the table, we can form three linear equations:
1. For Commodity A:
\( x + 2y + 3z = 11 \)
2. For Commodity B:
\( 2x + 4y + 5z = 21 \)
3. For Commodity C:
\( 3x + 5y + 6z = 27 \)

Now, we use Cramer's Rule to solve this system.
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{array}\right| \)
\( = 1((4 \times 6) - (5 \times 5)) - 2((2 \times 6) - (5 \times 3)) + 3((2 \times 5) - (4 \times 3)) \)
\( = 1(24 - 25) - 2(12 - 15) + 3(10 - 12) \)
\( = 1(-1) - 2(-3) + 3(-2) \)
\( = -1 + 6 - 6 = -1 \)
Since \( \Delta = -1 \neq 0 \), we can apply Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ccc} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{array}\right| \)
\( = 11((4 \times 6) - (5 \times 5)) - 2((21 \times 6) - (5 \times 27)) + 3((21 \times 5) - (4 \times 27)) \)
\( = 11(24 - 25) - 2(126 - 135) + 3(105 - 108) \)
\( = 11(-1) - 2(-9) + 3(-3) \)
\( = -11 + 18 - 9 = -2 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ccc} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{array}\right| \)
\( = 1((21 \times 6) - (5 \times 27)) - 11((2 \times 6) - (5 \times 3)) + 3((2 \times 27) - (21 \times 3)) \)
\( = 1(126 - 135) - 11(12 - 15) + 3(54 - 63) \)
\( = 1(-9) - 11(-3) + 3(-9) \)
\( = -9 + 33 - 27 = -3 \)

Next, calculate \( \Delta_z \) by replacing the z-coefficients with the constant terms:
\( \Delta_z = \left|\begin{array}{ccc} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{array}\right| \)
\( = 1((4 \times 27) - (21 \times 5)) - 2((2 \times 27) - (21 \times 3)) + 11((2 \times 5) - (4 \times 3)) \)
\( = 1(108 - 105) - 2(54 - 63) + 11(10 - 12) \)
\( = 1(3) - 2(-9) + 11(-2) \)
\( = 3 + 18 - 22 = -1 \)

Finally, apply Cramer's Rule to find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{-2}{-1} = 2 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-3}{-1} = 3 \)
\( \implies \) \( z = \frac{\Delta_z}{\Delta} = \frac{-1}{-1} = 1 \)
Therefore, the weights assigned to variety I, II, and III are 2, 3, and 1 respectively. In economics, matrices and determinants are essential tools for analyzing market data and making predictions.
In simple words: We take the numbers from the table to make three math equations. Then, we use Cramer's rule to find the 'weights' for each variety (I, II, and III) by calculating and dividing determinants.

🎯 Exam Tip: Carefully set up the equations from the table, ensuring each variable (x, y, z) correctly corresponds to its variety (I, II, III) and that the constant terms are the total weights.

Question 6. A total of Rs 8,500 was invested in three interest-earning accounts. The interest rates were 2%, 3%, and 6%, if the total simple interest for one year was Rs 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer's rule)
Answer: Let 'x' be the amount invested at 2%, 'y' be the amount invested at 3%, and 'z' be the amount invested at 6%.
From the problem, we can form three linear equations:
1. Total investment is Rs 8,500:
\( x + y + z = 8500 \) (Equation 1)
2. Total simple interest for one year is Rs 380:
\( 0.02x + 0.03y + 0.06z = 380 \)
Multiply by 100 to clear decimals:
\( 2x + 3y + 6z = 38000 \) (Equation 2)
3. Amount invested at 6% (z) was equal to the sum of the amounts in the other two accounts (x + y):
\( z = x + y \)
Rearrange this equation:
\( x + y - z = 0 \) (Equation 3)

Now, we use Cramer's Rule to solve the system:
\( x + y + z = 8500 \)
\( 2x + 3y + 6z = 38000 \)
\( x + y - z = 0 \)
First, calculate the determinant of the coefficient matrix \( \Delta \):
\( \Delta = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{array}\right| \)
\( = 1((3 \times -1) - (6 \times 1)) - 1((2 \times -1) - (6 \times 1)) + 1((2 \times 1) - (3 \times 1)) \)
\( = 1(-3 - 6) - 1(-2 - 6) + 1(2 - 3) \)
\( = 1(-9) - 1(-8) + 1(-1) \)
\( = -9 + 8 - 1 = -2 \)
Since \( \Delta = -2 \neq 0 \), we can apply Cramer's Rule.

Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constant terms:
\( \Delta_x = \left|\begin{array}{ccc} 8500 & 1 & 1 \\ 38000 & 3 & 6 \\ 0 & 1 & -1 \end{array}\right| \)
\( = 8500((3 \times -1) - (6 \times 1)) - 1((38000 \times -1) - (6 \times 0)) + 1((38000 \times 1) - (3 \times 0)) \)
\( = 8500(-3 - 6) - 1(-38000 - 0) + 1(38000 - 0) \)
\( = 8500(-9) + 38000 + 38000 \)
\( = -76500 + 76000 = -500 \)

Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constant terms:
\( \Delta_y = \left|\begin{array}{ccc} 1 & 8500 & 1 \\ 2 & 38000 & 6 \\ 1 & 0 & -1 \end{array}\right| \)
\( = 1((38000 \times -1) - (6 \times 0)) - 8500((2 \times -1) - (6 \times 1)) + 1((2 \times 0) - (38000 \times 1)) \)
\( = 1(-38000 - 0) - 8500(-2 - 6) + 1(0 - 38000) \)
\( = -38000 - 8500(-8) - 38000 \)
\( = -38000 + 68000 - 38000 = -8000 \)

Next, calculate \( \Delta_z \) by replacing the z-coefficients with the constant terms:
\( \Delta_z = \left|\begin{array}{ccc} 1 & 1 & 8500 \\ 2 & 3 & 38000 \\ 1 & 1 & 0 \end{array}\right| \)
\( = 1((3 \times 0) - (38000 \times 1)) - 1((2 \times 0) - (38000 \times 1)) + 8500((2 \times 1) - (3 \times 1)) \)
\( = 1(0 - 38000) - 1(0 - 38000) + 8500(2 - 3) \)
\( = -38000 + 38000 + 8500(-1) \)
\( = -8500 \)

Finally, apply Cramer's Rule to find x, y, and z:
\( x = \frac{\Delta_x}{\Delta} = \frac{-500}{-2} = 250 \)
\( \implies \) \( y = \frac{\Delta_y}{\Delta} = \frac{-8000}{-2} = 4000 \)
\( \implies \) \( z = \frac{\Delta_z}{\Delta} = \frac{-8500}{-2} = 4250 \)
Therefore, the amount invested at 2% is Rs 250, at 3% is Rs 4,000, and at 6% is Rs 4,250. Financial planning often involves solving systems of equations to manage investments and returns.
In simple words: We make three equations from the problem details, covering total money, total interest, and how the highest interest investment relates to the others. Then, we use Cramer's rule to find out how much money was put into each of the three accounts.

🎯 Exam Tip: Be extra careful when converting percentage rates into decimals or fractions for the interest equation to avoid errors in your calculations.