Samacheer Kalvi Class 11 Business Maths Solutions Chapter 1 Matrices and Determinants Exercise 1.3

Question 1. Solve by matrix inversion method: 2x + 3y – 5 = 0; x – 2y + 1 = 0.
Answer: The given equations are:
\( 2x + 3y = 5 \)
\( x - 2y = -1 \)
This system can be written in matrix form as \( AX = B \):
\[ \begin{pmatrix} 2 & 3 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 2 & 3 \\ 1 & -2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \end{pmatrix} \), and \( B = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \).
First, we calculate the determinant of A:
\( |A| = \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (3)(1) = -4 - 3 = -7 \)
Since \( |A| = -7 \neq 0 \), the inverse of A, \( A^{-1} \), exists.
Next, find the adjoint of A:
\( \text{adj A} = \begin{pmatrix} -2 & -3 \\ -1 & 2 \end{pmatrix} \)
Now, calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-7} \begin{pmatrix} -2 & -3 \\ -1 & 2 \end{pmatrix} \)
To solve for X, we use the formula \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-7} \begin{pmatrix} -2 & -3 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ -1 \end{pmatrix} \]
Multiply the matrices:
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-7} \begin{pmatrix} (-2)(5) + (-3)(-1) \\ (-1)(5) + (2)(-1) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-7} \begin{pmatrix} -10 + 3 \\ -5 - 2 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{-7} \begin{pmatrix} -7 \\ -7 \end{pmatrix} \]
Divide by -7:
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]
So, the solution is \( x = 1 \) and \( y = 1 \). This method helps find unknown values in a system of linear equations by using matrix operations.
In simple words: We changed the two equations into matrices. Then we found the inverse of the first matrix and multiplied it by the constant matrix to find the values of x and y, which are both 1.

🎯 Exam Tip: Always check that the determinant of matrix A is not zero before attempting to find the inverse, as a zero determinant means the inverse does not exist and the system cannot be solved by this method.

 

Question 2. Solve by matrix inversion method:
(i) 3x − y + 2z = 13; 2x + y − z = 3; x + 3y – 5z = -8
(ii) x − y + 2z = 3; 2x + z = 1; 3x + 2y + z = 4
(iii) 2x − z = 0; 5x + y = 4; y + 3z = 5
Answer:
(i) The given system of equations is:
\( 3x - y + 2z = 13 \)
\( 2x + y - z = 3 \)
\( x + 3y - 5z = -8 \)
We write this in matrix form \( AX = B \):
\[ \begin{pmatrix} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 13 \\ 3 \\ -8 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 13 \\ 3 \\ -8 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{vmatrix} \)
\( = 3((1)(-5) - (-1)(3)) - (-1)((2)(-5) - (-1)(1)) + 2((2)(3) - (1)(1)) \)
\( = 3(-5 + 3) + 1(-10 + 1) + 2(6 - 1) \)
\( = 3(-2) + 1(-9) + 2(5) \)
\( = -6 - 9 + 10 \)
\( = -5 \)
Since \( |A| = -5 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 1 & -1 \\ 3 & -5 \end{vmatrix} = -5 - (-3) = -2 \]
\[ A_{12} = - \begin{vmatrix} 2 & -1 \\ 1 & -5 \end{vmatrix} = -(-10 - (-1)) = -(-9) = 9 \]
\[ A_{13} = \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 6 - 1 = 5 \]
\[ A_{21} = - \begin{vmatrix} -1 & 2 \\ 3 & -5 \end{vmatrix} = -(5 - 6) = -(-1) = 1 \]
\[ A_{22} = \begin{vmatrix} 3 & 2 \\ 1 & -5 \end{vmatrix} = -15 - 2 = -17 \]
\[ A_{23} = - \begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix} = -(9 - (-1)) = -(10) = -10 \]
\[ A_{31} = \begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} = 1 - 2 = -1 \]
\[ A_{32} = - \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = -(-3 - 4) = -(-7) = 7 \]
\[ A_{33} = \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 - (-2) = 5 \]
So, \( [A_{ij}] = \begin{pmatrix} -2 & 9 & 5 \\ 1 & -17 & -10 \\ -1 & 7 & 5 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-5} \begin{pmatrix} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-5} \begin{pmatrix} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{pmatrix} \begin{pmatrix} 13 \\ 3 \\ -8 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-5} \begin{pmatrix} (-2)(13) + (1)(3) + (-1)(-8) \\ (9)(13) + (-17)(3) + (7)(-8) \\ (5)(13) + (-10)(3) + (5)(-8) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-5} \begin{pmatrix} -26 + 3 + 8 \\ 117 - 51 - 56 \\ 65 - 30 - 40 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-5} \begin{pmatrix} -15 \\ 10 \\ -5 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \]
Therefore, \( x = 3, y = -2, z = 1 \).

