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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 8

Q1 jee_main_2024_30_january_evening System of Linear Equations
Consider the system of linear equations x + y + z = 5, x + 2 y + lambda^ 2 z = 9, x + 3y + lambda z = mu , where lambda ,mu in mathbbR . Then, which of the following statement is NOT correct?
  • A. textSystem has infinite number of solution if lambda = 1 text and mu = 13
  • B. textSystem is inconsistent if lambda = 1 text and mu neq 13
  • C. textSystem is consistent if lambda neq 1 text and mu = 13
  • D. textSystem has unique solution if lambda neq 1 text and mu neq 13

Solution

### Related Formula Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix ### Core Logic For the given system, we find the determinant of the coefficient matrix: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & lambda^2 \\ 1 & 3 & lambda endvmatrix Expanding along the first row: Delta = 1(2lambda - 3lambda^2) - 1(lambda - lambda^2) + 1(3 - 2) Delta = 2lambda - 3lambda^2 - lambda + lambda^2 + 1 Delta = -2lambda^2 + lambda + 1 = 0 Rightarrow 2lambda^2 - lambda - 1 = 0 Rightarrow (2lambda + 1)(lambda - 1) = 0 Rightarrow lambda = 1, -frac12 ### Step 1: Analyzing Cases For infinite solutions or inconsistency, we check Delta_z (or others): Delta_z = beginvmatrix 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & mu endvmatrix = 0 Evaluating the determinant: 1(2mu - 27) - 1(mu - 9) + 5(3 - 2) = 0 2mu - 27 - mu + 9 + 5 = 0 Rightarrow mu - 13 = 0 Rightarrow mu = 13 Thus, if lambda = 1 and mu = 13, the system has an infinite number of solutions. If lambda = 1 and mu neq 13, the system is inconsistent. ### Step 2: Checking Options Option 1 is correct (True statement). Option 2 is correct (True statement). Option 3: If lambda neq 1 and lambda neq -frac12, the system has a unique solution (which is consistent) regardless of mu. If lambda = -frac12 and mu = 13, it may have infinite solutions or be inconsistent, but typically it is consistent for unique solutions if lambda neq 1 and lambda neq -1/2. Wait, the question option 4 says "System has unique solution if lambda neq 1 and mu neq 13". This is NOT correct because if lambda = -frac12 and mu neq 13, the determinant is zero and it does not have a unique solution. ### Pattern Recognition Look for the roots of Delta = 0. Since there are two roots (lambda=1, -1/2), a blanket statement about "unique solution if lambda neq 1" ignores the other root and is inherently false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants
Q20 jee_main_2024_30_january_evening Properties of Matrices
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:
  • A. textBoth (I) and (II) are true
  • B. textNeither (I) nor (II) is true.
  • C. textOnly (II) is true.
  • D. textOnly (I) is true.

Solution

### 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
Q17 jee_main_2024_30_jan_morning Properties of Determinants
If f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ 3 + 2 cos^4 x & 2 sin^4 x & sin^2 2x \\ 2 cos^4 x & 3 + 2 sin^4 x & sin^2 2x endvmatrix then frac15 f'(0) is equal to
  • A. 0
  • B. 1
  • C. 2
  • D. 6

Solution

### Related Formula Row Operations in Determinants: R_i to R_i + kR_j text does not change the determinant. ### Core Logic Apply row operations to simplify the determinant: R_2 to R_2 - R_1 and R_3 to R_3 - R_1 f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ (3 + 2cos^4 x) - 2cos^4 x & 2sin^4 x - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) \\ 2cos^4 x - 2cos^4 x & (3 + 2sin^4 x) - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) endvmatrix ### Step 1: Expanding the simplified determinant f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 endvmatrix Expand along the 2nd row: f(x) = -3 [ 2sin^4 x(-3) - 3(3 + sin^2 2x) ] + (-3) [ 2cos^4 x(3) - 0 ] Wait, let's expand properly: f(x) = -3 beginvmatrix 2sin^4 x & 3 + sin^2 2x \\ 3 & -3 endvmatrix - (-3) beginvmatrix 2cos^4 x & 2sin^4 x \\ 0 & 3 endvmatrix f(x) = -3 [ -6sin^4 x - 9 - 3sin^2 2x ] + 3 [ 6cos^4 x ] f(x) = 18sin^4 x + 27 + 9sin^2 2x + 18cos^4 x f(x) = 18(sin^4 x + cos^4 x) + 9(4sin^2 x cos^2 x) + 27 f(x) = 18(sin^2 x + cos^2 x)^2 - 36sin^2 x cos^2 x + 36sin^2 x cos^2 x + 27 f(x) = 18(1) + 27 = 45 ### Step 2: Differentiating constant Since f(x) = 45 is a constant function, f'(x) = 0 Therefore, frac15 f'(0) = 0. ### Pattern Recognition If row subtractions reduce variable rows to pure constants, the entire determinant expansion often resolves into a pure trigonometric identity producing a global constant, immediately rendering its derivative 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants Class 12 Maths: Continuity and Differentiability
Q19 jee_main_2024_31_jan_evening System of Linear Equations
Let A be a 3times3 real matrix such that Abeginbmatrix 1 \\ 0 \\ 1 endbmatrix=2beginbmatrix 1 \\ 0 \\ 1 endbmatrix, Abeginbmatrix -1 \\ 0 \\ 1 endbmatrix=4beginbmatrix -1 \\ 0 \\ 1 endbmatrix, Abeginbmatrix 0 \\ 1 \\ 0 endbmatrix=2beginbmatrix 0 \\ 1 \\ 0 endbmatrix. Then, the system (A-3I)beginbmatrix x \\ y \\ z endbmatrix=beginbmatrix 1 \\ 2 \\ 3 endbmatrix has
  • A. textunique solution
  • B. textexactly two solutions
  • C. textno solution
  • D. textinfinitely many solutions

Solution

### Core Logic Let A = beginbmatrix x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 endbmatrix. Using the given column transformations: 1) Abeginbmatrix 1 \\ 0 \\ 1 endbmatrix = beginbmatrix 2 \\ 0 \\ 2 endbmatrix implies x_1+z_1=2,\; x_2+z_2=0,\; x_3+z_3=2. 2) Abeginbmatrix -1 \\ 0 \\ 1 endbmatrix = beginbmatrix -4 \\ 0 \\ 4 endbmatrix implies -x_1+z_1=-4,\; -x_2+z_2=0,\; -x_3+z_3=4. 3) Abeginbmatrix 0 \\ 1 \\ 0 endbmatrix = beginbmatrix 0 \\ 2 \\ 0 endbmatrix implies y_1=0,\; y_2=2,\; y_3=0. Solving pairs from 1) and 2): x_1+z_1=2 and -x_1+z_1=-4 implies 2z_1=-2 implies z_1=-1, x_1=3. x_2+z_2=0 and -x_2+z_2=0 implies x_2=0, z_2=0. x_3+z_3=2 and -x_3+z_3=4 implies 2z_3=6 implies z_3=3, x_3=-1. So, A = beginbmatrix 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 endbmatrix. Now, compute A-3I: A-3I = beginbmatrix 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 endbmatrix. Determinant |A-3I| = -1(-1)(-1) = 1 neq 0. Since the determinant is non-zero, the system (A-3I)X = B has a unique solution. ### Pattern Recognition Whenever independent eigenvectors are given, the matrix can be uniquely constructed. If det != 0 for (A-3I), uniqueness is guaranteed unconditionally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices

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