Let A be a 3 times 3 matrix such that A + A^T = O. If mathrmA left[ beginarrayl 1 \\ - 1 \\ 0 endarray right] = left[ beginarrayl 3 \\ 3 \\ 2 endarray right], mathrmA ^ 2 left[ beginarrayl 1 \\ - 1 \\ 0 endarray right] = left[ beginarrayl - 3 \\ 1 9 \\ - 2 4 endarray right] and det(textadj(2textadj(A + I))) = (2)^alpha. (3)^beta. (11)^gamma, alpha, beta, gamma are non-negative integers, then alpha + beta + gamma is equal to ____

Numerical Answer Type:
Enter a numerical value Answer: 18 to 18 +4 marks

Solution & Explanation

### Related Formula A = -A^T implies A text is a skew-symmetric matrix. |textadj(kA)| = k^n(n-1)|A|^(n-1) |textadj(M)| = |M|^n-1 ### Core Logic Since A is a 3 times 3 skew-symmetric matrix, let A = beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix. Given A beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix 3 \\ 3 \\ 2 endbmatrix: beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix -a \\ -a \\ -b+c endbmatrix Wait, (-a)(1) + (0)(-1) + (c)(0) = -a. But wait, -b(1) - c(-1) + 0 = -b+c. Equating the result: -a = 3 implies a = -3 -b + c = 2 quad dots(1) ### Step 1: Finding Matrix A Using Second Given Condition Given A^2 beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix. We know A beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix 3 \\ 3 \\ 2 endbmatrix. Therefore, A beginbmatrix 3 \\ 3 \\ 2 endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix. beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix beginbmatrix 3 \\ 3 \\ 2 endbmatrix = beginbmatrix 3a + 2b \\ -3a + 2c \\ -3b - 3c endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix Substitute a = -3: 3(-3) + 2b = -3 implies -9 + 2b = -3 implies 2b = 6 implies b = 3 Use equation (1): -b + c = 2 implies -3 + c = 2 implies c = 5. So, a = -3, b = 3, c = 5. Matrix A = beginbmatrix 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 endbmatrix. ### Step 2: Calculate Determinant of A + I A + I = beginbmatrix 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 endbmatrix |A + I| = 1(1 - (-25)) - (-3)(3 - (-15)) + 3(-15 - (-3)) = 1(26) + 3(18) + 3(-12) = 26 + 54 - 36 = 44 ### Step 3: Calculating Adjoint Target Expression Target: det(textadj(2textadj(A + I))). Let M = A+I. For a 3 times 3 matrix, |textadj(X)| = |X|^2. det(textadj(2textadj M)) = |2textadj M|^2 We know |kY| = k^3|Y| for 3 times 3. So |2textadj M| = 2^3 |textadj M| = 8|M|^2. (8|M|^2)^2 = 64|M|^4 Substitute |M| = 44: 64(44)^4 = 2^6 cdot (4 cdot 11)^4 = 2^6 cdot (2^2 cdot 11)^4 = 2^6 cdot 2^8 cdot 11^4 = 2^14 cdot 3^0 cdot 11^4 Comparing to 2^alpha cdot 3^beta cdot 11^gamma: alpha = 14, beta = 0, gamma = 4. alpha + beta + gamma = 14 + 0 + 4 = 18. ### Pattern Recognition Skew-symmetric matrices only have 3 unknowns. Matrix multiplication acts linearly; A^2 x = A(Ax) provides immediate simultaneous equations without having to compute the heavy A^2 explicitly. Determinant adjoint loops always reduce to |A|^(n-1)^k modulated by constant pullouts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: 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
Q14 jee_main_2026_21_jan_evening Matrix Algebra
For the matrices A=beginbmatrix3 & -4 \\ 1 & -1endbmatrix and B=beginbmatrix-29 & 49 \\ -13 & 18endbmatrix, if (A^15+B)beginbmatrixx\\ yendbmatrix=beginbmatrix0\\ 0endbmatrix, then among the following which one is true?
  • A. x = 5, y = 7
  • B. x = 18, y = 11
  • C. x = 11, y = 2
  • D. x = 16, y = 3

