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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 6

Q70 jee_main_2025_29_jan_morning Cofactors and Determinant Value
Let mathrmA = left[mathbfa_mathrmijright] = beginbmatrix log_5128 & log_45 \\ log_58 & log_425 endbmatrix . If A_ij is the cofactor of a_ij , C_ij = sum_k=1^2 a_ik A_jk , 1 leq i, j leq 2 , and C = [C_ij] , then 8|C| is equal to:
  • A. 262
  • B. 288
  • C. 242
  • D. 222

Solution

### Related Formula sum_k a_ik A_jk = delta_ij |A| implies C = beginbmatrix |A| & 0 \\ 0 & |A| endbmatrix implies |C| = |A|^2 ### Core Logic Evaluate the determinant of matrix A: |A| = (log_5 128)(log_4 25) - (log_4 5)(log_5 8) Using change of base rules: |A| = left(7log_5 2right)left(2log_4 5right) - left(frac12log_2 5right)left(3log_5 2right) |A| = 14left(log_5 2 cdot frac12log_2 5right) - frac32 = 7 - 1.5 = 5.5 = frac112 ### Step 1: Compute |C| and evaluate response target Since matrix properties dictate |C| = |A|^2: |C| = left(frac112right)^2 = frac1214 Evaluate targeted multiplier: 8|C| = 8 times frac1214 = 2 times 121 = 242 ### Pattern Recognition Recognize the core cofactor theorem identity instantly: multiplying rows by cofactors of other rows creates zero elements, yielding basic diagonal scalar structures matching matrix attributes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q72 jee_main_2025_29_jan_morning Matrix Powers
Let S = left| m in Z : A^m^2 + A^m = 3I - A^-6 right| , where A = beginbmatrix 2 & -1 \\ 1 & 0 endbmatrix . Then n(S) is equal to
Numerical Answer. Answer: 2

Solution

### Related Formula textInductive formulation for exponent powers of a pattern matrix ### Core Logic Evaluate lower power forms of A to establish inductive sequence patterns: A = beginbmatrix 2 & -1 \\ 1 & 0 endbmatrix, quad A^2 = beginbmatrix 3 & -2 \\ 2 & -1 endbmatrix, quad A^3 = beginbmatrix 4 & -3 \\ 3 & -2 endbmatrix This cleanly establishes general power state rule configuration expression: A^m = beginbmatrix m+1 & -m \\ m & -m+1 endbmatrix ### Step 1: Setup Matrix Power Equation Using the pattern, find expressions for terms: A^6 = beginbmatrix 7 & -6 \\ 6 & -5 endbmatrix, quad A^-6 = (A^6)^-1 = beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix Substitute into the targeted equation block: A^m^2 + A^m = 3beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix - beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix = beginbmatrix 8 & -6 \\ 6 & -4 endbmatrix ### Step 2: Equate Corresponding Elements From the bottom-\left entry [2,1]: m^2 + m = 6 implies m^2 + m - 6 = 0 (m+3)(m-2) = 0 implies m = -3, 2 Both integer states satisfy all matrix components seamlessly. Therefore, the number of elements n(S) = 2. ### Pattern Recognition Always calculate first 2–3 matrix powers to spot linear arithmetic trends across specific cell values (m+1, -m), avoiding tedious full Cayley-Hamilton character expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices
Q3 jee_main_2024_01_february_morning Properties of Determinants
If A=beginbmatrixsqrt2 & 1 \\ -1 & sqrt2endbmatrix, B=beginbmatrix1 & 0 \\ 1 & 1endbmatrix, C=ABA^T and X=A^TC^2A, then det X is equal to:
  • A. 243
  • B. 729
  • C. 27
  • D. 891

