Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.3

Question 1. Solve the following system of linear equations by matrix inversion method.
(i) \( 2x + 5y = -2 \), \( x + 2y = -3 \)
(ii) \( 2x - y = 8 \), \( 3x + 2y = -2 \)
(iii) \( 2x + 3y - z = 9 \), \( x + y + z = 9 \), \( 3x - y - z = -1 \)
(iv) \( x + y + z - 2 = 0 \), \( 6x - 4y + 5z - 31 = 0 \), \( 5x + 2y + 2z = 13 \)
Answer:
(i) We need to solve the system of equations: \( 2x + 5y = -2 \) and \( x + 2y = -3 \).
First, we write these equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \end{pmatrix} \) and \( B = \begin{pmatrix} -2 \\ -3 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = (2 \times 2) - (5 \times 1) = 4 - 5 = -1 \). Since \( |A| = -1 \neq 0 \), the inverse matrix \( A^{-1} \) exists, which means there is a unique solution to the system.
Now, we find the adjoint of A:
\( \text{adj } A = \begin{pmatrix} 2 & -5 \\ -1 & 2 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1} \begin{pmatrix} 2 & -5 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} -2 & 5 \\ 1 & -2 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 & 5 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} -2 \\ -3 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (-2)(-2) + (5)(-3) \\ (1)(-2) + (-2)(-3) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 - 15 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} -11 \\ 4 \end{pmatrix} \] So, the solution is \( x = -11 \) and \( y = 4 \).
In simple words: First, turn the given math problems into a matrix. Then, find a special inverse matrix for the 'A' part. After that, multiply this inverse matrix by the 'B' part to get your answers for x and y.

🎯 Exam Tip: Always double-check your determinant calculation; if it's zero, the matrix inversion method can't be used directly.

(ii) We need to solve the system of equations: \( 2x - y = 8 \) and \( 3x + 2y = -2 \).
First, we write these equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ -2 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 2 & -1 \\ 3 & 2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \end{pmatrix} \) and \( B = \begin{pmatrix} 8 \\ -2 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = (2 \times 2) - (-1 \times 3) = 4 - (-3) = 4 + 3 = 7 \). Since \( |A| = 7 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A:
\( \text{adj } A = \begin{pmatrix} 2 & 1 \\ -3 & 2 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{7} \begin{pmatrix} 2 & 1 \\ -3 & 2 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 2 & 1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} 8 \\ -2 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{7} \begin{pmatrix} (2)(8) + (1)(-2) \\ (-3)(8) + (2)(-2) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 16 - 2 \\ -24 - 4 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 14 \\ -28 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} \] So, the solution is \( x = 2 \) and \( y = -4 \).
In simple words: Turn the two equations into a matrix form. Find the determinant and the adjoint of the first matrix. Use these to get the inverse matrix. Then, multiply the inverse by the constant matrix to find the values of x and y.

🎯 Exam Tip: When dealing with negative signs in matrix multiplication, be extra careful to avoid simple arithmetic errors.

