Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.6

Question 1. Test for consistency and if possible, solve the following systems of equations by rank method.
(i) \( x - y + 2z = 2, 2x + y + 4z = 7, 4x - y + z = 4 \)
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 4 \\ 4 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ 4 \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} 1 & -1 & 2 & 2 \\ 2 & 1 & 4 & 7 \\ 4 & -1 & 1 & 4 \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - 4R_1 \)
\[ \begin{pmatrix} 1 & -1 & 2 & 2 \\ 0 & 3 & 0 & 3 \\ 0 & 3 & -7 & -4 \end{pmatrix} \]
\( R_3 \rightarrow R_3 - R_2 \)
\[ \begin{pmatrix} 1 & -1 & 2 & 2 \\ 0 & 3 & 0 & 3 \\ 0 & 0 & -7 & -7 \end{pmatrix} \] From this echelon form, we find the ranks:
The rank of matrix A, \( \rho(A) = 3 \).
The rank of the augmented matrix \( [A|B] \), \( \rho[A|B] = 3 \).
The number of unknowns, \( n = 3 \).
Since \( \rho(A) = \rho[A|B] = n = 3 \), the system of equations is consistent and has a unique solution. A unique solution means there is exactly one set of values for the variables.
We can write the equivalent equations from the echelon form:
\( -7z = -7 \)
\( \implies z = 1 \)
\( 3y = 3 \)
\( \implies y = 1 \)
\( x - y + 2z = 2 \)
\( \implies x - 1 + 2(1) = 2 \)
\( \implies x - 1 + 2 = 2 \)
\( \implies x + 1 = 2 \)
\( \implies x = 1 \)
Thus, the unique solution is \( x = 1, y = 1, z = 1 \).
In simple words: We changed the equations into a matrix and then simplified it step by step. We found that the ranks of the main part and the whole matrix were the same as the number of unknowns. This means there is only one correct answer for x, y, and z, which we found by working backwards from the simplified equations.

🎯 Exam Tip: Always clearly state the rank of A and [A|B] and compare them to the number of unknowns to justify your conclusion about consistency and the type of solution.

Question 1. Test for consistency and if possible, solve the following systems of equations by rank method.
(ii) \( 3x + y + z = 2, x - 3y + 2z = 1, 7x - y + 4z = 5 \)
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} 3 & 1 & 1 \\ 1 & -3 & 2 \\ 7 & -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} 3 & 1 & 1 & 2 \\ 1 & -3 & 2 & 1 \\ 7 & -1 & 4 & 5 \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_1 \leftrightarrow R_2 \) (Swap rows to get a leading 1)
\[ \begin{pmatrix} 1 & -3 & 2 & 1 \\ 3 & 1 & 1 & 2 \\ 7 & -1 & 4 & 5 \end{pmatrix} \]
\( R_2 \rightarrow R_2 - 3R_1 \)
\( R_3 \rightarrow R_3 - 7R_1 \)
\[ \begin{pmatrix} 1 & -3 & 2 & 1 \\ 0 & 10 & -5 & -1 \\ 0 & 20 & -10 & -2 \end{pmatrix} \]
\( R_3 \rightarrow R_3 - 2R_2 \)
\[ \begin{pmatrix} 1 & -3 & 2 & 1 \\ 0 & 10 & -5 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \] From this echelon form, we find the ranks:
The rank of matrix A, \( \rho(A) = 2 \).
The rank of the augmented matrix \( [A|B] \), \( \rho[A|B] = 2 \).
The number of unknowns, \( n = 3 \).
Since \( \rho(A) = \rho[A|B] = 2 < n \), the system of equations is consistent and has infinitely many solutions. This happens because one of the original equations was a multiple of another, making it redundant.
We can write the equivalent equations from the echelon form:
\( x - 3y + 2z = 1 \) (1)
\( 10y - 5z = -1 \) (2)
Let \( z = t \), where \( t \) is any real number.
Substitute \( z=t \) into equation (2):
\( 10y - 5t = -1 \)
\( \implies 10y = 5t - 1 \)
\( \implies y = \frac{5t-1}{10} \)
Substitute \( y \) and \( z \) into equation (1):
\( x - 3\left(\frac{5t-1}{10}\right) + 2t = 1 \)
\( \implies x = 1 + 3\left(\frac{5t-1}{10}\right) - 2t \)
\( \implies x = \frac{10}{10} + \frac{15t-3}{10} - \frac{20t}{10} \)
\( \implies x = \frac{10 + 15t - 3 - 20t}{10} \)
\( \implies x = \frac{7 - 5t}{10} \)
Thus, the infinitely many solutions are given by \( (x, y, z) = \left(\frac{7-5t}{10}, \frac{5t-1}{10}, t\right) \), for any real number \( t \).
In simple words: After simplifying the matrices, we found that the number of useful equations was less than the number of variables. This means there are many possible answers. We can pick any number for 't' (which represents z), and then calculate x and y based on that choice.

