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

Solution & Explanation

### 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

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Q24 jee_main_2026_21_jan_morning Cayley-Hamilton Theorem Applications
For some alpha, beta in mathbbR , let A = beginbmatrix alpha & 2 \\ 1 & 2 endbmatrix and B = beginbmatrix 1 & 1 \\ 1 & beta endbmatrix be such that A^2 - 4A + 2I = B^2 - 3B + I = O . Then (det (operatornameadj(A^3 - B^3)))^2 is equal to ....
Numerical Answer. Answer: 225 to 225

Solution

### Related Formula For a 2 times 2 matrix M, Cayley-Hamilton equation states: M^2 - operatornameTr(M)M + operatornamedet(M)I = O Also, |operatornameadj(M)| = |M|^n-1, and for a 2 times 2 matrix, |operatornameadj(M)| = |M|. ### Core Logic Using the characteristic equation A^2 - operatornameTr(A)A + det(A)I = O: Comparing with A^2 - 4A + 2I = O: operatornameTr(A) = 4 Rightarrow alpha + 2 = 4 Rightarrow alpha = 2 Comparing with B^2 - 3B + I = O: operatornameTr(B) = 3 Rightarrow 1 + beta = 3 Rightarrow beta = 2 ### Step 1: Compute A^3 and B^3 via reduction A = beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix A^2 = 4A - 2I A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I A^3 = 14 beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix - 8 beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix = beginbmatrix 28 & 28 \\ 14 & 28 endbmatrix - beginbmatrix 8 & 0 \\ 0 & 8 endbmatrix = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix Similarly, B = beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix B^2 = 3B - I B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I B^3 = 8 beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 8 & 8 \\ 8 & 16 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix ### Step 2: Difference and Determinant A^3 - B^3 = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix - beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix = beginbmatrix 15 & 20 \\ 6 & 7 endbmatrix det(A^3 - B^3) = (15 times 7) - (20 times 6) = 105 - 120 = -15 ### Step 3: Final Answer For a 2 times 2 matrix, |operatornameadj(M)| = |M|. Thus, det(operatornameadj(A^3 - B^3)) = -15. (det(operatornameadj(A^3 - B^3)))^2 = (-15)^2 = 225 ### Pattern Recognition Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M^3 down to a linear combination cM + dI. Expanding this requires only basic scalar arithmetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices
Q58 jee_main_2025_02_april_evening System of Linear Equations
If the system of equations beginaligned 2x + lambda y + 3z &= 5 \\ 3x + 2y - z &= 7 \\ 4x + 5y + mu z &= 9 endaligned has infinitely many solutions, then (lambda^2 + mu^2) is equal to:
  • A. 22
  • B. 18
  • C. 26
  • D. 30

Solution

### Related Formula textFor infinitely many solutions: Delta = 0 quad textand quad Delta_i = 0 ### Core Logic For a system of 3 linear equations to have infinitely many solutions, the determinant of coefficients and all Cramer determinants must equal zero. ### Step 1: Set up determinant equations The determinant of coefficients is: Delta = beginvmatrix 2 & lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & mu endvmatrix = 0 2(2mu + 5) - lambda(3mu + 4) + 3(15 - 8) = 0 4mu + 10 - 3lambdamu - 4lambda + 21 = 0 4mu - 3lambdamu - 4lambda + 31 = 0 quad text--- (1) Now, set Delta_3 = 0: Delta_3 = beginvmatrix 2 & lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 endvmatrix = 0 2(18 - 35) - lambda(27 - 28) + 5(15 - 8) = 0 -34 + lambda + 35 = 0 implies lambda = -1 ### Step 2: Solve for mu and compute the sum of squares Substitute lambda = -1 into equation (1): 4mu - 3(-1)mu - 4(-1) + 31 = 0 4mu + 3mu + 4 + 31 = 0 7mu = -35 implies mu = -5 Now calculate the sum of squares: lambda^2 + mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 ### Pattern Recognition Whenever you need to solve for two variables in Cramer's theorem, identifying which determinant lacks the complex variable (like Delta_3 which lacks mu) is the fastest way to solve for one variable independently. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q70 jee_main_2025_02_april_evening Properties of Matrices
Let A be a 3 times 3 real matrix such that mathrmA^2 (mathrmA - 2mathrmI) - 4(mathrmA - mathrmI) = mathrmO, where I and O are the identity and null matrices, respectively. If mathrmA^5 = alpha mathrmA^2 + beta mathrmA + gamma mathrmI, where alpha, beta and gamma are real constants, then alpha + beta + gamma is equal to:
  • A. 12
  • B. 20
  • C. 76
  • D. 4

