### Related Formula The fundamental trigonometric limit is given by: lim_x to 0 fracsin xx = 1$\lim_{x \to 0} \frac{\sin x}{x} = 1$ ### Core Logic Apply the limit inside each element of the matrix determinant : lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{\vmatrix}$ ### Step 1: Simplify the Determinant via Row Operations Perform rows reductions R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ to create zeros: lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{\vmatrix}$ Expand across the first row : = (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$= (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$ = (a+1)(1) + 1 + b = a + b + 2$= (a+1)(1) + 1 + b = a + b + 2$ [cite: 3999, 4000] ### Step 2: Match Coefficients Equate this outcome with the target parameter template lambda + mu a + nu b$\lambda + \mu a + \nu b$ [cite: 3377, 4000]: lambda = 2, quad mu = 1, quad nu = 1$\lambda = 2, \quad \mu = 1, \quad \nu = 1$ Calculate (lambda + mu + nu)^2$(\lambda + \mu + \nu)^2$ [cite: 3378, 4001]: (2 + 1 + 1)^2 = 4^2 = 16$(2 + 1 + 1)^2 = 4^2 = 16$ ### Pattern Recognition Standard row manipulations on identity-shifted arrays quickly eliminate complex parameter symbols, reducing determinant calculations into straightforward polynomial expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Limits and Derivatives

### Related Formula According to Cramer's Rule, for a system of linear equations to have infinitely many solutions, the main determinant Delta$\Delta$ and all directional determinants Delta_1, Delta_2, Delta_3$\Delta_1, \Delta_2, \Delta_3$ must equal zero simultaneously. ### Core Logic Set up the directional determinant equation Delta_3 = 0$\Delta_3 = 0$ by replacing the third column with the constant vector: Delta_3 = left| beginmatrix 2 & -1 & 4 \\ 5 & lambda & 12 \\ 100 & -47 & 212 endmatrix right| = 0$\Delta_3 = \left| \begin{matrix} 2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212 \end{matrix} \right| = 0$ Expand the determinant along the first row: 2[212lambda - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100lambda] = 0$2[212\lambda - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100\lambda] = 0$ 2[212lambda + 564] + 1[1060 - 1200] + 4[-235 - 100lambda] = 0$2[212\lambda + 564] + 1[1060 - 1200] + 4[-235 - 100\lambda] = 0$ 424lambda + 1128 - 140 - 940 - 400lambda = 0$424\lambda + 1128 - 140 - 940 - 400\lambda = 0$ 24lambda + 48 = 0 implies lambda = -2$24\lambda + 48 = 0 \implies \lambda = -2$ ### Step 1: Solve for Mu using the main determinant Set the primary coefficient matrix determinant Delta = 0$\Delta = 0$ and substitute lambda = -2$\lambda = -2$: Delta = left| beginmatrix 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & mu endmatrix right| = 0$\Delta = \left| \begin{matrix} 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & \mu \end{matrix} \right| = 0$ Expand the determinant along the first row: 2[-2mu - 3(-47)] - (-1)[5mu - 3(100)] + 1[5(-47) - (-2)(100)] = 0$2[-2\mu - 3(-47)] - (-1)[5\mu - 3(100)] + 1[5(-47) - (-2)(100)] = 0$ 2[-2mu + 141] + [5mu - 300] + [-235 + 200] = 0$2[-2\mu + 141] + [5\mu - 300] + [-235 + 200] = 0$ -4mu + 282 + 5mu - 300 - 35 = 0$-4\mu + 282 + 5\mu - 300 - 35 = 0$ mu - 53 = 0 implies mu = 53$\mu - 53 = 0 \implies \mu = 53$ ### Step 2: Calculate the Target Value Substitute the values of mu$\mu$ and lambda$\lambda$ into the expression: mu - 2lambda = 53 - 2(-2) = 53 + 4 = 57$\mu - 2\lambda = 53 - 2(-2) = 53 + 4 = 57$ ### Pattern Recognition When solving systems of equations for infinite solution parameters, choosing a directional determinant that excludes one of the variables simplifies the problem into two separate single-variable equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula The quadratic form condition mathbfX^mathrmTmathbfAmathbfX = mathbf0$\mathbf{X}^{\mathrm{T}}\mathbf{A}\mathbf{X} = \mathbf{0}$ holds for all non-zero vectors mathbfX$\mathbf{X}$ if and only if A$A$ is a skew-symmetric matrix. For a 3 times 3$3 \times 3$ skew-symmetric matrix, the components satisfy: A = beginbmatrix 0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0 endbmatrix$A = \begin{\bmatrix} 0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0 \end{\bmatrix}$ ### Core Logic Let's define the matrix A$A$ using the parameters of a standard skew-symmetric form: A = beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix$A = \begin{\bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{\bmatrix}$ Apply the first given matrix multiplication vector condition: A beginbmatrix 1 \\ 1 \\ 1 endbmatrix = beginbmatrix a+b \\ -a+c \\ -b-c endbmatrix = beginbmatrix 1 \\ 4 \\ -5 endbmatrix$A \begin{\bmatrix} 1 \\ 1 \\ 1 \end{\bmatrix} = \begin{\bmatrix} a+b \\ -a+c \\ -b-c \end{\bmatrix} = \begin{\bmatrix} 1 \\ 4 \\ -5 \end{\bmatrix}$ This gives the system of linear equations: a + b = 1 quad dots (1)$a + b = 1 \quad \dots (1)$ -a + c = 4 quad dots (2)$-a + c = 4 \quad \dots (2)$ -b - c = -5 implies b + c = 5 quad dots (3)$-b - c = -5 \implies b + c = 5 \quad \dots (3)$ Apply the second given matrix multiplication vector condition: A beginbmatrix 1 \\ 2 \\ 1 endbmatrix = beginbmatrix 2a+b \\ -a+c \\ -b-2c endbmatrix = beginbmatrix 0 \\ 4 \\ -8 endbmatrix$A \begin{\bmatrix} 1 \\ 2 \\ 1 \end{\bmatrix} = \begin{\bmatrix} 2a+b \\ -a+c \\ -b-2c \end{\bmatrix} = \begin{\bmatrix} 0 \\ 4 \\ -8 \end{\bmatrix}$ This gives the equation: 2a + b = 0 quad dots (4)$2a + b = 0 \quad \dots (4)$ ### Step 1: Solve for Matrix Elements Subtract equation (1) from equation (4): (2a + b) - (a + b) = 0 - 1 implies a = -1$(2a + b) - (a + b) = 0 - 1 \implies a = -1$ Substitute a = -1$a = -1$ back into equation (1): -1 + b = 1 implies b = 2$-1 + b = 1 \implies b = 2$ Substitute a = -1$a = -1$ into equation (2): -(-1) + c = 4 implies 1 + c = 4 implies c = 3$-(-1) + c = 4 \implies 1 + c = 4 \implies c = 3$ Thus, the explicit matrix A$A$ is: A = beginbmatrix 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 endbmatrix$A = \begin{\bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{\bmatrix}$ ### Step 2: Compute Target Matrix Determinant Construct the modified target matrix 2(A+I)$2(A+I)$: A + I = beginbmatrix 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 endbmatrix$A + I = \begin{\bmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{\bmatrix}$ 2(A + I) = beginbmatrix 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 endbmatrix$2(A + I) = \begin{\bmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{\bmatrix}$ Calculate its determinant value: det(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)]$\det(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)]$ det(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120$\det(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120$ ### Step 3: Analyze Adjoint Power and Prime Factors Using the standard determinant identity for adjoints, det(mathrmadj(M)) = (det M)^n-1$\det(\mathrm{adj}(M)) = (\det M)^{n-1}$ where n=3$n=3$: det(mathrmadj(2(A+I))) = (120)^3-1 = 120^2$\det(\mathrm{adj}(2(A+I))) = (120)^{3-1} = 120^2$ Find the prime factorization of the result: 120 = 2^3 cdot 3^1 cdot 5^1 implies 120^2 = (2^3 cdot 3^1 cdot 5^1)^2 = 2^6 cdot 3^2 cdot 5^2$120 = 2^3 \cdot 3^1 \cdot 5^1 \implies 120^2 = (2^3 \cdot 3^1 \cdot 5^1)^2 = 2^6 \cdot 3^2 \cdot 5^2$ This maps the exponents directly to our target variables: alpha = 6, quad beta = 2, quad gamma = 2$\alpha = 6, \quad \beta = 2, \quad \gamma = 2$ Finally, calculate the \sum of their squares: alpha^2 + beta^2 + gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$\alpha^2 + \beta^2 + \gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$ ### Pattern Recognition The