🎯 Exam Tip: When dealing with infinitely many solutions, remember to introduce a parameter (like \( t \) or \( s \)) for the free variable(s) and express the other variables in terms of this parameter.

Question 1. Test for consistency and if possible, solve the following systems of equations by rank method.
(iii) \( 2x + 2y + z = 5, x - y + z = 1, 3x + y + 2z = 4 \)
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} 2 & 2 & 1 \\ 1 & -1 & 1 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} 2 & 2 & 1 & 5 \\ 1 & -1 & 1 & 1 \\ 3 & 1 & 2 & 4 \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_1 \leftrightarrow R_2 \) (Swap rows for a leading 1)
\[ \begin{pmatrix} 1 & -1 & 1 & 1 \\ 2 & 2 & 1 & 5 \\ 3 & 1 & 2 & 4 \end{pmatrix} \]
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - 3R_1 \)
\[ \begin{pmatrix} 1 & -1 & 1 & 1 \\ 0 & 4 & -1 & 3 \\ 0 & 4 & -1 & 1 \end{pmatrix} \]
\( R_3 \rightarrow R_3 - R_2 \)
\[ \begin{pmatrix} 1 & -1 & 1 & 1 \\ 0 & 4 & -1 & 3 \\ 0 & 0 & 0 & -2 \end{pmatrix} \] From this echelon form, we find the ranks:
The rank of matrix A, \( \rho(A) = 2 \).
The rank of the augmented matrix \( [A|B] \), \( \rho[A|B] = 3 \).
The number of unknowns, \( n = 3 \).
Since \( \rho(A) \neq \rho[A|B] \), the system of equations is inconsistent and has no solution. This means there is a contradiction in the equations, so no values of x, y, and z can satisfy all three simultaneously.
In simple words: When we simplified the matrix, we found a row that looked like "0x + 0y + 0z = -2". This means 0 equals -2, which is impossible. So, there is no answer that can make all three equations true at the same time.

🎯 Exam Tip: An inconsistent system will always result in a row in the echelon form where the coefficients are all zero, but the constant term is non-zero, indicating a contradiction.

Question 1. Test for consistency and if possible, solve the following systems of equations by rank method.
(iv) \( 2x - y + z = 2, 6x - 3y + 3z = 6, 4x - 2y + 2z = 4 \)
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 4 \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} 2 & -1 & 1 & 2 \\ 6 & -3 & 3 & 6 \\ 4 & -2 & 2 & 4 \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_2 \rightarrow R_2 - 3R_1 \)
\( R_3 \rightarrow R_3 - 2R_1 \)
\[ \begin{pmatrix} 2 & -1 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \] From this echelon form, we find the ranks:
The rank of matrix A, \( \rho(A) = 1 \).
The rank of the augmented matrix \( [A|B] \), \( \rho[A|B] = 1 \).
The number of unknowns, \( n = 3 \).
Since \( \rho(A) = \rho[A|B] = 1 < n \), the system of equations is consistent and has infinitely many solutions. This indicates that all three original equations were essentially the same, just multiplied by different constants.
We can write the equivalent equation from the echelon form:
\( 2x - y + z = 2 \)
Let \( y = s \) and \( z = t \), where \( s \) and \( t \) are any real numbers.
Substitute \( y=s \) and \( z=t \) into the equation:
\( 2x - s + t = 2 \)
\( \implies 2x = 2 + s - t \)
\( \implies x = \frac{2 + s - t}{2} \)
Thus, the infinitely many solutions are given by \( (x, y, z) = \left(\frac{2+s-t}{2}, s, t\right) \), for any real numbers \( s \) and \( t \).
In simple words: All the equations given are just different versions of the same basic equation. So, we only have one actual equation for three variables (x, y, z). This means we can choose any values for two of the variables (like y and z), and then calculate the value for the remaining variable (x). There are endless possible combinations that work.

