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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 5

Q65 jee_main_2025_24_jan_evening Evaluation of Determinants using Limits
For some a, b, let f(x)=beginvmatrix a+fracsin xx & 1 & b \\ a & 1+fracsin xx & b \\ a & 1 & b+fracsin xx endvmatrix, xne0 [cite: 3370, 3371, 3372, 3374, 3375] lim_xrightarrow0f(x)=lambda+mu a+vb [cite: 3376, 3377] Then (lambda+mu+nu)^2 is equal to:
  • A. 25
  • B. 9
  • C. 36
  • D. 16

Solution

### Related Formula The fundamental trigonometric limit is given by: lim_x to 0 fracsin xx = 1 ### Core Logic Apply the limit inside each element of the matrix determinant : lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 endvmatrix ### Step 1: Simplify the Determinant via Row Operations Perform rows reductions R_2 to R_2 - R_1 and R_3 to R_3 - R_1 to create zeros: lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 endvmatrix Expand across the first row : = (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)] = (a+1)(1) + 1 + b = a + b + 2 [cite: 3999, 4000] ### Step 2: Match Coefficients Equate this outcome with the target parameter template lambda + mu a + nu b [cite: 3377, 4000]: lambda = 2, quad mu = 1, quad nu = 1 Calculate (lambda + mu + nu)^2 [cite: 3378, 4001]: (2 + 1 + 1)^2 = 4^2 = 16 ### Pattern Recognition Standard row manipulations on identity-shifted arrays quickly eliminate complex parameter symbols, reducing determinant calculations into straightforward polynomial expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Limits and Derivatives
Q70 jee_main_2025_24_jan_morning System of Linear Equations
If the system of equations 2x - y + z = 4 5x + lambda y + 3z = 12 100x - 47y + mu z = 212 has infinitely many solutions, then mu - 2lambda is equal to :
  • A. 56
  • B. 59
  • C. 55
  • D. 57

Solution

### Related Formula According to Cramer's Rule, for a system of linear equations to have infinitely many solutions, the main determinant Delta and all directional determinants Delta_1, Delta_2, Delta_3 must equal zero simultaneously. ### Core Logic Set up the directional determinant equation Delta_3 = 0 by replacing the third column with the constant vector: Delta_3 = left| beginmatrix 2 & -1 & 4 \\ 5 & lambda & 12 \\ 100 & -47 & 212 endmatrix right| = 0 Expand the determinant along the first row: 2[212lambda - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100lambda] = 0 2[212lambda + 564] + 1[1060 - 1200] + 4[-235 - 100lambda] = 0 424lambda + 1128 - 140 - 940 - 400lambda = 0 24lambda + 48 = 0 implies lambda = -2 ### Step 1: Solve for Mu using the main determinant Set the primary coefficient matrix determinant Delta = 0 and substitute lambda = -2: Delta = left| beginmatrix 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & mu endmatrix right| = 0 Expand the determinant along the first row: 2[-2mu - 3(-47)] - (-1)[5mu - 3(100)] + 1[5(-47) - (-2)(100)] = 0 2[-2mu + 141] + [5mu - 300] + [-235 + 200] = 0 -4mu + 282 + 5mu - 300 - 35 = 0 mu - 53 = 0 implies mu = 53 ### Step 2: Calculate the Target Value Substitute the values of mu and lambda into the expression: mu - 2lambda = 53 - 2(-2) = 53 + 4 = 57 ### Pattern Recognition When solving systems of equations for infinite solution parameters, choosing a directional determinant that excludes one of the variables simplifies the problem into two separate single-variable equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q74 jee_main_2025_24_jan_morning Properties of Matrices and Adjoints
Let A be a 3 times 3 matrix such that mathbfX^mathrmTmathbfAmathbfX = mathbf0 for all nonzero 3 times 1 matrices mathbfX = beginbmatrix x \\ y \\ z endbmatrix . If A beginbmatrix 1\\ 1\\ 1 endbmatrix = beginbmatrix 1\\ 4\\ -5 endbmatrix, Abeginbmatrix 1\\ 2\\ 1 endbmatrix = beginbmatrix 0\\ 4\\ -8 endbmatrix, and det(mathrmadj(2(A + I))) = 2^alpha3^beta5^gamma for alpha, beta, gamma in mathbbN, then alpha^2 + beta^2 + gamma^2 is ________.
Numerical Answer. Answer: 44