condition mathbfX^T A mathbfX = 0$\mathbf{X}^T A \mathbf{X} = 0$ always implies that A$A$ is a skew-symmetric matrix, which instantly forces the diagonal entries to be zero, reducing the number of unknown parameters from 9 down to 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For orthogonal matrix P$P$, P^T P = P P^T = I$P^T P = P P^T = I$. If B = PAP^T$B = PAP^T$, then: B^k = (PAP^T)(PAP^T)dots(PAP^T) = PA^kP^T$B^k = (PAP^T)(PAP^T)\dots(PAP^T) = PA^kP^T$ ### Core Logic Given C = P^T B^10 P$C = P^T B^{10} P$. Substitute B^10 = P A^10 P^T$B^{10} = P A^{10} P^T$ into the expression: C = P^T (P A^10 P^T) P$C = P^T (P A^{10} P^T) P$ C = (P^T P) A^10 (P^T P)$C = (P^T P) A^{10} (P^T P)$ Since P$P$ is an orthogonal rotation matrix, P^T P = I$P^T P = I$, meaning: C = I cdot A^10 cdot I = A^10$C = I \cdot A^{10} \cdot I = A^{10}$ Therefore, the sum of diagonal elements of C$C$ is simply the trace of A^10$A^{10}$. ### Step 1: Analyze Powers of Upper Triangular Matrix A Matrix A$A$ is upper triangular: A = left[beginmatrixfrac1sqrt2&-2\\ 0&1endmatrixright]$A = \left[\begin{matrix}\frac{1}{\sqrt{2}}&-2\\ 0&1\end{matrix}\right]$ For any upper triangular matrix, any integer power k$k$ preserves the main diagonal entries as simply the powers of the individual diagonal elements: A^10 = left[beginmatrixleft(frac1sqrt2right)^10&*\\ 0&1^10endmatrixright] = left[beginmatrixfrac132&*\\ 0&1endmatrixright]$A^{10} = \left[\begin{matrix}\left(\frac{1}{\sqrt{2}}\right)^{10}&*\\ 0&1^{10}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{32}&*\\ 0&1\end{matrix}\right]$ ### Step 2: Calculate the Trace and sum m+n textSum of diagonal elements = textTrace(C) = textTrace(A^10) = frac132 + 1 = frac3332$\text{Sum of diagonal elements } = \text{Trace}(C) = \text{Trace}(A^{10}) = \frac{1}{32} + 1 = \frac{33}{32}$ Given fracmn = frac3332$\frac{m}{n} = \frac{33}{32}$ with textgcd(33, 32) = 1$\text{\gcd}(33, 32) = 1$: m = 33, quad n = 32$m = 33, \quad n = 32$ m + n = 33 + 32 = 65$m + n = 33 + 32 = 65$ ### Pattern Recognition Traces of matrices are invariant under cyclic permutations, so textTr(P^T B^10 P) = textTr(P P^T B^10) = textTr(B^10)$\text{Tr}(P^T B^{10} P) = \text{Tr}(P P^T B^{10}) = \text{Tr}(B^{10})$. Furthermore, textTr(P A^10 P^T) = textTr(A^10)$\text{Tr}(P A^{10} P^T) = \text{Tr}(A^{10})$. This identity bypasses the need to evaluate any outer matrix multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$ -1 le sin 4x le 1$-1 \le \sin 4x \le 1$ ### Core Logic Apply the row operations R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ to simplify the determinant: f(x) = left| beginarrayc c c 1 + sin^ 2 x & cos^ 2 x & 4 sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 endarray right|$f(x) = \left| \begin{array}{c c c} 1 + \sin^ {2} x & \cos^ {2} x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right|$ ### Step 1: Expand the Determinant Expanding along the first row: f(x) = (1 + sin^2 x)(1 - 0) - cos^2 x(-1 - 0) + 4sin 4x(0 - (-1))$f(x) = (1 + \sin^2 x)(1 - 0) - \cos^2 x(-1 - 0) + 4\sin 4x(0 - (-1))$ f(x) = 1 + sin^2 x + cos^2 x + 4sin 4x$f(x) = 1 + \sin^2 x + \cos^2 x + 4\sin 4x$ Since sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$, we get: f(x) = 2 + 4sin 4x$f(x) = 2 + 4\sin 4x$ ### Step 2: Find Maximum and Minimum Values The range of sin 4x$\sin 4x$ is [-1, 1]$[-1, 1]$. M = 2 + 4(1) = 6$M = 2 + 4(1) = 6$ m = 2 + 4(-1) = -2$m = 2 + 4(-1) = -2$ ### Step 3: Calculate M^4 - m^4$M^4 - m^4$ M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280$M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280$ ### Pattern Recognition Look for repeated structures or cyclic additions in rows. Subtracting rows quickly creates zeros, reducing complex trigonometric matrices into elementary algebraic expressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Trigonometric Functions