🎯 Exam Tip: When the rank is much smaller than the number of variables, it implies that many variables can be chosen freely (as parameters), leading to infinitely many solutions.

Question 2. Find the value of k for which the equations \( kx - 2y + z = 1, x - 2ky + z = -2, x - 2y + kz = 1 \), have
(i) no solution
(ii) unique solution
(iii) infinitely many solution.
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} k & -2 & 1 & 1 \\ 1 & -2k & 1 & -2 \\ 1 & -2 & k & 1 \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_1 \leftrightarrow R_3 \) (Swap R1 and R3 for an easier leading 1)
\[ \begin{pmatrix} 1 & -2 & k & 1 \\ 1 & -2k & 1 & -2 \\ k & -2 & 1 & 1 \end{pmatrix} \]
\( R_2 \rightarrow R_2 - R_1 \)
\( R_3 \rightarrow R_3 - kR_1 \)
\[ \begin{pmatrix} 1 & -2 & k & 1 \\ 0 & -2k+2 & 1-k & -3 \\ 0 & -2+2k & 1-k^2 & 1-k \end{pmatrix} \] We can simplify the terms: \( -2k+2 = 2(1-k) \), \( -2+2k = -2(1-k) \), and \( 1-k^2 = (1-k)(1+k) \).
The matrix becomes:
\[ \begin{pmatrix} 1 & -2 & k & 1 \\ 0 & 2(1-k) & 1-k & -3 \\ 0 & -2(1-k) & (1-k)(1+k) & 1-k \end{pmatrix} \]
\( R_3 \rightarrow R_3 + R_2 \)
\[ \begin{pmatrix} 1 & -2 & k & 1 \\ 0 & 2(1-k) & 1-k & -3 \\ 0 & 0 & (1-k)(1+k) + (1-k) & (1-k) - 3 \end{pmatrix} \] Simplify the terms in the third row:
\( (1-k)(1+k) + (1-k) = (1-k)(1+k+1) = (1-k)(k+2) \)
\( (1-k) - 3 = -k - 2 = -(k+2) \)
The final echelon form is:
\[ \begin{pmatrix} 1 & -2 & k & 1 \\ 0 & 2(1-k) & 1-k & -3 \\ 0 & 0 & (1-k)(k+2) & -(k+2) \end{pmatrix} \] Now, we analyze the conditions for \( k \):
(i) **No solution:**
This occurs when \( \rho(A) \neq \rho[A|B] \). For this to happen, the last row of A must be zero while the corresponding element in [A|B] is non-zero.
We need \( (1-k)(k+2) = 0 \) and \( -(k+2) \neq 0 \).
From \( (1-k)(k+2) = 0 \), we get \( k=1 \) or \( k=-2 \).
From \( -(k+2) \neq 0 \), we get \( k \neq -2 \).
Combining these, the only possibility is \( k=1 \).
When \( k=1 \):
\( \rho(A) = 2 \) (because the third row of A becomes zero).
\( \rho[A|B] = 3 \) (because the element \( -(1+2) = -3 \) is non-zero).
Since \( \rho(A) \neq \rho[A|B] \), the system has **no solution** when \( k=1 \).
(ii) **Unique solution:**
This occurs when \( \rho(A) = \rho[A|B] = n \) (where \( n=3 \)). This means all three rows must be non-zero.
We need \( (1-k)(k+2) \neq 0 \).
This implies \( k \neq 1 \) and \( k \neq -2 \).
When \( k \neq 1 \) and \( k \neq -2 \):
\( \rho(A) = 3 \).
\( \rho[A|B] = 3 \).
Since \( \rho(A) = \rho[A|B] = n = 3 \), the system has a **unique solution** when \( k \neq 1 \) and \( k \neq -2 \).
(iii) **Infinitely many solutions:**
This occurs when \( \rho(A) = \rho[A|B] < n \). This means the last row of both A and [A|B] must be zero.
We need \( (1-k)(k+2) = 0 \) and \( -(k+2) = 0 \).
From \( -(k+2) = 0 \), we get \( k=-2 \). This condition also satisfies \( (1-k)(k+2) = 0 \).
When \( k=-2 \):
\( \rho(A) = 2 \).
\( \rho[A|B] = 2 \).
Since \( \rho(A) = \rho[A|B] = 2 < n = 3 \), the system has **infinitely many solutions** when \( k=-2 \).
In simple words: We used matrix operations to simplify the system of equations. By looking at the last row of the simplified matrix, we can tell what kind of solution exists. If k is 1, there's a contradiction. If k is -2, one equation becomes useless, giving many answers. For any other value of k, there is exactly one specific answer.

