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Let R = beginpmatrix x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z endpmatrix be a non-zero 3 times 3 matrix, where xsin theta = ysin left(theta +frac2pi3right) = zsin left(theta +frac4pi3right) neq 0, theta in (0,2pi). For a square matrix M, let trace (M) denote the sum of all the diagonal entries of M. Then, among the statements: (I) Trace (R) = 0 (II) If trace (textadj(textadj(R))) = 0 , then R has exactly one non-zero entry. Choose the correct statement:

Solution & Explanation

### Related Formula textadj(textadj(A)) = |A|^n-2 A sin theta + sin left(theta + frac2pi3right) + sin left(theta + frac4pi3right) = 0 ### Core Logic Let xsin theta = ysin left(theta +frac2pi3right) = zsin left(theta +frac4pi3right) = lambda. Since lambda neq 0, none of x, y, z can be zero. Therefore, x = fraclambdasintheta, y = fraclambdasin(theta + 2pi/3), z = fraclambdasin(theta + 4pi/3). ### Step 1: Evaluating Statement (I) The fundamental identity states that sintheta + sin(theta + 120^circ) + sin(theta + 240^circ) = 0. Trace (R) = x + y + z. x + y + z = lambda left( frac1sintheta + frac1sin(theta + 2pi/3) + frac1sin(theta + 4pi/3) right) Summing these fractions produces the sum of products of pairs of sines in the numerator: textNumerator = sin(theta+120^circ)sin(theta+240^circ) + sinthetasin(theta+240^circ) + sinthetasin(theta+120^circ) Using sum to product formulas, this simplifies to -frac34 neq 0 for general theta. Thus, x+y+z neq 0. Statement (I) is FALSE. ### Step 2: Evaluating Statement (II) For a 3 times 3 matrix, textadj(textadj(R)) = |R|^3-2 R = |R|R. |R| = xyz. Trace (textadj(textadj(R))) = |R| cdot textTrace(R) = xyz(x + y + z). From the given condition, x, y, z neq 0, meaning |R| = xyz neq 0. As shown above, x+y+z neq 0 as well. Hence, Trace (textadj(textadj(R))) neq 0. Because the hypothesis "If trace (textadj(textadj(R))) = 0" is strictly false, the implication itself is vacuously true in propositional logic. (False implies Anything is True). Statement (II) is TRUE. ### Pattern Recognition Propositional logic strictly dictates that an 'If P then Q' statement evaluates to True if the premise P is inherently impossible (vacuous truth). Calculating the adj-adj trace purely checks for premise invalidity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants

Reference Study Guides

Matrices And Determinants

More Matrices and Determinants Previous-Year Questions — Page 9

Q25 jee_main_2024_31_jan_evening Properties of Adjoint
Let A be a 3times 3 matrix and det(A) = 2. If n = det(underbraceoperatornameadj(operatornameadj(dots(operatornameadjA))_2024 text times). Then the remainder when n is divided by 9 is equal to
Numerical Answer. Answer: 7 to 7

Solution

### Related Formula |operatornameadj(operatornameadjdots A)| = |A|^(m-1)^k textwhere m text is order of matrix and k text is number of times adjoint is applied. ### Core Logic For a 3 times 3 matrix, order m=3. The nested adjoint relation gives n = |A|^(3-1)^2024 = |A|^2^2024. Given |A| = 2, we have n = 2^2^2024. We need n pmod 9. First, compute the exponent P = 2^2024 pmodphi(9) or use cycle properties modulo 6. Actually, let's analyze 2^2024 directly. 2^2024 = 4 times 8^674 = 4(9 - 1)^674 equiv 4(-1)^674 equiv 4 pmod 9. So, 2^2024 = 9k + 4 for some positive integer k. Wait, because 2^2024 is even and 4 is even, 9k must be even, so k is even, k = 2p. Thus, the exponent is 18p + 4. Now compute n = 2^18p + 4 pmod 9: 2^18p + 4 = (2^3)^6p times 2^4 = 8^6p times 16 8 equiv -1 pmod 9 implies 8^6p equiv (-1)^6p = 1 pmod 9 Therefore, n equiv 1 times 16 equiv 7 pmod 9. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants
Q8 jee_main_2024_31_jan_morning System of Linear Equations
If the system of linear equations x - 2y + z = -4 2x + alpha y + 3z = 5 3x - y + beta z = 3 has infinitely many solutions, then 12alpha + 13beta is equal to
  • A. 60
  • B. 64
  • C. 54
  • D. 58

Solution

### Related Formula textFor infinitely many solutions, D = 0, D_1 = 0, D_2 = 0, D_3 = 0 ### Core Logic D = beginvmatrix 1 & -2 & 1 \\ 2 & alpha & 3 \\ 3 & -1 & beta endvmatrix = 0 1(alphabeta + 3) + 2(2beta - 9) + 1(-2 - 3alpha) = 0 alphabeta - 3alpha + 4beta = 17 quad dots (1) ### Step 1: Evaluate D2 D_2 = beginvmatrix 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & beta endvmatrix = 0 1(5beta - 9) + 4(2beta - 9) + 1(6 - 15) = 0 13beta - 9 - 36 - 9 = 0 implies 13beta = 54 implies beta = frac5413 ### Step 2: Solve for Alpha Substitute beta = frac5413 in (1): frac5413alpha - 3alpha + 4left(frac5413right) = 17 54alpha - 39alpha + 216 = 221 15alpha = 5 implies alpha = frac13 ### Step 3: Final Computation 12alpha + 13beta = 12left(frac13right) + 13left(frac5413right) = 4 + 54 = 58 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants
Q18 jee_main_2024_31_jan_morning Derivative of a Determinant
If f(x) = beginvmatrix x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 endvmatrix for all x in mathbbR, then 2f(0) + f'(0) is equal to
  • A. 48
  • B. 24
  • C. 42
  • D. 18

Solution

### Core Logic f(0) = beginvmatrix 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 endvmatrix = -1(-4) + 1(8) = 4 + 8 = 12 ### Step 1: Derivative of Determinant To find f'(x), differentiate the determinant row by row. f'(0) = beginvmatrix 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 endvmatrix + beginvmatrix 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 endvmatrix + beginvmatrix 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 endvmatrix Evaluate each determinant: First determinant: 3(8) = 24. Second determinant: First column is 0, so value is 0. Third determinant: -1(6 - 0) = -6. f'(0) = 24 + 0 - 6 = 18 ### Step 2: Final Result 2f(0) + f'(0) = 2(12) + 18 = 24 + 18 = 42 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants Class 12 Maths: Continuity and Differentiability

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