(iii) We need to solve the system of equations: \( 2x + 3y - z = 9 \), \( x + y + z = 9 \), and \( 3x - y - z = -1 \).
First, we write these equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 9 \\ 9 \\ -1 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 9 \\ 9 \\ -1 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = 2((-1)(1) - (1)(-1)) - 3((1)(-1) - (1)(3)) + (-1)((1)(-1) - (1)(3)) \)
\( |A| = 2(-1+1) - 3(-1-3) - 1(-1-3) \)
\( |A| = 2(0) - 3(-4) - 1(-4) = 0 + 12 + 4 = 16 \). Since \( |A| = 16 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A. First, we find the cofactor matrix:
\( C_{11} = 0 \), \( C_{12} = 4 \), \( C_{13} = -4 \)
\( C_{21} = 2 \), \( C_{22} = 1 \), \( C_{23} = 11 \)
\( C_{31} = 4 \), \( C_{32} = -3 \), \( C_{33} = -1 \)
The cofactor matrix is \( \begin{pmatrix} 0 & 4 & -4 \\ 2 & 1 & 11 \\ 4 & -3 & -1 \end{pmatrix} \). The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj } A = \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{16} \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{16} \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{pmatrix} \begin{pmatrix} 9 \\ 9 \\ -1 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{16} \begin{pmatrix} (0)(9) + (2)(9) + (4)(-1) \\ (4)(9) + (1)(9) + (-3)(-1) \\ (-4)(9) + (11)(9) + (-1)(-1) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{16} \begin{pmatrix} 0 + 18 - 4 \\ 36 + 9 + 3 \\ -36 + 99 + 1 \end{pmatrix} = \frac{1}{16} \begin{pmatrix} 14 \\ 48 \\ 64 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 14/16 \\ 48/16 \\ 64/16 \end{pmatrix} = \begin{pmatrix} 7/8 \\ 3 \\ 4 \end{pmatrix} \] So, the solution is \( x = 7/8 \), \( y = 3 \) and \( z = 4 \).
In simple words: For three equations, first set them up as a 3x3 matrix equation. Calculate the determinant and the adjoint matrix. Use these to find the inverse matrix. Then, multiply the inverse matrix by the column matrix of constants to find the values of x, y, and z.

🎯 Exam Tip: Calculating the adjoint matrix for 3x3 systems requires careful attention to signs and individual 2x2 determinants for each cofactor.

(iv) We need to solve the system of equations: \( x + y + z - 2 = 0 \), \( 6x - 4y + 5z - 31 = 0 \), and \( 5x + 2y + 2z = 13 \).
First, we rewrite the equations as:
\( x + y + z = 2 \)
\( 6x - 4y + 5z = 31 \)
\( 5x + 2y + 2z = 13 \)
Now, we write these equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 31 \\ 13 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 2 \\ 31 \\ 13 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = 1((-4)(2) - (5)(2)) - 1((6)(2) - (5)(5)) + 1((6)(2) - (-4)(5)) \)
\( |A| = 1(-8-10) - 1(12-25) + 1(12+20) \)
\( |A| = 1(-18) - 1(-13) + 1(32) = -18 + 13 + 32 = 27 \). Since \( |A| = 27 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A. First, we find the cofactor matrix:
\( C_{11} = -18 \), \( C_{12} = 13 \), \( C_{13} = 32 \)
\( C_{21} = 0 \), \( C_{22} = -3 \), \( C_{23} = 3 \)
\( C_{31} = 9 \), \( C_{32} = 1 \), \( C_{33} = -10 \)
The cofactor matrix is \( \begin{pmatrix} -18 & 13 & 32 \\ 0 & -3 & 3 \\ 9 & 1 & -10 \end{pmatrix} \). The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj } A = \begin{pmatrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{27} \begin{pmatrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{27} \begin{pmatrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{pmatrix} \begin{pmatrix} 2 \\ 31 \\ 13 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{27} \begin{pmatrix} (-18)(2) + (0)(31) + (9)(13) \\ (13)(2) + (-3)(31) + (1)(13) \\ (32)(2) + (3)(31) + (-10)(13) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{27} \begin{pmatrix} -36 + 0 + 117 \\ 26 - 93 + 13 \\ 64 + 93 - 130 \end{pmatrix} = \frac{1}{27} \begin{pmatrix} 81 \\ -54 \\ 27 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \] So, the solution is \( x = 3 \), \( y = -2 \) and \( z = 1 \).
In simple words: First, rearrange the equations so that the constants are on the right side. Then, set up the matrix equation. Calculate the determinant and the adjoint matrix. Use these to find the inverse matrix, and then multiply it by the constant matrix to find the x, y, and z values.

🎯 Exam Tip: Always convert the given equations into the standard \( Ax + By + Cz = D \) form before forming the matrices to avoid errors.