🎯 Exam Tip: The key to solving these types of problems is correctly identifying the conditions on the parameter (like \( k \)) that make the determinant of the coefficient matrix zero, and then analyzing the ranks of A and [A|B] for those specific values.

Question 3. Investigate the values of \( \lambda \) and \( \mu \) the system of linear equations \( 2x + 3y + 5z = 9, 7x + 3y - 5z = 8, 2x + 3y + \lambda z = \mu \), have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
Answer: First, we write the given system of equations in matrix form \( AX=B \):
\[ \begin{pmatrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 9 \\ 8 \\ \mu \end{pmatrix} \] Next, we form the augmented matrix \( [A|B] \):
\[ [A|B] = \begin{pmatrix} 2 & 3 & 5 & 9 \\ 7 & 3 & -5 & 8 \\ 2 & 3 & \lambda & \mu \end{pmatrix} \] Now, we perform elementary row operations to transform the augmented matrix into an echelon form:
\( R_2 \rightarrow 2R_2 - 7R_1 \)
\( R_3 \rightarrow R_3 - R_1 \)
\[ \begin{pmatrix} 2 & 3 & 5 & 9 \\ 0 & 2(3)-7(3) & 2(-5)-7(5) & 2(8)-7(9) \\ 0 & 3-3 & \lambda-5 & \mu-9 \end{pmatrix} \] Calculate the elements for the second row:
\( 6-21 = -15 \)
\( -10-35 = -45 \)
\( 16-63 = -47 \)
The augmented matrix in echelon form is:
\[ \begin{pmatrix} 2 & 3 & 5 & 9 \\ 0 & -15 & -45 & -47 \\ 0 & 0 & \lambda-5 & \mu-9 \end{pmatrix} \] Now, we analyze the conditions for \( \lambda \) and \( \mu \):
(i) **No solution:**
This occurs when \( \rho(A) \neq \rho[A|B] \). This happens if the last coefficient of A is zero, but the corresponding constant term in [A|B] is not zero.
So, we need \( \lambda - 5 = 0 \) and \( \mu - 9 \neq 0 \).
This means \( \lambda = 5 \) and \( \mu \neq 9 \).
In this case, \( \rho(A) = 2 \) and \( \rho[A|B] = 3 \). Since \( \rho(A) \neq \rho[A|B] \), the system has **no solution**.
(ii) **A unique solution:**
This occurs when \( \rho(A) = \rho[A|B] = n \) (where \( n=3 \)). This means all three rows in the echelon form must be non-zero.
So, we need \( \lambda - 5 \neq 0 \).
This means \( \lambda \neq 5 \). (The value of \( \mu \) can be any real number in this case).
In this case, \( \rho(A) = 3 \) and \( \rho[A|B] = 3 \). Since \( \rho(A) = \rho[A|B] = n = 3 \), the system has a **unique solution**.
(iii) **An infinite number of solutions:**
This occurs when \( \rho(A) = \rho[A|B] < n \). This means the last row of both A and [A|B] must be zero.
So, we need \( \lambda - 5 = 0 \) and \( \mu - 9 = 0 \).
This means \( \lambda = 5 \) and \( \mu = 9 \).
In this case, \( \rho(A) = 2 \) and \( \rho[A|B] = 2 \). Since \( \rho(A) = \rho[A|B] = 2 < n = 3 \), the system has **infinitely many solutions**.
In simple words: We changed the equations into a matrix and simplified it. By looking at the last row of the simplified matrix, we can determine how many solutions the system has. If \( \lambda \) is 5 and \( \mu \) is not 9, there's no answer. If \( \lambda \) is not 5, there's always one specific answer. If \( \lambda \) is 5 and \( \mu \) is 9, there are countless answers.

🎯 Exam Tip: Carefully observe the last row of the echelon form. If it becomes \( [0 \quad 0 \quad 0 \quad C] \) where \( C \neq 0 \), there's no solution. If it's \( [0 \quad 0 \quad 0 \quad 0] \), you have infinite solutions (provided other conditions are met), and if no row reduces to all zeros, it's a unique solution.