Let alpha, beta ( alpha neq beta ) be the values of m, for which the equations x + y + z = 1 ; x + 2y + 4z = m and x + 4y + 10z = m^2 have infinitely many solutions. Then the value of sum_n=1^10 (n^alpha + n^beta) is equal to:

Solution & Explanation

### Related Formula Cramer's rule for infinite solutions in a 3 variable system requires: Delta = Delta_x = Delta_y = Delta_z = 0 ### Core Logic Set up the primary matrix determinant Delta: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 endvmatrix = 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0 Since Delta = 0 is true independent of m, analyze secondary delta constraints to maintain consistency for infinite paths. ### Step 1: Compute Dependent Variable Constraints Evaluate Delta_x = 0: Delta_x = beginvmatrix 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 endvmatrix = 0 1(20 - 16) - 1(10m - 4m^2) + 1(4m - 2m^2) = 0 4 - 10m + 4m^2 + 4m - 2m^2 = 0 2m^2 - 6m + 4 = 0 implies m^2 - 3m + 2 = 0 Thus, m = 1, 2, which gives alpha = 1, beta = 2. ### Step 2: Calculate Sigma Expression sum_n=1^10 (n^1 + n^2) = sum_n=1^10 n + sum_n=1^10 n^2 = frac10(11)2 + frac10(11)(21)6 = 55 + 385 = 440 ### Pattern Recognition When infinitely many solutions are required, solving the determinant created by replacing one column with the constant vector provides parameter roots directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

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More Matrices and Determinants Previous-Year Questions — Page 9

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
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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