(ii) The given system of equations is:
\( x - y + 2z = 3 \)
\( 2x + z = 1 \)
\( 3x + 2y + z = 4 \)
We write this in matrix form \( AX = B \):
\[ \begin{pmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{vmatrix} \)
\( = 1((0)(1) - (1)(2)) - (-1)((2)(1) - (1)(3)) + 2((2)(2) - (0)(3)) \)
\( = 1(0 - 2) + 1(2 - 3) + 2(4 - 0) \)
\( = 1(-2) + 1(-1) + 2(4) \)
\( = -2 - 1 + 8 \)
\( = 5 \)
Since \( |A| = 5 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = 0 - 2 = -2 \]
\[ A_{12} = - \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = -(2 - 3) = -(-1) = 1 \]
\[ A_{13} = \begin{vmatrix} 2 & 0 \\ 3 & 2 \end{vmatrix} = 4 - 0 = 4 \]
\[ A_{21} = - \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} = -(-1 - 4) = -(-5) = 5 \]
\[ A_{22} = \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1 - 6 = -5 \]
\[ A_{23} = - \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = -(2 - (-3)) = -(5) = -5 \]
\[ A_{31} = \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -1 - 0 = -1 \]
\[ A_{32} = - \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -(1 - 4) = -(-3) = 3 \]
\[ A_{33} = \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} = 0 - (-2) = 2 \]
So, \( [A_{ij}] = \begin{pmatrix} -2 & 1 & 4 \\ 5 & -5 & -5 \\ -1 & 3 & 2 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{5} \begin{pmatrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{5} \begin{pmatrix} (-2)(3) + (5)(1) + (-1)(4) \\ (1)(3) + (-5)(1) + (3)(4) \\ (4)(3) + (-5)(1) + (2)(4) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -6 + 5 - 4 \\ 3 - 5 + 12 \\ 12 - 5 + 8 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -5 \\ 10 \\ 15 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} \]
Therefore, \( x = -1, y = 2, z = 3 \).

(iii) The given system of equations is:
\( 2x - z = 0 \)
\( 5x + y = 4 \)
\( y + 3z = 5 \)
We write this in matrix form \( AX = B \):
\[ \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 4 \\ 5 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 0 \\ 4 \\ 5 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix} \)
\( = 2((1)(3) - (0)(1)) - 0((5)(3) - (0)(0)) - 1((5)(1) - (1)(0)) \)
\( = 2(3 - 0) - 0(15 - 0) - 1(5 - 0) \)
\( = 2(3) - 0(15) - 1(5) \)
\( = 6 - 0 - 5 \)
\( = 1 \)
Since \( |A| = 1 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} = 3 - 0 = 3 \]
\[ A_{12} = - \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} = -(15 - 0) = -15 \]
\[ A_{13} = \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} = 5 - 0 = 5 \]
\[ A_{21} = - \begin{vmatrix} 0 & -1 \\ 1 & 3 \end{vmatrix} = -(0 - (-1)) = -(1) = -1 \]
\[ A_{22} = \begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} = 6 - 0 = 6 \]
\[ A_{23} = - \begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -(2 - 0) = -2 \]
\[ A_{31} = \begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} = 0 - (-1) = 1 \]
\[ A_{32} = - \begin{vmatrix} 2 & -1 \\ 5 & 0 \end{vmatrix} = -(0 - (-5)) = -(5) = -5 \]
\[ A_{33} = \begin{vmatrix} 2 & 0 \\ 5 & 1 \end{vmatrix} = 2 - 0 = 2 \]
So, \( [A_{ij}] = \begin{pmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{1} \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \begin{pmatrix} 0 \\ 4 \\ 5 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} (3)(0) + (-1)(4) + (1)(5) \\ (-15)(0) + (6)(4) + (-5)(5) \\ (5)(0) + (-2)(4) + (2)(5) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 - 4 + 5 \\ 0 + 24 - 25 \\ 0 - 8 + 10 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]
Therefore, \( x = 1, y = -1, z = 2 \).
In simple words: For each set of equations, we turn them into a matrix problem. We then find the determinant, calculate the matrix of cofactors, find its transpose (the adjoint), and then the inverse of the main matrix. Finally, we multiply this inverse by the constant matrix to get the values for x, y, and z. This systematic approach ensures accurate solutions for all variables.

