Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.7

Question 1. Solve the following system of homogeneous equations.
(i) 3x + 2y + 7z = 0; 4x - 3y - 2z = 0; 5x + 9y + 23z = 0
Answer:
To solve the given homogeneous system, we first write it in matrix form \( AX = B \): \[ \begin{pmatrix} 3 & 2 & 7 \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] Next, we form the augmented matrix \( [A | B] \): \[ [A | B] = \begin{pmatrix} 3 & 2 & 7 & 0 \\ 4 & -3 & -2 & 0 \\ 5 & 9 & 23 & 0 \end{pmatrix} \] We apply elementary row operations to transform the augmented matrix into row echelon form: Applying \( R_2 \rightarrow 3R_2 - 4R_1 \) and \( R_3 \rightarrow 3R_3 - 5R_1 \): \[ \begin{pmatrix} 3 & 2 & 7 & 0 \\ 0 & -17 & -34 & 0 \\ 0 & 17 & 34 & 0 \end{pmatrix} \] Applying \( R_3 \rightarrow R_3 + R_2 \): \[ \begin{pmatrix} 3 & 2 & 7 & 0 \\ 0 & -17 & -34 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \] From the echelon form, the rank of the coefficient matrix \( A \) is \( \rho(A) = 2 \), and the rank of the augmented matrix \( [A | B] \) is \( \rho[A | B] = 2 \). The number of unknowns \( n = 3 \). Since \( \rho(A) = \rho[A | B] = 2 < n = 3 \), the system is consistent and has infinitely many non-trivial solutions. Now, we write the equivalent equations from the echelon form: 1) \( 3x + 2y + 7z = 0 \) 2) \( -17y - 34z = 0 \) From equation (2), let \( z = t \), where \( t \) is any real number. \( -17y - 34t = 0 \)
\( \implies -17y = 34t \)
\( \implies y = \frac{34t}{-17} = -2t \) Substitute \( y = -2t \) and \( z = t \) into equation (1): \( 3x + 2(-2t) + 7t = 0 \)
\( \implies 3x - 4t + 7t = 0 \)
\( \implies 3x + 3t = 0 \)
\( \implies 3x = -3t \)
\( \implies x = -t \) Thus, the general solution is \( (x, y, z) = (-t, -2t, t) \) for any \( t \in \mathbb{R} \). This means there are many solutions depending on the value of 't'.
In simple words: We changed the equations into a matrix and used row operations to simplify it. Because the rank was less than the number of variables, it means there are many solutions, which we expressed using a variable 't'.

🎯 Exam Tip: For homogeneous systems, if the rank of the coefficient matrix is less than the number of unknowns, you will always have infinitely many non-trivial solutions, which should be expressed using a parameter like 't'.

Question 1. (ii) 2x + 3y - z = 0, x - y - 2z = 0, 3x + y + 3z = 0
Answer:
First, we write the given system of homogeneous equations in matrix form \( AX = B \): \[ \begin{pmatrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] Next, we form the augmented matrix \( [A | B] \): \[ [A | B] = \begin{pmatrix} 2 & 3 & -1 & 0 \\ 1 & -1 & -2 & 0 \\ 3 & 1 & 3 & 0 \end{pmatrix} \] We apply elementary row operations to transform the augmented matrix into row echelon form: Swap \( R_1 \) and \( R_2 \) to get 1 in the leading position: \( R_1 \leftrightarrow R_2 \) \[ \begin{pmatrix} 1 & -1 & -2 & 0 \\ 2 & 3 & -1 & 0 \\ 3 & 1 & 3 & 0 \end{pmatrix} \] Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \): \[ \begin{pmatrix} 1 & -1 & -2 & 0 \\ 0 & 5 & 3 & 0 \\ 0 & 4 & 9 & 0 \end{pmatrix} \] Applying \( R_3 \rightarrow 5R_3 - 4R_2 \): \[ \begin{pmatrix} 1 & -1 & -2 & 0 \\ 0 & 5 & 3 & 0 \\ 0 & 0 & 33 & 0 \end{pmatrix} \] From the echelon form, the rank of the coefficient matrix \( A \) is \( \rho(A) = 3 \), and the rank of the augmented matrix \( [A | B] \) is \( \rho[A | B] = 3 \). The number of unknowns \( n = 3 \). Since \( \rho(A) = \rho[A | B] = n = 3 \), the system is consistent and has a unique solution. For a homogeneous system, a unique solution is always the trivial solution: \( x = 0, y = 0, z = 0 \). This signifies that all three equations are only satisfied when x, y, and z are all zero.
In simple words: We converted the equations to a matrix and simplified it. Since the matrix rank matched the number of variables, it means there's only one way to solve it, which is when x, y, and z are all zero.