Question 2. Find the products AB and BA and hence solve the system of equations \( x + y + 2z = 1 \), \( 3x + 2y + z = 7 \), \( 2x + y + 3z = 2 \)
Answer:
We are given the matrices \( A = \begin{pmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \).
First, we calculate the product AB: \[ AB = \begin{pmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \] \[ AB = \begin{pmatrix} (-5)(1)+1(3)+3(2) & (-5)(1)+1(2)+3(1) & (-5)(2)+1(1)+3(3) \\ 7(1)+1(3)+(-5)(2) & 7(1)+1(2)+(-5)(1) & 7(2)+1(1)+(-5)(3) \\ 1(1)+(-1)(3)+1(2) & 1(1)+(-1)(2)+1(1) & 1(2)+(-1)(1)+1(3) \end{pmatrix} \] \[ AB = \begin{pmatrix} -5+3+6 & -5+2+3 & -10+1+9 \\ 7+3-10 & 7+2-5 & 14+1-15 \\ 1-3+2 & 1-2+1 & 2-1+3 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = 4I_3 \] Next, we calculate the product BA: \[ BA = \begin{pmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{pmatrix} \] \[ BA = \begin{pmatrix} 1(-5)+1(7)+2(1) & 1(1)+1(1)+2(-1) & 1(3)+1(-5)+2(1) \\ 3(-5)+2(7)+1(1) & 3(1)+2(1)+1(-1) & 3(3)+2(-5)+1(1) \\ 2(-5)+1(7)+3(1) & 2(1)+1(1)+3(-1) & 2(3)+1(-5)+3(1) \end{pmatrix} \] \[ BA = \begin{pmatrix} -5+7+2 & 1+1-2 & 3-5+2 \\ -15+14+1 & 3+2-1 & 9-10+1 \\ -10+7+3 & 2+1-3 & 6-5+3 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = 4I_3 \] So, we have \( AB = BA = 4I_3 \).
Now, we use this result to solve the system of equations:
\( x + y + 2z = 1 \)
\( 3x + 2y + z = 7 \)
\( 2x + y + 3z = 2 \)
We can write this system in matrix form \( BX = C \), where: \[ B = \begin{pmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( C = \begin{pmatrix} 1 \\ 7 \\ 2 \end{pmatrix} \] From our product calculation, we know \( BA = 4I_3 \).
\( BA = 4I_3 \) We can multiply both sides by \( \frac{1}{4} \).
\( \frac{1}{4} BA = I_3 \) This means \( B^{-1} = \frac{1}{4} A \). Now, we solve \( BX = C \) for X:
\( X = B^{-1} C \)
\( X = \frac{1}{4} A C \) \[ X = \frac{1}{4} \begin{pmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 7 \\ 2 \end{pmatrix} \] \[ X = \frac{1}{4} \begin{pmatrix} (-5)(1) + (1)(7) + (3)(2) \\ (7)(1) + (1)(7) + (-5)(2) \\ (1)(1) + (-1)(7) + (1)(2) \end{pmatrix} \] \[ X = \frac{1}{4} \begin{pmatrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 8 \\ 4 \\ -4 \end{pmatrix} \] \[ X = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \] So, the solution is \( x = 2 \), \( y = 1 \), and \( z = -1 \).
In simple words: First, multiply the given matrices A and B in both orders (AB and BA) to see if they give a special identity matrix. If they do, use that relationship to find the inverse of B. Then, write the system of equations as a matrix problem and use the inverse of B to solve for x, y, and z.

🎯 Exam Tip: When \( AB = kI \), you can quickly find \( B^{-1} = \frac{1}{k} A \) (or \( A^{-1} = \frac{1}{k} B \)), which simplifies solving linear equations.

