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:

Solution & Explanation

### 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

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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
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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