🎯 Exam Tip: For a homogeneous system, if the rank equals the number of unknowns, the only solution is always the trivial solution where all variables are zero.

Question 2. Determine the values of \( \lambda \) for which the following system of equations. x + y + 3z = 0; 4x + 3y + \( \lambda \)z = 0, 2x + y + 2z = 0 has
(i) a unique solution
Answer:
First, we write the given system of homogeneous equations in matrix form \( AX = B \): \[ \begin{pmatrix} 1 & 1 & 3 \\ 4 & 3 & \lambda \\ 2 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] Next, we form the augmented matrix \( [A | B] \): \[ [A | B] = \begin{pmatrix} 1 & 1 & 3 & 0 \\ 4 & 3 & \lambda & 0 \\ 2 & 1 & 2 & 0 \end{pmatrix} \] We apply elementary row operations to transform the augmented matrix into row echelon form: Applying \( R_2 \rightarrow R_2 - 4R_1 \) and \( R_3 \rightarrow R_3 - 2R_1 \): \[ \begin{pmatrix} 1 & 1 & 3 & 0 \\ 0 & -1 & \lambda - 12 & 0 \\ 0 & -1 & -4 & 0 \end{pmatrix} \] Applying \( R_3 \rightarrow R_3 - R_2 \): \[ \begin{pmatrix} 1 & 1 & 3 & 0 \\ 0 & -1 & -4 & 0 \\ 0 & 0 & \lambda - 8 & 0 \end{pmatrix} \] (i) **For a unique solution:** For a homogeneous system to have a unique solution, the rank of the coefficient matrix \( A \) must be equal to the number of unknowns \( n \). In this case, \( n = 3 \). From the echelon form, for \( \rho(A) = 3 \), the last entry in the third row, \( \lambda - 8 \), must not be zero.
\( \implies \lambda - 8 \neq 0 \)
\( \implies \lambda \neq 8 \) If \( \lambda \neq 8 \), then \( \rho(A) = 3 \) and \( \rho[A | B] = 3 \). Since \( \rho(A) = \rho[A | B] = n = 3 \), the system is consistent and has a unique solution, which is the trivial solution \( x = 0, y = 0, z = 0 \). This occurs because the determinant of the coefficient matrix is non-zero.
**Case (ii) For a non-trivial solution (Implicitly asked by "determine values for \( \lambda \)" and shown in source):** If \( \lambda = 8 \), then the augmented matrix becomes: \[ \begin{pmatrix} 1 & 1 & 3 & 0 \\ 0 & -1 & -4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \] In this case, \( \rho(A) = 2 \) and \( \rho[A | B] = 2 \). Since \( \rho(A) = \rho[A | B] = 2 < n = 3 \), the system is consistent and has infinitely many non-trivial solutions. Therefore, the system has a unique solution when \( \lambda \neq 8 \).
In simple words: We used matrix operations to find out when the system has only one solution. This happens when a special value, \( \lambda \), is not equal to 8. If \( \lambda \) is 8, there are many solutions, not just one.

🎯 Exam Tip: Remember that for a homogeneous system, a unique solution always means the trivial solution (all variables are zero). The condition for this is that the rank of the coefficient matrix must equal the number of variables.

