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Consider the system of linear equation x + y + z = 4mu, x + 2y + 2lambda z = 10mu, x + 3y + 4lambda^2 z = mu^2 + 15, where lambda, mu in mathbbR. Which one of the following statements is NOT correct?

Solution & Explanation

### Related Formula Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix If Delta neq 0, unique solution. If Delta = 0 and Delta_x = Delta_y = Delta_z = 0, infinitely many solutions. If Delta = 0 and at least one of Delta_x, Delta_y, Delta_z neq 0, inconsistent (no solution). ### Core Logic Given system: x + y + z = 4mu x + 2y + 2lambda z = 10mu x + 3y + 4lambda^2 z = mu^2 + 15 Compute the main determinant Delta: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 2lambda \\ 1 & 3 & 4lambda^2 endvmatrix Apply operations: R_2 to R_2 - R_1, R_3 to R_3 - R_1 Delta = beginvmatrix 1 & 1 & 1 \\ 0 & 1 & 2lambda-1 \\ 0 & 2 & 4lambda^2-1 endvmatrix = 1 cdot (4lambda^2 - 1 - 2(2lambda - 1)) = 4lambda^2 - 1 - 4lambda + 2 = 4lambda^2 - 4lambda + 1 = (2lambda - 1)^2 ### Step 1: Analyzing Unique Solution For a unique solution, Delta neq 0 Rightarrow 2lambda - 1 neq 0 Rightarrow lambda neq frac12. Note: For unique solution, mu can be anything. Option (4) states the system is consistent if lambda neq frac12, which is purely correct. Option (1) says "unique solution if lambda neq frac12 and mu neq 1, 15". While true that it has a unique solution under those conditions, it also has a unique solution for mu = 1, 15. Let's check consistency conditions. ### Step 2: Checking Delta components Let Delta = 0, so lambda = frac12. Compute Delta_x and Delta_z (or Delta_y): Wait, substituting lambda = 1/2, the equations become: x + y + z = 4mu x + 2y + z = 10mu x + 3y + z = mu^2 + 15 From the first two, (x+2y+z) - (x+y+z) = 10mu - 4mu Rightarrow y = 6mu. From the second and third, (x+3y+z) - (x+2y+z) = mu^2 + 15 - 10mu Rightarrow y = mu^2 - 10mu + 15. For the system to be consistent (infinite solutions since Delta = 0), the two values of y must match: 6mu = mu^2 - 10mu + 15 mu^2 - 16mu + 15 = 0 (mu - 1)(mu - 15) = 0 So, if lambda = frac12, the system is consistent (infinite solutions) ONLY when mu = 1 or mu = 15. If lambda = frac12 and mu neq 1, 15, it is inconsistent. ### Step 3: Checking Options Option (2) states: "The system is inconsistent if lambda = frac12 and mu neq 1". If mu = 15 (which is neq 1), the system is actually CONSISTENT (infinite solutions). Therefore, Option (2) is NOT strictly correct because mu=15 makes it consistent. Thus, statement (2) is the incorrect statement. ### Pattern Recognition Cramer's rule dependencies can be quickly identified using algebraic elimination. When variables align symmetrically (like z mapping identically), subtracting equations exposes the consistency constraint directly without resolving full 3times3 determinants. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants

Reference Study Guides

More 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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