Question 3. A man is appointed in a job with a monthly salary of a certain amount and a fixed amount of annual increment. If his salary was Rs 19,800 per month at the end of the first month after 3 years of service and Rs 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use the matrix inversion method to solve the problem.)
Answer:
Let the man's starting monthly salary be Rs \( x \) and his annual increment be Rs \( y \).
After 3 years of service, his salary is \( x + 3y \). We are given that this is Rs 19,800.
So, \( x + 3y = 19800 \) . . . (1)
After 9 years of service, his salary is \( x + 9y \). We are given that this is Rs 23,400.
So, \( x + 9y = 23400 \) . . . (2)
We can write these two equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 1 & 3 \\ 1 & 9 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 19800 \\ 23400 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 1 & 3 \\ 1 & 9 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \end{pmatrix} \) and \( B = \begin{pmatrix} 19800 \\ 23400 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = (1 \times 9) - (3 \times 1) = 9 - 3 = 6 \). Since \( |A| = 6 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A:
\( \text{adj } A = \begin{pmatrix} 9 & -3 \\ -1 & 1 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{6} \begin{pmatrix} 9 & -3 \\ -1 & 1 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 9 & -3 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 19800 \\ 23400 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{6} \begin{pmatrix} (9)(19800) + (-3)(23400) \\ (-1)(19800) + (1)(23400) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 178200 - 70200 \\ -19800 + 23400 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 108000 \\ 3600 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 18000 \\ 600 \end{pmatrix} \] Therefore, the starting monthly salary is Rs 18,000, and the annual increment is Rs 600.
In simple words: We set up equations for the salary over time, then turned them into a matrix problem. By finding the inverse of the matrix and multiplying it, we could figure out the starting salary and how much it increased each year.

🎯 Exam Tip: Define variables clearly (like x for starting salary, y for increment) before setting up the equations to avoid confusion. Always ensure units (like Rs) are consistent.

Question 4. 4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by Using the matrix inversion method.
Answer:
Let \( x \) be the number of days one man takes to finish the work alone.
Let \( y \) be the number of days one woman takes to finish the work alone.
So, in one day:
Work done by 1 man = \( \frac{1}{x} \)
Work done by 1 woman = \( \frac{1}{y} \)
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \).
According to the problem:
"4 men and 4 women can finish a piece of work jointly in 3 days":
\( 4u + 4v = \frac{1}{3} \) . . . (1)
"2 men and 5 women can finish the same work jointly in 4 days":
\( 2u + 5v = \frac{1}{4} \) . . . (2)
We can write these two equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 4 & 4 \\ 2 & 5 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 1/3 \\ 1/4 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 4 & 4 \\ 2 & 5 \end{pmatrix} \), \( X = \begin{pmatrix} u \\ v \end{pmatrix} \) and \( B = \begin{pmatrix} 1/3 \\ 1/4 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = (4 \times 5) - (4 \times 2) = 20 - 8 = 12 \). Since \( |A| = 12 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A:
\( \text{adj } A = \begin{pmatrix} 5 & -4 \\ -2 & 4 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{12} \begin{pmatrix} 5 & -4 \\ -2 & 4 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} u \\ v \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 5 & -4 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 1/3 \\ 1/4 \end{pmatrix} \] \[ \begin{pmatrix} u \\ v \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 5(1/3) + (-4)(1/4) \\ (-2)(1/3) + (4)(1/4) \end{pmatrix} \] \[ \begin{pmatrix} u \\ v \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 5/3 - 1 \\ -2/3 + 1 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} 2/3 \\ 1/3 \end{pmatrix} \] \[ \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} (2/3) \times (1/12) \\ (1/3) \times (1/12) \end{pmatrix} = \begin{pmatrix} 2/36 \\ 1/36 \end{pmatrix} = \begin{pmatrix} 1/18 \\ 1/36 \end{pmatrix} \] So, \( u = \frac{1}{18} \) and \( v = \frac{1}{36} \).
Since \( u = \frac{1}{x} \), we have \( \frac{1}{x} = \frac{1}{18} \implies x = 18 \).
Since \( v = \frac{1}{y} \), we have \( \frac{1}{y} = \frac{1}{36} \implies y = 36 \).
Therefore, one man alone can complete the work in 18 days, and one woman alone can complete the work in 36 days.
In simple words: We first set up work-rate equations, where \(1/x\) is the work a man does in a day and \(1/y\) is for a woman. Then, we solved these equations using matrices to find x and y, which represent the total days each person takes alone.