🎯 Exam Tip: When dealing with 3x3 matrices, pay close attention to the signs when calculating cofactors. A single sign error can change the entire result. Double-check your determinant and adjoint calculations carefully.

 

Question 3. A salesperson Ravi has the following record of sales for the month of January, February, and March 2009 for three products A, B, and C. He has been paid a commission at a fixed rate per unit but at varying rates for products A, B and C. Find the rate of commission payable on A, B and C per unit sold using matrix inversion method.
Answer: Let \( x, y \), and \( z \) be the rate of commission (in Rs) for the three products A, B, and C, respectively.
The given sales and commission data can be represented in a table:

Based on the table, we can form a system of linear equations:
\( 9x + 10y + 2z = 800 \)
\( 15x + 5y + 4z = 900 \)
\( 6x + 10y + 3z = 850 \)
This system can be written in matrix form \( AX = B \):
\[ \begin{pmatrix} 9 & 10 & 2 \\ 15 & 5 & 4 \\ 6 & 10 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 800 \\ 900 \\ 850 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 9 & 10 & 2 \\ 15 & 5 & 4 \\ 6 & 10 & 3 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 800 \\ 900 \\ 850 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 9 & 10 & 2 \\ 15 & 5 & 4 \\ 6 & 10 & 3 \end{vmatrix} \)
\( = 9((5)(3) - (4)(10)) - 10((15)(3) - (4)(6)) + 2((15)(10) - (5)(6)) \)
\( = 9(15 - 40) - 10(45 - 24) + 2(150 - 30) \)
\( = 9(-25) - 10(21) + 2(120) \)
\( = -225 - 210 + 240 \)
\( = -195 \)
Since \( |A| = -195 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 5 & 4 \\ 10 & 3 \end{vmatrix} = 15 - 40 = -25 \]
\[ A_{12} = - \begin{vmatrix} 15 & 4 \\ 6 & 3 \end{vmatrix} = -(45 - 24) = -21 \]
\[ A_{13} = \begin{vmatrix} 15 & 5 \\ 6 & 10 \end{vmatrix} = 150 - 30 = 120 \]
\[ A_{21} = - \begin{vmatrix} 10 & 2 \\ 10 & 3 \end{vmatrix} = -(30 - 20) = -10 \]
\[ A_{22} = \begin{vmatrix} 9 & 2 \\ 6 & 3 \end{vmatrix} = 27 - 12 = 15 \]
\[ A_{23} = - \begin{vmatrix} 9 & 10 \\ 6 & 10 \end{vmatrix} = -(90 - 60) = -30 \]
\[ A_{31} = \begin{vmatrix} 10 & 2 \\ 5 & 4 \end{vmatrix} = 40 - 10 = 30 \]
\[ A_{32} = - \begin{vmatrix} 9 & 2 \\ 15 & 4 \end{vmatrix} = -(36 - 30) = -6 \]
\[ A_{33} = \begin{vmatrix} 9 & 10 \\ 15 & 5 \end{vmatrix} = 45 - 150 = -105 \]
So, \( [A_{ij}] = \begin{pmatrix} -25 & -21 & 120 \\ -10 & 15 & -30 \\ 30 & -6 & -105 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} -25 & -10 & 30 \\ -21 & 15 & -6 \\ 120 & -30 & -105 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-195} \begin{pmatrix} -25 & -10 & 30 \\ -21 & 15 & -6 \\ 120 & -30 & -105 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-195} \begin{pmatrix} -25 & -10 & 30 \\ -21 & 15 & -6 \\ 120 & -30 & -105 \end{pmatrix} \begin{pmatrix} 800 \\ 900 \\ 850 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-195} \begin{pmatrix} (-25)(800) + (-10)(900) + (30)(850) \\ (-21)(800) + (15)(900) + (-6)(850) \\ (120)(800) + (-30)(900) + (-105)(850) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-195} \begin{pmatrix} -20000 - 9000 + 25500 \\ -16800 + 13500 - 5100 \\ 96000 - 27000 - 89250 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-195} \begin{pmatrix} -3500 \\ -8400 \\ -20250 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \approx \begin{pmatrix} 17.9487 \\ 43.0769 \\ 103.8461 \end{pmatrix} \]
Rounding to two decimal places:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 17.95 \\ 43.08 \\ 103.85 \end{pmatrix} \]
Therefore, the rate of commission for products A, B, and C are Rs 17.95, Rs 43.08, and Rs 103.85 respectively. This type of matrix calculation is very useful for solving systems with many variables, like in business or economics.
In simple words: We used the sales data to make three equations. Then we solved these equations using matrices to find out how much commission Ravi gets for each unit of product A, B, and C sold. The commission rates are approximately Rs 17.95 for A, Rs 43.08 for B, and Rs 103.85 for C.