Solution

### Related Formula The quadratic form condition mathbfX^mathrmTmathbfAmathbfX = mathbf0 holds for all non-zero vectors mathbfX if and only if A is a skew-symmetric matrix. For a 3 times 3 skew-symmetric matrix, the components satisfy: A = beginbmatrix 0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0 endbmatrix ### Core Logic Let's define the matrix A using the parameters of a standard skew-symmetric form: A = beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix Apply the first given matrix multiplication vector condition: A beginbmatrix 1 \\ 1 \\ 1 endbmatrix = beginbmatrix a+b \\ -a+c \\ -b-c endbmatrix = beginbmatrix 1 \\ 4 \\ -5 endbmatrix This gives the system of linear equations: a + b = 1 quad dots (1) -a + c = 4 quad dots (2) -b - c = -5 implies b + c = 5 quad dots (3) Apply the second given matrix multiplication vector condition: A beginbmatrix 1 \\ 2 \\ 1 endbmatrix = beginbmatrix 2a+b \\ -a+c \\ -b-2c endbmatrix = beginbmatrix 0 \\ 4 \\ -8 endbmatrix This gives the equation: 2a + b = 0 quad dots (4) ### Step 1: Solve for Matrix Elements Subtract equation (1) from equation (4): (2a + b) - (a + b) = 0 - 1 implies a = -1 Substitute a = -1 back into equation (1): -1 + b = 1 implies b = 2 Substitute a = -1 into equation (2): -(-1) + c = 4 implies 1 + c = 4 implies c = 3 Thus, the explicit matrix A is: A = beginbmatrix 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 endbmatrix ### Step 2: Compute Target Matrix Determinant Construct the modified target matrix 2(A+I): A + I = beginbmatrix 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 endbmatrix 2(A + I) = beginbmatrix 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 endbmatrix Calculate its determinant value: det(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)] det(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120 ### Step 3: Analyze Adjoint Power and Prime Factors Using the standard determinant identity for adjoints, det(mathrmadj(M)) = (det M)^n-1 where n=3: det(mathrmadj(2(A+I))) = (120)^3-1 = 120^2 Find the prime factorization of the result: 120 = 2^3 cdot 3^1 cdot 5^1 implies 120^2 = (2^3 cdot 3^1 cdot 5^1)^2 = 2^6 cdot 3^2 cdot 5^2 This maps the exponents directly to our target variables: alpha = 6, quad beta = 2, quad gamma = 2 Finally, calculate the \sum of their squares: alpha^2 + beta^2 + gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44 ### Pattern Recognition The condition mathbfX^T A mathbfX = 0 always implies that A is a skew-symmetric matrix, which instantly forces the diagonal entries to be zero, reducing the number of unknown parameters from 9 down to 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q60 jee_main_2025_28_jan_evening Properties of Matrices and Powers
Let A=left[beginmatrixfrac1sqrt2&-2\\ 0&1endmatrixright] and P=left[beginmatrixcostheta&-sintheta\\ sintheta&costhetaendmatrixright], theta>0. If B=PAP^T, C=P^TB^10P and the sum of the diagonal elements of C is fracmn where textgcd(m,n)=1, then m+n is:
  • A. 65
  • B. 127
  • C. 258
  • D. 2049

Solution

### Related Formula For orthogonal matrix P, P^T P = P P^T = I. If B = PAP^T, then: B^k = (PAP^T)(PAP^T)dots(PAP^T) = PA^kP^T ### Core Logic Given C = P^T B^10 P. Substitute B^10 = P A^10 P^T into the expression: C = P^T (P A^10 P^T) P C = (P^T P) A^10 (P^T P) Since P is an orthogonal rotation matrix, P^T P = I, meaning: C = I cdot A^10 cdot I = A^10 Therefore, the sum of diagonal elements of C is simply the trace of A^10. ### Step 1: Analyze Powers of Upper Triangular Matrix A Matrix A is upper triangular: A = left[beginmatrixfrac1sqrt2&-2\\ 0&1endmatrixright] For any upper triangular matrix, any integer power k preserves the main diagonal entries as simply the powers of the individual diagonal elements: A^10 = left[beginmatrixleft(frac1sqrt2right)^10&*\\ 0&1^10endmatrixright] = left[beginmatrixfrac132&*\\ 0&1endmatrixright] ### Step 2: Calculate the Trace and sum m+n textSum of diagonal elements = textTrace(C) = textTrace(A^10) = frac132 + 1 = frac3332 Given fracmn = frac3332 with textgcd(33, 32) = 1: m = 33, quad n = 32 m + n = 33 + 32 = 65 ### Pattern Recognition Traces of matrices are invariant under cyclic permutations, so textTr(P^T B^10 P) = textTr(P P^T B^10) = textTr(B^10). Furthermore, textTr(P A^10 P^T) = textTr(A^10). This identity bypasses the need to evaluate any outer matrix multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q jee_main_2025_29_jan_morning Properties of Determinants
Let M and m respectively be the maximum and the minimum values of f (x) = left| beginarrayc c c 1 + sin^ 2 x & cos^ 2 x & 4 sin 4 x \\ sin^ 2 x & 1 + cos^ 2 x & 4 sin 4 x \\ sin^ 2 x & cos^ 2 x & 1 + 4 sin 4 x endarray right|, x in R Then mathbfM^4 -mathbfm^4 is equal to :
  • A. 1280
  • B. 1295
  • C. 1040
  • D. 1215

Solution

### Related Formula sin^2 x + cos^2 x = 1 -1 le sin 4x le 1 ### Core Logic Apply the row operations R_2 to R_2 - R_1 and R_3 to R_3 - R_1 to simplify the determinant: f(x) = left| beginarrayc c c 1 + sin^ 2 x & cos^ 2 x & 4 sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 endarray right| ### Step 1: Expand the Determinant Expanding along the first row: f(x) = (1 + sin^2 x)(1 - 0) - cos^2 x(-1 - 0) + 4sin 4x(0 - (-1)) f(x) = 1 + sin^2 x + cos^2 x + 4sin 4x Since sin^2 x + cos^2 x = 1, we get: f(x) = 2 + 4sin 4x ### Step 2: Find Maximum and Minimum Values The range of sin 4x is [-1, 1]. M = 2 + 4(1) = 6 m = 2 + 4(-1) = -2 ### Step 3: Calculate M^4 - m^4 M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280 ### Pattern Recognition Look for repeated structures or cyclic additions in rows. Subtracting rows quickly creates zeros, reducing complex trigonometric matrices into elementary algebraic expressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Trigonometric Functions

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