Solution

### Related Formula textCharacteristic equation reduction: mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI ### Core Logic We use the given cubic matrix equation recursively to express the fifth power of matrix A solely in terms of quadratic and linear terms. ### Step 1: Simplify the cubic matrix equation The given equation is: mathrmA^2 (mathrmA - 2mathrmI) - 4(mathrmA - mathrmI) = mathrmO mathrmA^3 - 2mathrmA^2 - 4mathrmA + 4mathrmI = mathrmO implies mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI Multiply by matrix A to find the fourth power: mathrmA^4 = 2mathrmA^3 + 4mathrmA^2 - 4mathrmA ### Step 2: Reduce the fourth power term Substitute the expression for mathrmA^3 into our formula for mathrmA^4: mathrmA^4 = 2left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 4mathrmA mathrmA^4 = 4mathrmA^2 + 8mathrmA - 8mathrmI + 4mathrmA^2 - 4mathrmA = 8mathrmA^2 + 4mathrmA - 8mathrmI Multiply by matrix A to find the fifth power: mathrmA^5 = 8mathrmA^3 + 4mathrmA^2 - 8mathrmA ### Step 3: Reduce the fifth power term and solve Substitute the expression for mathrmA^3 again: mathrmA^5 = 8left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 8mathrmA mathrmA^5 = 16mathrmA^2 + 32mathrmA - 32mathrmI + 4mathrmA^2 - 8mathrmA = 20mathrmA^2 + 24mathrmA - 32mathrmI Comparing this with mathrmA^5 = alpha mathrmA^2 + beta mathrmA + gamma mathrmI, we find: - alpha = 20 - beta = 24 - gamma = -32 Sum the coefficients: alpha + beta + gamma = 20 + 24 - 32 = 12 ### Pattern Recognition Cayley-Hamilton reduction: For any polynomial equation of a matrix, higher powers A^k can always be reduced down to polynomials of order less than the degree of the characteristic equation by recursive substitution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q jee_main_2025_02_april_morning Idempotent Matrices
Let A = beginbmatrix alpha & -1 \\ 6 & beta endbmatrix, alpha > 0, such that det(A) = 0 and alpha + beta = 1. If I denotes the 2 times 2 identity matrix, then the matrix (I + A)^8 is:
  • A. beginbmatrix 4 & -1 \\ 6 & -1 endbmatrix
  • B. beginbmatrix 257 & -64 \\ 514 & -127 endbmatrix
  • C. beginbmatrix 1025 & -511 \\ 2024 & -1024 endbmatrix
  • D. beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix

Solution

### Related Formula For a matrix satisfying A^2 = A (Idempotent Matrix): (I+A)^n = I + (2^n - 1)A ### Core Logic Given det(A) = alphabeta + 6 = 0 implies alphabeta = -6 and alpha + beta = 1. Solving these gives alpha = 3, beta = -2 (since alpha > 0). ### Step 1: Check Powers of A Substitute values into A: A = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix Compute A^2: A^2 = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 9-6 & -3+2 \\ 18-12 & -6+4 endbmatrix = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = A ### Step 2: Expand Matrix Expression Since A^2 = A, it follows that A^n = A for all integers n ge 1. (I + A)^8 = I + sum_k=1^8 binom8k A^k = I + A sum_k=1^8 binom8k = I + (2^8 - 1)A = I + 255A ### Step 3: Construct the Final Matrix (I + A)^8 = beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix + 255 beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 1 + 765 & -255 \\ 1530 & 1 - 510 endbmatrix = beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix ### Pattern Recognition Whenever texttr(A) = 1 and det(A) = 0 for a 2 times 2 matrix, Cayley-Hamilton theorem gives A^2 - texttr(A)A + det(A)I = 0 implies A^2 = A. Thus A is idempotent, simplifying polynomial expansions exponentially. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants
Q jee_main_2025_02_april_morning System of Linear Equations
If the system of linear equations 3x + y + beta z = 3 2x + alpha y - z = -3 x + 2y + z = 4 has infinitely many solutions, then the value of 22beta - 9alpha is:
  • A. 49
  • B. 31
  • C. 43
  • D. 37

Solution

### Related Formula Cramer's Rule for infinite solutions specifies that the main determinant and all component determinants must vanish: Delta = 0 quad textand quad Delta_1 = Delta_2 = Delta_3 = 0 ### Core Logic Set the key system determinants to zero to form equations linking alpha and beta, then isolate the constants. ### Step 1: Set Main Determinant to Zero Delta = beginvmatrix 3 & 1 & beta \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0 Expand along the first row: 3(alpha + 2) - 1(2 + 1) + beta(4 - alpha) = 0 3alpha + 6 - 3 + 4beta - alphabeta = 0 implies 3alpha + 4beta - alphabeta + 3 = 0 quad dots (1) ### Step 2: Set Subsidiary Determinant to Zero Using Delta_3 = 0 by substituting the constants vector into the third column: Delta_3 = beginvmatrix 3 & 1 & 3 \\ 2 & alpha & -3 \\ 1 & 2 & 4 endvmatrix = 0 Expand along the first row: 3(4alpha + 6) - 1(8 + 3) + 3(4 - alpha) = 0 12alpha + 18 - 11 + 12 - 3alpha = 0 implies 9alpha + 19 = 0 implies alpha = -frac199 ### Step 3: Solve for Beta and Final Expression Substitute alpha = -frac199 into equation (1): 3left(-frac199right) + 4beta - left(-frac199right)beta + 3 = 0 -frac193 + 3 + betaleft(4 + frac199right) = 0 implies -frac103 + betaleft(frac559right) = 0 frac559beta = frac103 implies beta = frac103 cdot frac955 = frac611 Now compute 22beta - 9alpha: 22left(frac611right) - 9left(-frac199 ight) = 12 + 19 = 31 ### Pattern Recognition Choosing Delta_3 over Delta_1 or Delta_2 eliminates beta entirely because the variable parameters are localized in specific positions. This yields alpha directly without requiring a coupled system solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

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