Question 3. By using Gaussian elimination method, balance the chemical -reaction equation: C2H6 + O2 → H2O + CO2.
Answer:
Let the balanced chemical equation be: \( x_1 C_2H_6 + x_2 O_2 \rightarrow x_3 H_2O + x_4 CO_2 \) (1) Now, we balance the atoms for each element on both sides of the equation to form a system of linear homogeneous equations: **For Carbon (C):** \( 2x_1 = x_4 \)
\( \implies 2x_1 - x_4 = 0 \) (2) **For Hydrogen (H):** \( 6x_1 = 2x_3 \)
\( \implies 6x_1 - 2x_3 = 0 \)
Dividing by 2 gives: \( 3x_1 - x_3 = 0 \) (3) **For Oxygen (O):** \( 2x_2 = x_3 + 2x_4 \)
\( \implies 2x_2 - x_3 - 2x_4 = 0 \) (4) We form the augmented matrix for this system of equations: \[ [A | B] = \begin{pmatrix} 2 & 0 & 0 & -1 & 0 \\ 3 & 0 & -1 & 0 & 0 \\ 0 & 2 & -1 & -2 & 0 \end{pmatrix} \] Now, we apply Gaussian elimination (row operations) to transform the augmented matrix into row echelon form: Applying \( R_2 \rightarrow 2R_2 - 3R_1 \): \[ \begin{pmatrix} 2 & 0 & 0 & -1 & 0 \\ 0 & 0 & -2 & 3 & 0 \\ 0 & 2 & -1 & -2 & 0 \end{pmatrix} \] Swap \( R_2 \) and \( R_3 \) to get a leading non-zero element in the second row: \( R_2 \leftrightarrow R_3 \) \[ \begin{pmatrix} 2 & 0 & 0 & -1 & 0 \\ 0 & 2 & -1 & -2 & 0 \\ 0 & 0 & -2 & 3 & 0 \end{pmatrix} \] From the echelon form, the rank of the coefficient matrix \( A \) is \( \rho(A) = 3 \), and the rank of the augmented matrix \( [A | B] \) is \( \rho[A | B] = 3 \). The number of unknowns \( n = 4 \). Since \( \rho(A) = \rho[A | B] = 3 < n = 4 \), the system is consistent and has infinitely many non-trivial solutions. We write the equivalent equations from the echelon form: 1) \( 2x_1 - x_4 = 0 \) 2) \( 2x_2 - x_3 - 2x_4 = 0 \) 3) \( -2x_3 + 3x_4 = 0 \) Let \( x_4 = t \), where \( t \) is a positive integer. From equation (3): \( -2x_3 + 3t = 0 \)
\( \implies -2x_3 = -3t \)
\( \implies x_3 = \frac{3t}{2} \) From equation (1): \( 2x_1 - t = 0 \)
\( \implies 2x_1 = t \)
\( \implies x_1 = \frac{t}{2} \) From equation (2): \( 2x_2 - x_3 - 2x_4 = 0 \)
\( \implies 2x_2 - \frac{3t}{2} - 2t = 0 \)
\( \implies 2x_2 = \frac{3t}{2} + 2t \)
\( \implies 2x_2 = \frac{3t + 4t}{2} = \frac{7t}{2} \)
\( \implies x_2 = \frac{7t}{4} \) To get the smallest positive integer values for \( x_1, x_2, x_3, x_4 \), we choose \( t \) to be the least common multiple (LCM) of the denominators (2, 4, 2, 1), which is 4. So, let \( t = 4 \): \( x_1 = \frac{4}{2} = 2 \) \( x_2 = \frac{7 \times 4}{4} = 7 \) \( x_3 = \frac{3 \times 4}{2} = 6 \) \( x_4 = 4 \) Therefore, the balanced chemical equation is: \( 2C_2H_6 + 7O_2 \rightarrow 6H_2O + 4CO_2 \). This method helps ensure atom conservation in chemical reactions.
In simple words: We turned the chemical balancing problem into math equations. By solving these equations using matrices, we found the right numbers for each part of the reaction to make sure all atoms are balanced.

🎯 Exam Tip: When balancing chemical equations with this method, ensure you correctly set up the conservation equations for each element. Always choose the smallest integer value for the parameter 't' to get the simplest whole-number coefficients.