Solution

### Related Formula Properties of Determinants: 1. det(AB) = det(A) cdot det(B) 2. det(A^T) = det(A) 3. det(A^n) = (det(A))^n ### Core Logic First, find the determinant of matrix A and matrix B: det(A) = |A| = beginvmatrixsqrt2 & 1 \\ -1 & sqrt2endvmatrix = (sqrt2)(sqrt2) - (1)(-1) = 2 + 1 = 3 det(B) = |B| = beginvmatrix1 & 0 \\ 1 & 1endvmatrix = (1)(1) - (0)(1) = 1 ### Step 1: Calculate determinant of C Given C = ABA^T, we calculate its determinant: det(C) = det(ABA^T) = det(A) cdot det(B) cdot det(A^T) Using det(A^T) = det(A) = 3: det(C) = 3 cdot 1 cdot 3 = 9 ### Step 2: Calculate determinant of X Given X = A^TC^2A, we determine |X|: det(X) = |A^TC^2A| = |A^T| cdot |C^2| cdot |A| det(X) = |A| cdot |C|^2 cdot |A| = |A|^2 cdot |C|^2 Substitute the values |A| = 3 and |C| = 9: det(X) = (3)^2 cdot (9)^2 = 9 cdot 81 = 729 ### Pattern Recognition Sees: Transpose and multiplication operations nested inside a determinant statement. Shortcut: Never explicitly compute product matrices like ABA^T or A^TC^2A. Apply the distributive identity of determinants |XYZ| = |X||Y||Z| entirely to work strictly with scalar multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q12 jee_main_2024_01_february_morning System of Linear Equations
If the system of equations 2x+3y-z=5 x+alpha y+3z=-4 3x-y+beta z=7 has infinitely many solutions, then 13 alpha beta is equal to:
  • A. 1110
  • B. 1120
  • C. 1210
  • D. 1220

Solution

### Related Formula For a system of linear equations representing planes to have infinitely many solutions, the planes must belong to a single family sharing a common line of intersection: P_1 = k_1 P_2 + k_2 P_3 ### Core Logic Let the planes be defined as: - P_1: 2x + 3y - z - 5 = 0 - P_2: x + alpha y + 3z + 4 = 0 - P_3: 3x - y + beta z - 7 = 0 Expressing P_1 as a linear combination of P_2 and P_3: 2x + 3y - z - 5 = k_1(x + alpha y + 3z + 4) + k_2(3x - y + beta z - 7) ### Step 1: Evaluate parameters k1 and k2 Comparing the coefficients of x and the constant terms on both sides: - For x: k_1 + 3k_2 = 2 quad implies (1) - For the constants: 4k_1 - 7k_2 = -5 quad implies (2) Multiplying equation (1) by 4 gives 4k_1 + 12k_2 = 8. Subtracting equation (2) from this result: (4k_1 + 12k_2) - (4k_1 - 7k_2) = 8 - (-5) 19k_2 = 13 implies k_2 = frac1319 Substituting k_2 back into equation (1): k_1 + 3left(frac1319right) = 2 implies k_1 = 2 - frac3919 = -frac119 ### Step 2: Calculate alpha, beta and final product Comparing coefficients for y and z: - For y: k_1alpha - k_2 = 3 implies -frac119alpha - frac1319 = 3 -alpha - 13 = 57 implies alpha = -70 - For z: 3k_1 + k_2beta = -1 implies 3left(-frac119right) + frac1319beta = -1 -3 + 13beta = -19 implies 13beta = -16 implies beta = -frac1613 Now, compute 13 alpha beta: 13 alpha beta = 13 times (-70) times left(-frac1613right) = 70 times 16 = 1120 ### Pattern Recognition Sees: Infinite solution framework for three linear planes. Shortcut: Using the family of planes equation is significantly less prone to fractional algebraic mistakes compared to establishing Cramer's rule determinants (D = D_x = D_y = D_z = 0). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q1 jee_main_2024_29_january_evening Properties of Determinants
Let A = beginbmatrix 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 endbmatrix and P = beginbmatrix 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 endbmatrix. The sum of the prime factors of |P^-1AP - 2I| is equal to
  • A. 26
  • B. 27
  • C. 66
  • D. 23

Solution

### Related Formula |P^-1AP - 2I| = |P^-1(A - 2I)P| = |P^-1| cdot |A - 2I| cdot |P| = |A - 2I| ### Core Logic Since |P^-1| cdot |P| = 1, the expression simplifies completely to the determinant of A - 2I. First, let us construct the matrix A - 2I: A - 2I = beginbmatrix 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 endbmatrix = beginbmatrix 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 endbmatrix Now, evaluating the determinant: |A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69 ### Step 1: Finding Prime Factors The number obtained is 69. Let us find its prime factorization: 69 = 3 times 23 Both 3 and 23 are prime numbers. Their sum is: textSum = 3 + 23 = 26 ### Pattern Recognition Whenever you encounter a matrix expression of the form P^-1AP - kI, always factor out P^-1 and P to simplify it to |A - kI|. This saves tremendous time over computing matrix multiplications. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

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