Samacheer Kalvi Class 12 Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Choose the most suitable answer from the given four alternatives:

 

Question 1. A = (1, 2, 3), then the rank of \( AA^T \) is
(a) 0
(b) 1
(c) 3
(d) 1
Answer: (d) 1
In simple words: We multiply the matrix A by its flipped version \( A^T \). This gives us the number 14. Since 14 is a non-zero number, the rank of this new matrix is 1.

🎯 Exam Tip: For a 1x3 matrix A, the product \( AA^T \) will be a 1x1 matrix. Its rank is 1 if the single element is non-zero.

 

Question 2. The rank of m x n matrix whose elements are unity is
(a) 0
(b) 1
(c) m
(d) n
Answer: (b) 1
In simple words: If every number in a matrix is 1, its rank is always 1. This means all rows are just copies of each other, so only one row is unique.

🎯 Exam Tip: A matrix with all identical non-zero rows or columns will always have a rank of 1.

 

Question 3. If \( T = \begin{pmatrix} 0.4 & 0.6 \\ 0.2 & 0.8 \end{pmatrix} \) is a transition probability matrix, then at equilibrium A is equal to
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{8} \)
Answer: (a) \( \frac{1}{4} \)
In simple words: At a stable point (equilibrium) for this kind of matrix, we use an equation that involves A and B. Since A and B must add up to 1, we replace B with \( 1 - A \). Solving this equation gives us the value of A as \( \frac{1}{4} \).

🎯 Exam Tip: Remember that for a transition matrix at equilibrium, the state vector times the transition matrix equals the state vector itself, and the sum of probabilities in the state vector is always 1.

 

Question 4. If \( A = \begin{pmatrix} 2 & 0 \\ 0 & 8 \end{pmatrix} \), then \( \rho(A) \) is
(a) 0
(b) 1
(c) 2
(d) n
Answer: (c) 2
In simple words: We count the rows that are not all zeros. Here, both rows have numbers that are not zero, so the rank is 2.

🎯 Exam Tip: For a diagonal matrix, its rank is equal to the number of non-zero elements on its main diagonal.

 

Question 5. The rank of the matrix \( \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{pmatrix} \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (d) 3
In simple words: We change the matrix step-by-step to make it simpler. We see that after changing it, all three rows still have numbers that are not zero. This means its rank is 3.

🎯 Exam Tip: Remember that row operations like subtracting a multiple of one row from another do not change the rank of the matrix. The goal is to get it into row-echelon form to easily count non-zero rows.

 

Question 6. The rank of the unit matrix of order n is
(a) n - 1
(b) n
(c) n + 1
(d) \( n^2 \)
Answer: (b) n
In simple words: A unit matrix has ones along its middle line and zeros everywhere else. Because every row has a '1' in a different spot, all 'n' rows are unique. So, its rank is 'n'.

🎯 Exam Tip: The rank of an identity matrix (or unit matrix) of any order 'n' is always 'n' because all its rows are linearly independent.

 

Question 7. If \( \rho(A) = r \) then which of the following is correct?
(a) all the minors of order r which does not vanish
(b) A has at least one minor of order r which does not vanish
(c) A has at least one \( (r + 1) \) order minor which vanishes
(d) all \( (r + 1) \) and higher order minors should not vanish
Answer: (b) A has at least one minor of order r which does not vanish.
In simple words: If a matrix has a rank of 'r', it means we can find at least one smaller square part of size 'r' by 'r' inside it, where the determinant of that part is not zero. We don't need all such parts to be non-zero, just one.

🎯 Exam Tip: The definition of matrix rank states two conditions: existence of at least one non-zero minor of order 'r', and all minors of order \( r+1 \) or greater being zero.

 

Question 8. If \( A = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) then the rank of \( AA^T \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
In simple words: We multiply the given matrix A by its flipped version \( A^T \). This creates a new matrix. After some row changes, we see that only one row in the new matrix has numbers other than zero. So, its rank is 1.

🎯 Exam Tip: When multiplying a column matrix by a row matrix, the resulting matrix will have a rank of 1 if the original matrices are non-zero.

 

Question 9. If the rank of the matrix \( A = \begin{pmatrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{pmatrix} \) is 2, then \( \lambda \) is
(a) 1
(b) 2
(c) 3
(d) only real number
Answer: (a) 1
In simple words: Since the rank is 2, the determinant of the whole 3x3 matrix must be zero. When we calculate this determinant, we get the equation \( \lambda^3 - 1 = 0 \). If we solve this, we find that \( \lambda \) must be 1.

🎯 Exam Tip: If the rank of an 'n x n' matrix is less than 'n', then its determinant must be zero. This is a crucial property for finding unknown values in such matrices.