🎯 Exam Tip: When dealing with real-world problems involving money, remember to convert the final answer to the appropriate currency unit (Rs in this case) and round to a sensible number of decimal places, typically two for currency.

 

Question 4. The prices of three commodities A, B, and C are x, y, and z per unit respectively. P purchases 4 units of C and sells 3 units of A and 5 units of B. Q purchases 3 units of D and sells 2 units of A and 1 unit of C. R purchases 1 unit of A and sells 4 units of B and 6 units of C. In the process P, Q and R earn ₹ 6,000, ₹ 5,000 and ₹ 13,000 respectively. By using the matrix inversion method, find the prices per unit of A, B, and C.
Answer: Let \( x, y, \) and \( z \) be the prices per unit of commodities A, B, and C, respectively.
The problem states that selling units gives positive earnings, while buying units gives negative earnings.
The information can be summarized in the following table:

(Note: The question has a small typo, "Q purchases 3 units of D" - it is assumed to be A, B, or C, and based on the matrix, it's for B. Also, "sells 2 units of A" and "1 unit of C" leads to 2x + (-3)y + 1z. The table correctly reflects values for A, B, C.)
From the given earnings, we form the system of equations:
\( 3x + 5y - 4z = 6000 \)
\( 2x - 3y + z = 5000 \)
\( -1x + 4y + 6z = 13000 \)
This system can be written in matrix form \( AX = B \):
\[ \begin{pmatrix} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6000 \\ 5000 \\ 13000 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 6000 \\ 5000 \\ 13000 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{vmatrix} \)
\( = 3((-3)(6) - (1)(4)) - 5((2)(6) - (1)(-1)) - 4((2)(4) - (-3)(-1)) \)
\( = 3(-18 - 4) - 5(12 + 1) - 4(8 - 3) \)
\( = 3(-22) - 5(13) - 4(5) \)
\( = -66 - 65 - 20 \)
\( = -151 \)
Since \( |A| = -151 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} -3 & 1 \\ 4 & 6 \end{vmatrix} = -18 - 4 = -22 \]
\[ A_{12} = - \begin{vmatrix} 2 & 1 \\ -1 & 6 \end{vmatrix} = -(12 - (-1)) = -(13) = -13 \]
\[ A_{13} = \begin{vmatrix} 2 & -3 \\ -1 & 4 \end{vmatrix} = 8 - 3 = 5 \]
\[ A_{21} = - \begin{vmatrix} 5 & -4 \\ 4 & 6 \end{vmatrix} = -(30 - (-16)) = -(46) = -46 \]
\[ A_{22} = \begin{vmatrix} 3 & -4 \\ -1 & 6 \end{vmatrix} = 18 - 4 = 14 \]
\[ A_{23} = - \begin{vmatrix} 3 & 5 \\ -1 & 4 \end{vmatrix} = -(12 - (-5)) = -(17) = -17 \]
\[ A_{31} = \begin{vmatrix} 5 & -4 \\ -3 & 1 \end{vmatrix} = 5 - 12 = -7 \]
\[ A_{32} = - \begin{vmatrix} 3 & -4 \\ 2 & 1 \end{vmatrix} = -(3 - (-8)) = -(11) = -11 \]
\[ A_{33} = \begin{vmatrix} 3 & 5 \\ 2 & -3 \end{vmatrix} = -9 - 10 = -19 \]
So, \( [A_{ij}] = \begin{pmatrix} -22 & -13 & 5 \\ -46 & 14 & -17 \\ -7 & -11 & -19 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-151} \begin{pmatrix} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-151} \begin{pmatrix} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{pmatrix} \begin{pmatrix} 6000 \\ 5000 \\ 13000 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-151} \begin{pmatrix} (-22)(6000) + (-46)(5000) + (-7)(13000) \\ (-13)(6000) + (14)(5000) + (-11)(13000) \\ (5)(6000) + (-17)(5000) + (-19)(13000) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-151} \begin{pmatrix} -132000 - 230000 - 91000 \\ -78000 + 70000 - 143000 \\ 30000 - 85000 - 247000 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-151} \begin{pmatrix} -453000 \\ -151000 \\ -302000 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3000 \\ 1000 \\ 2000 \end{pmatrix} \]
Therefore, the prices per unit of A, B, and C are Rs 3000, Rs 1000, and Rs 2000 respectively. Matrices are widely used in business to solve complex problems involving multiple variables and transactions.
In simple words: We turned the buying and selling activities and earnings into equations, then put them into matrices. By finding the inverse of the main matrix and multiplying it, we figured out the cost of each item. Product A costs Rs 3000, B costs Rs 1000, and C costs Rs 2000.

