If the system of linear equations 3x + y + beta z = 3$$3x + y + \beta z = 3$$ 2x + alpha y - z = -3$$2x + \alpha y - z = -3$$ x + 2y + z = 4$x + 2y + z = 4$ has infinitely many solutions, then the value of 22beta - 9alpha$22\beta - 9\alpha$ is:

### Related Formula Cramer\'s rule states that a system of non-homogeneous linear equations has infinitely many solutions if the main determinant D = 0$D = 0$ and the variable determinants D_1 = D_2 = D_3 = 0$D_1 = D_2 = D_3 = 0$. ### Core Logic Set the main determinant D$D$ to zero : D = beginvmatrix 1 & 2 & -3 \\ 2 & lambda & 5 \\ 14 & 3 & mu endvmatrix = 0$$D = \begin{\vmatrix} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{\vmatrix} = 0$$ 1(lambdamu - 15) - 2(2mu - 70) - 3(6 - 14lambda) = 0$$1(\lambda\mu - 15) - 2(2\mu - 70) - 3(6 - 14\lambda) = 0$$ lambdamu - 15 - 4mu + 140 - 18 + 42lambda = 0$$\lambda\mu - 15 - 4\mu + 140 - 18 + 42\lambda = 0$$ lambdamu + 42lambda - 4mu + 107 = 0$$\lambda\mu + 42\lambda - 4\mu + 107 = 0$$ ### Step 1: Use D_2 = 0$D_2 = 0$ to solve for mu$\mu$ Form determinant D_2$D_2$ by replacing the second column with the constant terms vector : D_2 = beginvmatrix 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & mu endvmatrix = 0$$D_2 = \begin{\vmatrix} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{\vmatrix} = 0$$ 1(5mu - 165) - 2(2mu - 70) - 3(66 - 70) = 0$$1(5\mu - 165) - 2(2\mu - 70) - 3(66 - 70) = 0$$ 5mu - 165 - 4mu + 140 + 12 = 0 Rightarrow mu - 13 = 0 Rightarrow mu = 13$$5\mu - 165 - 4\mu + 140 + 12 = 0 \Rightarrow \mu - 13 = 0 \Rightarrow \mu = 13$$ [cite: 3979, 3982] ### Step 2: Solve for lambda$\lambda$ Substitute mu = 13$\mu = 13$ back into the first equation derived from D=0$D=0$ : 13lambda + 42lambda - 4(13) + 107 = 0$$13\lambda + 42\lambda - 4(13) + 107 = 0$$ 55lambda - 52 + 107 = 0 Rightarrow 55lambda + 55 = 0 Rightarrow lambda = -1$$55\lambda - 52 + 107 = 0 \Rightarrow 55\lambda + 55 = 0 \Rightarrow \lambda = -1$$ Thus, lambda + mu = -1 + 13 = 12$\lambda + \mu = -1 + 13 = 12$. ### Pattern Recognition When evaluating infinite solutions, look for columns that are easily solvable using D_i = 0$D_i = 0$ forms. Calculating D_2 = 0$D_2 = 0$ avoids dealing with any non-linear products of lambdamu$\lambda\mu$ directly at the start. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### 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