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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

Determinants

More Determinants Previous-Year Questions — Page 7

Q22 jee_main_2024_29_january_evening System of Linear Equations
Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax + by + c = 0 be concurrent at the point P and Q (alpha, beta) be a point such that the system of equations x + y + z = 6, 2 x + 5 y + alpha z = beta and x + 2y + 3z = 4 has infinitely many solutions. Then (PQ)^2 is equal to
Numerical Answer. Answer: 113 to 113

Solution

### Related Formula For infinite solutions of a system of equations, the main determinant D and auxiliary determinants D_1, D_2, D_3 must all equal 0. ### Core Logic Since a, b, c are in A.P., we have 2b = a + c implies a - 2b + c = 0. Comparing this identity with the line equation ax + by + c = 0, we immediately find that the lines always pass through the fixed point configuration (1, -2). Thus, P = (1, -2). ### Step 1: Evaluation of the Matrix for Infinite Solutions Let us set the system determinant D = 0: D = beginbmatrix 1 & 1 & 1 \\ 2 & 5 & alpha \\ 1 & 2 & 3 endbmatrix = 0 1(15 - 2alpha) - 1(6 - alpha) + 1(4 - 5) = 0 15 - 2alpha - 6 + alpha - 1 = 0 implies 8 - alpha = 0 implies alpha = 8 Now set D_1 = 0 by substituting columns: D_1 = beginbmatrix 6 & 1 & 1 \\ beta & 5 & 8 \\ 4 & 2 & 3 endbmatrix = 0 6(15 - 16) - 1(3beta - 32) + 1(2beta - 20) = 0 -6 - 3beta + 32 + 2beta - 20 = 0 implies 6 - beta = 0 implies beta = 6 Thus, Q = (8, 6). ### Step 2: Distance Metric Calculation Using the distance formula between P(1, -2) and Q(8, 6): (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 7^2 + 8^2 = 49 + 64 = 113 ### Pattern Recognition A.P. coefficients inside a standard linear equation reveal a fixed point of concurrency by mapping matching coefficient components (1, -2, 1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines
Q19 jee_main_2024_27_jan_morning Matrix Multiplication
Consider the matrix f(x) = beginbmatrix cos x & -sin x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 endbmatrix Given below are two statements: Statement I: f(-x) is the inverse of the matrix f(x) Statement II: f(x)f(y) = f(x+y). In the light of the above statements, choose the correct answer from the options given below
  • A. textStatement I is false but Statement II is true
  • B. textBoth Statement I and Statement II are false
  • C. textStatement I is true but Statement II is false
  • D. textBoth Statement I and Statement II are true

Solution

### Related Formula cos(-x) = cos x sin(-x) = -sin x sin(x+y) = sin x cos y + cos x sin y cos(x+y) = cos x cos y - sin x sin y ### Core Logic Evaluate f(-x): f(-x) = beginbmatrix cos(-x) & -sin(-x) & 0 \\ sin(-x) & cos(-x) & 0 \\ 0 & 0 & 1 endbmatrix = beginbmatrix cos x & sin x & 0 \\ -sin x & cos x & 0 \\ 0 & 0 & 1 endbmatrix Checking if f(-x) is the inverse by evaluating f(x) cdot f(-x): f(x) f(-x) = beginbmatrix cos^2 x + sin^2 x & cos x sin x - sin x cos x & 0 \\ sin x cos x - cos x sin x & sin^2 x + cos^2 x & 0 \\ 0 & 0 & 1 endbmatrix f(x) f(-x) = beginbmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 endbmatrix = I Thus, Statement I is true. ### Step 1: Checking Statement II Evaluate the matrix multiplication f(x) cdot f(y): f(x) f(y) = beginbmatrix cos x & -sin x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 endbmatrix beginbmatrix cos y & -sin y & 0 \\ sin y & cos y & 0 \\ 0 & 0 & 1 endbmatrix = beginbmatrix cos x cos y - sin x sin y & -cos x sin y - sin x cos y & 0 \\ sin x cos y + cos x sin y & -sin x sin y + cos x cos y & 0 \\ 0 & 0 & 1 endbmatrix Apply standard trigonometric compound angle formulas: = beginbmatrix cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 endbmatrix = f(x+y) Thus, Statement II is also true. ### Step 2: Final Conclusion Both Statement I and Statement II are true. ### Pattern Recognition This specific matrix represents a standard 2D rotation matrix embedded in 3D space. Rotation matrices naturally follow R(x)R(y) = R(x+y) (additive property of angles) and their inverse is always obtained by negating the angle R(-x) = R(x)^-1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices Class 11 Maths: Trigonometric Functions
Q29 jee_main_2024_27_jan_morning Matrix Inverse and Determinant
Let A=beginbmatrix 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 endbmatrix, B=[B_1, B_2, B_3], where B_1, B_2, B_3 are column matrices, and AB_1=beginbmatrix 1 \\ 0 \\ 0 endbmatrix, AB_2=beginbmatrix 2 \\ 3 \\ 0 endbmatrix, AB_3=beginbmatrix 3 \\ 2 \\ 1 endbmatrix. If alpha=|B| and beta is the sum of all the diagonal elements of B, then alpha^3+beta^3 is equal to:
Numerical Answer. Answer: 28 to 28