🎯 Exam Tip: When setting up equations for transactions, ensure that sales are recorded as positive values and purchases as negative values to correctly represent the earnings or expenditures.

 

Question 5. The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it, we get 46. By using the matrix inversion method find the numbers.
Answer: Let the three numbers be \( x, y \), and \( z \).
Based on the given information, we can form the following system of linear equations:
1. The sum of three numbers is 20:
\( x + y + z = 20 \)
2. If we multiply the first by 2, add the second, and subtract the third, we get 23:
\( 2x + y - z = 23 \)
3. If we multiply the first by 3, add the second, and add the third, we get 46:
\( 3x + y + z = 46 \)
This system can be written in matrix form \( AX = B \):
\[ \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 20 \\ 23 \\ 46 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 1 & 1 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 20 \\ 23 \\ 46 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 1 & 1 \end{vmatrix} \)
\( = 1((1)(1) - (-1)(1)) - 1((2)(1) - (-1)(3)) + 1((2)(1) - (1)(3)) \)
\( = 1(1 + 1) - 1(2 + 3) + 1(2 - 3) \)
\( = 1(2) - 1(5) + 1(-1) \)
\( = 2 - 5 - 1 \)
\( = -4 \)
Since \( |A| = -4 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = 1 - (-1) = 2 \]
\[ A_{12} = - \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = -(2 - (-3)) = -(5) = -5 \]
\[ A_{13} = \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 - 3 = -1 \]
\[ A_{21} = - \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = -(1 - 1) = 0 \]
\[ A_{22} = \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = 1 - 3 = -2 \]
\[ A_{23} = - \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = -(1 - 3) = -(-2) = 2 \]
\[ A_{31} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2 \]
\[ A_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -(-1 - 2) = -(-3) = 3 \]
\[ A_{33} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = 1 - 2 = -1 \]
So, \( [A_{ij}] = \begin{pmatrix} 2 & -5 & -1 \\ 0 & -2 & 2 \\ -2 & 3 & -1 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} 2 & 0 & -2 \\ -5 & -2 & 3 \\ -1 & 2 & -1 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-4} \begin{pmatrix} 2 & 0 & -2 \\ -5 & -2 & 3 \\ -1 & 2 & -1 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-4} \begin{pmatrix} 2 & 0 & -2 \\ -5 & -2 & 3 \\ -1 & 2 & -1 \end{pmatrix} \begin{pmatrix} 20 \\ 23 \\ 46 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-4} \begin{pmatrix} (2)(20) + (0)(23) + (-2)(46) \\ (-5)(20) + (-2)(23) + (3)(46) \\ (-1)(20) + (2)(23) + (-1)(46) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-4} \begin{pmatrix} 40 + 0 - 92 \\ -100 - 46 + 138 \\ -20 + 46 - 46 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{-4} \begin{pmatrix} -52 \\ -8 \\ -20 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 13 \\ 2 \\ 5 \end{pmatrix} \]
The numbers are 13, 2, and 5. This method helps quickly find unknown quantities when several conditions are given in the form of linear equations.
In simple words: We wrote down the three clues about the numbers as equations. Then, we used matrices to solve them. We found that the three numbers are 13, 2, and 5.

🎯 Exam Tip: Always set up your system of equations carefully from the word problem, ensuring each variable is correctly placed and signs are accurate, as this is the foundation for correct matrix calculation.