Solution

### Related Formula A = I + N implies A^n = (I+N)^n = I + nN (provided N^2 = 0, meaning N is nilpotent of order 2). ### Core Logic Evaluate powers of matrix A to detect a pattern. A = beginbmatrix3 & -4 \\ 1 & -1endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix2 & -4 \\ 1 & -2endbmatrix Let N = beginbmatrix2 & -4 \\ 1 & -2endbmatrix. Check N^2: N^2 = beginbmatrix2 & -4 \\ 1 & -2endbmatrixbeginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix4-4 & -8+8 \\ 2-2 & -4+4endbmatrix = beginbmatrix0 & 0 \\ 0 & 0endbmatrix ### Step 1: Compute A^15 Since N^2 = 0, A^n = I + nN. A^15 = I + 15N = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + 15beginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix30 & -60 \\ 15 & -30endbmatrix A^15 = beginbmatrix31 & -60 \\ 15 & -29endbmatrix ### Step 2: Evaluate Matrix Operation A^15 + B = beginbmatrix31 & -60 \\ 15 & -29endbmatrix + beginbmatrix-29 & 49 \\ -13 & 18endbmatrix = beginbmatrix2 & -11 \\ 2 & -11endbmatrix Given (A^15 + B)beginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix: beginbmatrix2 & -11 \\ 2 & -11endbmatrixbeginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix This yields a single independent equation: 2x - 11y = 0 implies 2x = 11y Checking the options, for x=11, y=2, 2(11) = 11(2) implies 22 = 22. Thus, Option 3 satisfies the condition. ### Pattern Recognition Whenever you need a high power of a 2times 2 matrix, check its trace and determinant. If |A-lambda I|=0 yields lambda=1,1, then A can be written as I+N where N is nilpotent. This bypasses lengthy matrix multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants
Q18 jee_main_2026_21_jan_evening System of Linear Equations
If the system of equations 3x + y + 4z = 3 2x + alpha y - z = -3 x + 2y + z = 4 has no solution, then the value of alpha is equal to:
  • A. 19
  • B. 4
  • C. 13
  • D. 23

Solution

### Related Formula textFor no solution in AX = B, text we must have |A| = 0 text (i.e. Delta = 0 text) text and at least one of Delta_x, Delta_y, Delta_z neq 0. ### Core Logic Evaluate the main determinant Delta to 0 to find alpha. Then quickly verify Delta_x neq 0 to confirm it produces no solution. ### Step 1: Calculate Delta Delta = beginvmatrix 3 & 1 & 4 \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0 Expand along the first row: 3(alpha + 2) - 1(2 - (-1)) + 4(4 - alpha) = 0 3alpha + 6 - 3 + 16 - 4alpha = 0 19 - alpha = 0 implies alpha = 19 ### Step 2: Verification of Delta_x For alpha = 19: Delta_x = beginvmatrix 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 endvmatrix Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76) Delta_x = 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 neq 0 Since Delta = 0 and Delta_x neq 0, the system has no solution for alpha = 19. ### Pattern Recognition Unless the problem specifically asks to check between infinite and no solution, setting the coefficient determinant Delta = 0 directly yields the unique required parameter value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants
Q13 jee_main_2026_22_january_morning Properties of Determinants
If A=beginbmatrix2 & 3 \\ 3 & 5endbmatrix, then the determinant of the matrix (A^2025-3A^2024+A^2023) is
  • A. 28
  • B. 12
  • C. 24
  • D. 16

Solution

### Related Formula |AB| = |A||B| |A^n| = |A|^n ### Core Logic Given A = beginbmatrix2 & 3 \\ 3 & 5endbmatrix. The determinant |A| = (2)(5) - (3)(3) = 10 - 9 = 1. We need to find the determinant of X = A^2025 - 3A^2024 + A^2023. Factor out the lowest power of A: X = A^2023(A^2 - 3A + I) ### Step 1: Determinant Factoring Using properties of determinants: |X| = |A^2023| cdot |A^2 - 3A + I| Since |A| = 1, |A|^2023 = 1. Therefore, |X| = |A^2 - 3A + I|. ### Step 2: Matrix Computation Calculate A^2: A^2 = beginbmatrix2 & 3 \\ 3 & 5endbmatrix beginbmatrix2 & 3 \\ 3 & 5endbmatrix = beginbmatrix4+9 & 6+15 \\ 6+15 & 9+25endbmatrix = beginbmatrix13 & 21 \\ 21 & 34endbmatrix Calculate -3A: -3A = beginbmatrix-6 & -9 \\ -9 & -15endbmatrix Assemble A^2 - 3A + I: A^2 - 3A + I = beginbmatrix13 & 21 \\ 21 & 34endbmatrix + beginbmatrix-6 & -9 \\ -9 & -15endbmatrix + beginbmatrix1 & 0 \\ 0 & 1endbmatrix = beginbmatrix13 - 6 + 1 & 21 - 9 + 0 \\ 21 - 9 + 0 & 34 - 15 + 1endbmatrix = beginbmatrix8 & 12 \\ 12 & 20endbmatrix ### Step 3: Final Determinant |A^2 - 3A + I| = beginvmatrix8 & 12 \\ 12 & 20endvmatrix = (8)(20) - (12)(12) = 160 - 144 = 16 ### Pattern Recognition High-powered matrix polynomials simplify radically by factoring out the lowest power. Since |A| = 1, the leading coefficient collapses completely, leaving only a basic A^2 evaluation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants

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