🎯 Exam Tip: Remember to use reciprocals (\( 1/x \) and \( 1/y \)) for work-rate problems before setting up the matrix equations, and convert back to days at the end.

Question 5. The prices of three commodities A, B, and C are Rs x,y, and z per unit respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B. Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn Rs 15,000, Rs 1,000 and 14,000 respectively. Find the prices per unit of A, B, and C. (Use the matrix inversion method to solve the problem.)
Answer:
Let the prices per unit of commodities A, B, and C be Rs \( x \), Rs \( y \), and Rs \( z \) respectively.
We set up equations based on the earnings of persons P, Q, and R.
"Sells" means positive earning, "purchases" means negative earning (cost).
For Person P: Sells 2 units of A, purchases 4 units of B, sells 5 units of C, earns Rs 15,000.
\( 2x - 4y + 5z = 15000 \) . . . (1)
For Person Q: Sells 3 units of A, purchases 1 unit of B, purchases 2 units of C, earns Rs 1,000.
\( 3x - y - 2z = 1000 \) . . . (2)
For Person R: Purchases 1 unit of A, sells 3 units of B, sells 1 unit of C, earns Rs 4,000.
\( -x + 3y + z = 4000 \) . . . (3)
We can write these equations in matrix form, \( AX = B \): \[ \begin{pmatrix} 2 & -4 & 5 \\ 3 & -1 & -2 \\ -1 & 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 15000 \\ 1000 \\ 4000 \end{pmatrix} \] Here, \( A = \begin{pmatrix} 2 & -4 & 5 \\ 3 & -1 & -2 \\ -1 & 3 & 1 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( B = \begin{pmatrix} 15000 \\ 1000 \\ 4000 \end{pmatrix} \).
Next, we find the determinant of matrix A:
\( |A| = 2(1+6) + 4(3-2) + 5(9+1) = 2(7) + 4(1) + 5(10) = 14 + 4 + 50 = 68 \). Since \( |A| = 68 \neq 0 \), the inverse matrix \( A^{-1} \) exists.
Now, we find the adjoint of A. First, we find the cofactor matrix:
\( C_{11} = 7 \), \( C_{12} = 1 \), \( C_{13} = 10 \)
\( C_{21} = 19 \), \( C_{22} = 7 \), \( C_{23} = -2 \)
\( C_{31} = 3 \), \( C_{32} = 19 \), \( C_{33} = 14 \)
The cofactor matrix is \( \begin{pmatrix} 7 & 1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{pmatrix} \). The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj } A = \begin{pmatrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{pmatrix} \). Using the adjoint, we calculate the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{68} \begin{pmatrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{pmatrix} \). Finally, we find X using the formula \( X = A^{-1} B \): \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{68} \begin{pmatrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{pmatrix} \begin{pmatrix} 15000 \\ 1000 \\ 4000 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{68} \begin{pmatrix} 7(15000) + 19(1000) + 3(4000) \\ -1(15000) + 7(1000) + 19(4000) \\ 10(15000) - 2(1000) + 14(4000) \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{68} \begin{pmatrix} 105000 + 19000 + 12000 \\ -15000 + 7000 + 76000 \\ 150000 - 2000 + 56000 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{68} \begin{pmatrix} 136000 \\ 68000 \\ 204000 \end{pmatrix} \] \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2000 \\ 1000 \\ 3000 \end{pmatrix} \] Therefore, the prices per unit are: A = Rs 2000, B = Rs 1000, and C = Rs 3000.
In simple words: We set up three equations from the problem details, representing how each person earned money. We then used these equations to form a matrix system. By calculating the determinant, adjoint, and inverse of the matrix, we found the individual prices of commodities A, B, and C.

🎯 Exam Tip: Pay close attention to keywords like "sells" (earning, positive coefficient) and "purchases" (cost, negative coefficient) when setting up the initial equations for such word problems.