 

Question 6. Weekly expenditure in an office for three weeks is given as follows. Assuming that the salary in all three weeks of different categories of staff did not vary, calculate the salary for each type of staff, using the matrix inversion method.
Answer: Let \( Rs \ x, Rs \ y, \) and \( Rs \ z \) be the salaries for each type of staff A, B, and C respectively.
The weekly expenditure data is given in the table:

From the table, we can form the system of linear equations:
\( 4x + 2y + 3z = 4900 \)
\( 3x + 3y + 2z = 4500 \)
\( 4x + 3y + 4z = 5800 \)
This system can be written in matrix form \( AX = B \):
\[ \begin{pmatrix} 4 & 2 & 3 \\ 3 & 3 & 2 \\ 4 & 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4900 \\ 4500 \\ 5800 \end{pmatrix} \]
Here, \( A = \begin{pmatrix} 4 & 2 & 3 \\ 3 & 3 & 2 \\ 4 & 3 & 4 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), and \( B = \begin{pmatrix} 4900 \\ 4500 \\ 5800 \end{pmatrix} \).
First, find the determinant of A:
\( |A| = \begin{vmatrix} 4 & 2 & 3 \\ 3 & 3 & 2 \\ 4 & 3 & 4 \end{vmatrix} \)
\( = 4((3)(4) - (2)(3)) - 2((3)(4) - (2)(4)) + 3((3)(3) - (3)(4)) \)
\( = 4(12 - 6) - 2(12 - 8) + 3(9 - 12) \)
\( = 4(6) - 2(4) + 3(-3) \)
\( = 24 - 8 - 9 \)
\( = 7 \)
Since \( |A| = 7 \neq 0 \), \( A^{-1} \) exists.
Next, find the matrix of cofactors \( [A_{ij}] \):
\[ A_{11} = \begin{vmatrix} 3 & 2 \\ 3 & 4 \end{vmatrix} = 12 - 6 = 6 \]
\[ A_{12} = - \begin{vmatrix} 3 & 2 \\ 4 & 4 \end{vmatrix} = -(12 - 8) = -4 \]
\[ A_{13} = \begin{vmatrix} 3 & 3 \\ 4 & 3 \end{vmatrix} = 9 - 12 = -3 \]
\[ A_{21} = - \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = -(8 - 9) = -(-1) = 1 \]
\[ A_{22} = \begin{vmatrix} 4 & 3 \\ 4 & 4 \end{vmatrix} = 16 - 12 = 4 \]
\[ A_{23} = - \begin{vmatrix} 4 & 2 \\ 4 & 3 \end{vmatrix} = -(12 - 8) = -4 \]
\[ A_{31} = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = 4 - 9 = -5 \]
\[ A_{32} = - \begin{vmatrix} 4 & 3 \\ 3 & 2 \end{vmatrix} = -(8 - 9) = -(-1) = 1 \]
\[ A_{33} = \begin{vmatrix} 4 & 2 \\ 3 & 3 \end{vmatrix} = 12 - 6 = 6 \]
So, \( [A_{ij}] = \begin{pmatrix} 6 & -4 & -3 \\ 1 & 4 & -4 \\ -5 & 1 & 6 \end{pmatrix} \)
Now, find the adjoint of A:
\( \text{adj A} = [A_{ij}]^T = \begin{pmatrix} 6 & 1 & -5 \\ -4 & 4 & 1 \\ -3 & -4 & 6 \end{pmatrix} \)
Calculate the inverse \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{7} \begin{pmatrix} 6 & 1 & -5 \\ -4 & 4 & 1 \\ -3 & -4 & 6 \end{pmatrix} \)
Finally, find X using \( X = A^{-1}B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 6 & 1 & -5 \\ -4 & 4 & 1 \\ -3 & -4 & 6 \end{pmatrix} \begin{pmatrix} 4900 \\ 4500 \\ 5800 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} (6)(4900) + (1)(4500) + (-5)(5800) \\ (-4)(4900) + (4)(4500) + (1)(5800) \\ (-3)(4900) + (-4)(4500) + (6)(5800) \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 29400 + 4500 - 29000 \\ -19600 + 18000 + 5800 \\ -14700 - 18000 + 34800 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 4900 \\ 4200 \\ 2100 \end{pmatrix} \]
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 700 \\ 600 \\ 300 \end{pmatrix} \]
Therefore, the salary for staff type A is Rs 700, staff type B is Rs 600, and staff type C is Rs 300. This shows how matrix inversion can be used to determine individual costs from combined expenditures.
In simple words: We used the office's weekly spending and the number of staff from each group to create equations. By solving these equations using matrices, we found that the salary for staff A is Rs 700, for staff B is Rs 600, and for staff C is Rs 300.

🎯 Exam Tip: Always clearly define your variables (like x, y, z for salaries) and ensure that the equations correctly represent the given information before performing matrix operations.

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