 

Question 10. The rank of the diagonal matrix \( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3 \end{pmatrix} \) is
(a) 0
(b) 2
(c) 3
(d) 5
Answer: (c) 3
In simple words: For a matrix that only has numbers down the middle, its rank is how many of those numbers are not zero. Here, 1, 2, and -3 are all non-zero, so the rank is 3.

🎯 Exam Tip: The rank of a diagonal matrix is equal to the number of non-zero diagonal entries.

 

Question 11. If \( T = \begin{pmatrix} 0.7 & 0.3 \\ 0.6 & x \end{pmatrix} \) is a transition probability matrix, then the value of x is
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.7
Answer: (c) 0.4
In simple words: In this special kind of matrix, the numbers in each row must add up to 1. For the second row, we know one number is 0.6. So, the other number, x, must be 1 minus 0.6, which is 0.4.

🎯 Exam Tip: Always remember that a key property of a transition probability matrix is that the sum of entries in each row must be 1, representing the total probability of moving from one state to any other.

 

Question 12. Which of the following is not an elementary transformation?
(a) \( R_i \leftrightarrow R_j \)
(b) \( R_i \to 2R_i + 2C_j \)
(c) \( R_i \to 2R_i + 4R_j \)
(d) \( C_i \to C_i + 5C_j \)
Answer: (b) \( R_i \to 2R_i + 2C_j \)
In simple words: Elementary transformations are simple changes you can make to a matrix's rows or columns, like swapping them, multiplying them by a number, or adding one to another. Option (b) tries to mix a row change with a column change, which is not a proper elementary transformation.

🎯 Exam Tip: Elementary transformations involve only rows or only columns at a time, not a mix of both in a single step (e.g., you cannot add a multiple of a column to a row).

 

Question 13. If \( \rho(A) = \rho(A,B)= \) then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Answer: (d) consistent
In simple words: When the rank of the main matrix is the same as the rank of the bigger matrix (which includes the answer part), it means the system of equations will always have a solution. So, the system is consistent.

🎯 Exam Tip: The first step in determining the nature of solutions for a system of linear equations is to check if \( \rho(A) = \rho(A,B) \). If they are equal, the system is consistent; if not, it is inconsistent.

 

Question 14. If \( \rho(A) = \rho(A,B) = \) the number of unknowns, then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Answer: (b) Consistent and has a unique solution
In simple words: When the rank of the main matrix is the same as the rank of the bigger matrix, and this number is also the same as how many unknowns there are, then the equations have one exact answer. The system is consistent and has a unique solution.

🎯 Exam Tip: For a system of linear equations, \( \rho(A) = \rho(A,B) = \text{number of unknowns} \) is the condition for a unique solution.

 

Question 15. If \( \rho(A) \ne \rho(A, B) \), then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Answer: (c) inconsistent
In simple words: If the rank of the main matrix is different from the rank of the bigger matrix, it means the equations don't work together and have no answer. So, the system is inconsistent.

🎯 Exam Tip: An inconsistent system of equations means there are no values for the unknowns that can satisfy all the equations simultaneously.

 

Question 16. In a transition probability matrix, all the entries are greater than or equal to
(a) 2
(b) 1
(c) 0
(d) 3
Answer: (c) 0
In simple words: All the numbers inside a transition probability matrix must be zero or positive. You can't have a negative chance of something happening.

🎯 Exam Tip: Remember the two main rules for a transition probability matrix: (1) all entries are non-negative (greater than or equal to 0), and (2) the sum of entries in each row is 1.

 

Question 17. If the number of variables in a non-homogeneous system \( AX = B \) is n, then the system possesses a unique solution only when
(a) \( \rho(A) = \rho(A, B) > n \)
(b) \( \rho(A) = \rho(A, B) = n \)
(c) \( \rho(A) = \rho(A, B) < n \)
(d) none of these
Answer: (b) \( \rho(A) = \rho(A, B) = n \)
In simple words: For the system to have just one exact answer, the rank of matrix A must be the same as the rank of the combined matrix (A,B), and this number must also be exactly equal to how many unknown variables 'n' there are.

🎯 Exam Tip: This is a direct application of the Rouché-Capelli theorem. Understand the conditions for unique, infinitely many, and no solutions.

 

Question 18. The system of equations \( 4x + 6y = 5, 6x + 9y = 7 \) has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these
Answer: (b) no solution
In simple words: We write the equations as a matrix and simplify it. We find that the rank of the main part of the matrix is 1, but the rank of the whole matrix is 2. Since these ranks are not the same, the equations have no solution.

🎯 Exam Tip: Always remember to compare the ranks of the coefficient matrix and the augmented matrix. If they are different, there's no solution.