Solution

### Related Formula |AB| = |A||B| Trace(B) = sum b_ii ### Core Logic Since B = [B_1, B_2, B_3], the matrix multiplication AB effectively applies A to each column of B: AB = [AB_1, AB_2, AB_3] = beginbmatrix 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 endbmatrix Let this target matrix be C. We know AB = C. Thus, taking determinants on both sides: |A| |B| = |C| ### Step 1: Finding Determinants Calculate determinant of A: |A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1 Calculate determinant of C. Since C is an upper triangular matrix, its determinant is simply the product of its main diagonal: |C| = 1 times 3 times 1 = 3 Therefore, 1 times |B| = 3 Rightarrow |B| = 3. So alpha = 3. ### Step 2: Finding Matrix B To find beta (the trace of B), we must explicitly find B = A^-1C. Rather than finding the full inverse, manually solve A B_i = C_i: For B_1 = [x, y, z]^T: 2x+z=1, x+y=0, x+z=0 Rightarrow x=1, z=-1, y=-1. Thus B_1 = [1, -1, -1]^T. For B_2: 2x+z=2, x+y=3, x+z=0 Rightarrow x=2, z=-2, y=1. Thus B_2 = [2, 1, -2]^T. For B_3: 2x+z=3, x+y=2, x+z=1 Rightarrow x=2, z=-1, y=0. Thus B_3 = [3, 0, -1]^T. ### Step 3: Calculating Trace and Final Output Constructing Matrix B: B = beginbmatrix 1 & 2 & 3 \\ -1 & 1 & 0 \\ -1 & -2 & -1 endbmatrix The diagonal elements are 1, 1, -1. Trace beta = 1 + 1 - 1 = 1. Finally compute alpha^3 + beta^3: 3^3 + 1^3 = 27 + 1 = 28 ### Pattern Recognition When given AX_i = Y_i for multiple columns, they collectively form A X = Y. Using |A||X| = |Y| bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices
Q15 jee_main_2024_29_jan_morning Properties of Determinants
Let A=beginbmatrix1&0&0\\ 0&alpha&beta\\ 0&beta&alphaendbmatrix and |2A|^3=2^21 where alpha, betain Z, Then a value of alpha is
  • A. 3
  • B. 5
  • C. 17
  • D. 9

Solution

### Related Formula |kA| = k^n |A| Where A is an n times n matrix, and k is a scalar. ### Core Logic Find the determinant of the 3 times 3 matrix A: |A| = 1(alpha cdot alpha - beta cdot beta) - 0 + 0 |A| = alpha^2 - beta^2 Given the condition |2A|^3 = 2^21. Since A is a 3 times 3 matrix, applying the scalar property |kA| = k^3|A|: |2A| = 2^3|A| = 8|A| Substitute this back into the original condition: (2^3|A|)^3 = 2^21 2^9 |A|^3 = 2^21 |A|^3 = frac2^212^9 = 2^12 Taking the cube root of both sides: |A| = 2^4 = 16 ### Step 1: Solve the Diophantine Equation We have: alpha^2 - beta^2 = 16 (alpha - beta)(alpha + beta) = 16 Since alpha and beta are integers, their sum and difference must also be integers. Also, (alpha + beta) and (alpha - beta) must share the same parity (both even or both odd) because their sum is 2alpha (an even number). Since their product is 16, the only valid integer factor pairs of 16 that share the same parity are (8, 2) and (-8, -2) and (4, 4) and (-4, -4). Case 1: (alpha + beta) = 8 and (alpha - beta) = 2 Adding them gives 2alpha = 10 Rightarrow alpha = 5. Thus beta = 3. Case 2: (alpha + beta) = 4 and (alpha - beta) = 4 Adding them gives 2alpha = 8 Rightarrow alpha = 4. Thus beta = 0. Looking at the options provided (3, 5, 17, 9), the value alpha = 5 is listed. ### Pattern Recognition Extracting scalar multipliers from determinants always depends on the dimension n of the matrix. For Diophantine equations like x^2 - y^2 = k, factoring into (x-y)(x+y) and analyzing parity constraints restricts the solution space instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Determinants
Q20 jee_main_2024_29_jan_morning Matrix Multiplication and Transpose
Let A be a square matrix such that AA^T=I. Then frac12A[(A+A^T)^2+(A-A^T)^2] is equal to
  • A. A^2+I
  • B. A^3+I
  • C. A^2+A^T
  • D. A^3+A^T

Solution

### Related Formula (X+Y)^2 = X^2 + XY + YX + Y^2 Because matrix multiplication is generally non-commutative, you cannot assume XY = YX unless explicitly proven. ### Core Logic Given A A^T = I. Since A is a square matrix, this implies A is orthogonal, so A^T A = I as well. Expand the inner binomials strictly respecting matrix order: (A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2 (A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2 Add both expanded expressions together. Notice that the cross terms perfectly cancel each other out: (A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2 = 2[A^2 + (A^T)^2] ### Step 1: Simplify Final Expression Substitute this simplified sum back into the original expression: frac12A left[ 2(A^2 + (A^T)^2) right] = A[A^2 + (A^T)^2] Distribute A from the left side: = A cdot A^2 + A cdot (A^T)^2 = A^3 + A A^T A^T Since we are given that A A^T = I, substitute I into the second term: = A^3 + I A^T = A^3 + A^T ### Pattern Recognition Identities like (X+Y)^2 + (X-Y)^2 = 2(X^2+Y^2) natively hold for matrices as well because the restrictive cross terms +XY+YX and -XY-YX cancel strictly linearly regardless of commutativity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices

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