 

Question 19. For the system of equations \( x + 2y + 3z = 1, 2x + y - z = 3, 3x + 2y + kz = 4 \)
(a) there is only one solution
(b) there exists infinitely many solutions
(c) there is no solution
(d) none of these
Answer: (a) there is only one solution
In simple words: We write the equations as a matrix and change its rows to simplify it. After simplifying, we see that if \( k \) is not a specific value that makes part of the matrix zero, then the rank of the matrix is 3. Since there are also 3 unknown variables, this means there is only one specific answer for x, y, and z.

🎯 Exam Tip: For a system of three linear equations with three unknowns, a unique solution exists if the determinant of the coefficient matrix is non-zero, which is equivalent to its rank being 3.

 

Question 20. If \( |A| \ne 0 \), then A is
(a) non-singular matrix
(b) singular matrix
(c) zero matrix
Answer: (a) non-singular matrix
In simple words: If the determinant of a matrix A is not zero, then A is called a non-singular matrix. This just means it's a special kind of matrix that can be inverted.

🎯 Exam Tip: Remember the fundamental definitions: \( |A| \ne 0 \) means A is non-singular and invertible, while \( |A| = 0 \) means A is singular and not invertible.

 

Question 21. The system of linear equations \( x = y + z = 2, 2x + y - z = 3, 3x + 2y + kz = 4 \) has unique solution, if k is not equal to
(a) 1
(b) 0
(c) 3
(d) 7
Answer: (b) 0
In simple words: We change the equations into a matrix and simplify it using row steps. For there to be one exact answer, the bottom-right number in the simplified matrix, which is 'k', cannot be zero. So, \( k \) must not be 0.

🎯 Exam Tip: For a system of equations to have a unique solution, after putting its augmented matrix into row-echelon form, the rank of the coefficient matrix (number of non-zero rows on the left side) must be equal to the number of variables.

 

Question 22. Cramer's rule is applicable only to get an unique solution when
(a) \( \Delta_z \ne 0 \)
(b) \( \Delta_x \ne 0 \)
(c) \( \Delta \ne 0 \)
(d) \( \Delta_y \ne 0 \)
Answer: (c) \( \Delta \ne 0 \)
In simple words: You can use Cramer's rule to find one exact answer for equations only if the main determinant, called \( \Delta \), is not zero. If \( \Delta \) is zero, Cramer's rule won't work for finding a single answer.

🎯 Exam Tip: The condition \( \Delta \ne 0 \) is critical for applying Cramer's rule and also for a square coefficient matrix to be non-singular and guarantee a unique solution.

 

Question 23. If \( \frac{a_1}{x} + \frac{b_1}{y} = c_1, \frac{a_2}{x} + \frac{b_2}{y} = c_2 \), then (x, y) is
(a) \( (\frac{\Delta_2}{\Delta_1}, \frac{\Delta_3}{\Delta_1}) \)
(b) \( (\frac{\Delta_3}{\Delta_1}, \frac{\Delta_2}{\Delta_1}) \)
(c) \( (\frac{\Delta_1}{\Delta_2}, \frac{\Delta_1}{\Delta_3}) \)
(d) \( (\frac{-\Delta_1}{\Delta_2}, \frac{-\Delta_1}{\Delta_3}) \)
Answer: (a) \( (\frac{\Delta_2}{\Delta_1}, \frac{\Delta_3}{\Delta_1}) \)
In simple words: We change the variables from \( x, y \) to \( 1/x, 1/y \) to make the equations simpler. Then we use Cramer's rule to find the values of \( 1/x \) and \( 1/y \). The values come out as fractions using \( \Delta \) (determinants).

🎯 Exam Tip: When solving equations with variables in the denominator (like \( 1/x \)), substitute new variables (e.g., \( u = 1/x \)) to convert them into a standard linear system, which can then be solved using methods like Cramer's rule. Pay attention to what the final answer is asking for: \( (x,y) \) or \( (1/x, 1/y) \).

 

Question 24. \( |A_{n \times n}| = 3 |adj A| = 243 = (3)^5 \) then the value n is
(b) 5
(c) 6
(d) 7
Answer: (c) 6
In simple words: We know a rule that relates the determinant of a matrix (A) to its adjoint (adj A) using 'n'. From the given numbers, we figure out that \( |A| \) is 3 and \( |adj A| \) is 243. We then put these numbers into the rule, \( 243 = 3^{n-1} \), and solve for 'n'. This gives \( n=6 \).

🎯 Exam Tip: Remember the important relationship \( |adj A| = |A|^{n-1} \). It is frequently used to find the order of a matrix or its determinant when other information is given.

 

Question 25. Rank of a null matrix is
(a) 0
(b) -1
(c) \( \infty \)
(d) 1
Answer: (a) 0
In simple words: A null matrix is a matrix where all the numbers are zero. Because there are no rows with any non-zero numbers, its rank is 0.

🎯 Exam Tip: The null matrix is the only matrix that has a rank of 0. Any other matrix with at least one non-zero element will have a rank of at least 1.

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