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 alpha^2 + alpha + 1 = 0 implies alpha = omega quad textwhere omega^3 = 1$$\alpha^2 + \alpha + 1 = 0 \implies \alpha = \omega \quad \text{where } \omega^3 = 1$$ ### Core Logic Perform row-matrix vector multiplication to generate a system of linear equations in a$a$ and b$b$. Solve for the powers and reduce the algebraic equation using complex roots of unity. ### Step 1: Solve Matrix Vector Multiplication 4 - a - 2b = 0$4 - a - 2b = 0$ 64 - a - 14b = 0$$64 - a - 14b = 0$$ 52 + 2a - 8b = 0$$52 + 2a - 8b = 0$$ From the first two equations, subtracting them gives: 60 - 12b = 0 implies b = 5$$60 - 12b = 0 \implies b = 5$$ Substituting b = 5$b = 5$ into the first equation: 4 - a - 10 = 0 implies a = -6$$4 - a - 10 = 0 \implies a = -6$$ ### Step 2: Evaluate Exponential Equation with Roots of Unity Substitute a = -6, b = 5$a = -6, b = 5$ into the given equation: frac4alpha^4 + fracmalpha^-6 + fracnalpha^5 = 3 implies frac4omega + m + fracnomega^2 = 3$$\frac{4}{\alpha^4} + \frac{m}{\alpha^{-6}} + \frac{n}{\alpha^5} = 3 \implies \frac{4}{\omega} + m + \frac{n}{\omega^2} = 3$$ 4omega^2 + m + nomega = 3$$4\omega^2 + m + n\omega = 3$$ ### Step 3: Resolve Real and Imaginary Components Substitute standard values omega = -frac12 + fracsqrt32i$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and omega^2 = -frac12 - fracsqrt32i$\omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$: 4left(-frac12 - fracsqrt32iright) + m + nleft(-frac12 + fracsqrt32iright) = 3$$4\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) + m + n\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 3$$ Equating the imaginary components: frac-4sqrt32 + fracnsqrt32 = 0 implies n = 4$$\frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \implies n = 4$$ Equating the real components: -2 + m - fracn2 = 3 implies -2 + m - 2 = 3 implies m = 7$$-2 + m - \frac{n}{2} = 3 \implies -2 + m - 2 = 3 \implies m = 7$$ m + n = 7 + 4 = 11$$m + n = 7 + 4 = 11$$ ### Pattern Recognition Whenever an expression satisfies Aomega^2 + Bomega + C = 0$A\omega^2 + B\omega + C = 0$, it directly maps to a comparison with the standard identity omega^2 + omega + 1 = 0$\omega^2 + \omega + 1 = 0$ up to a linear translation shift. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Complex Numbers

### Related Formula |operatornameadj(operatornameadj(M))| = |M|^(n-1)^2$$|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(n-1)^2}$$ |kM| = k^n|M|$|kM| = k^n|M|$ ### Core Logic Perform determinant row reduction transforms to decouple tracking metrics p$p$ and q$q$, leaving a baseline numerical determinant value behind. ### Step 1: Simplify the Determinant of Matrix A Execute columns adjustments: C_3 rightarrow C_3 - C_2 - C_1 times fracq2$C_3 \rightarrow C_3 - C_2 - C_1 \times \frac{q}{2}$, then C_2 rightarrow C_2 - C_1 times left(1 + fracp2right)$C_2 \rightarrow C_2 - C_1 \times \left(1 + \frac{p}{2}\right)$: |A| = beginvmatrix 2 & 0 & 0 \\ 4 & 2 & 2+p \\ 6 & 6 & 8+3p endvmatrix = 2left(16 + 6p - 12 - 6pright) = 8 = 2^3$$|A| = \begin{vmatrix} 2 & 0 & 0 \\ 4 & 2 & 2+p \\ 6 & 6 & 8+3p \end{vmatrix} = 2\left(16 + 6p - 12 - 6p\right) = 8 = 2^3$$ ### Step 2: Apply Adjoint Exponent Transforms For a matrix of dimensional profile size 3: |operatornameadj(operatornameadj(3A))| = |3A|^(3-1)^2 = |3A|^4$$|\operatorname{adj}(\operatorname{adj}(3A))| = |3A|^{(3-1)^2} = |3A|^4$$ ### Step 3: Resolve Exponential System Size |3A| = 3^3 |A| = 3^3 times 2^3$$|3A| = 3^3 |A| = 3^3 \times 2^3$$ |3A|^4 = (3^3 times 2^3)^4 = 2^12 times 3^12$$|3A|^4 = (3^3 \times 2^3)^4 = 2^{12} \times 3^{12}$$ Matching base parameter targets: m = 12, n = 12 implies m + n = 24$m = 12, n = 12 \implies m + n = 24$ ### Pattern Recognition Linear parameter shifts down secondary column profiles usually dissolve cleanly during forward element elimination column steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula Cramer's rule for infinite solutions in a 3 variable system requires: Delta = Delta_x = Delta_y = Delta_z = 0$$\Delta = \Delta_x = \Delta_y = \Delta_z = 0$$ ### Core Logic Set up the primary matrix determinant Delta$\Delta$: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 endvmatrix$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix}$$ = 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0$$= 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0$$ Since Delta = 0$\Delta = 0$ is true independent of m$m$, analyze secondary delta constraints to maintain consistency for infinite paths. ### Step 1: Compute Dependent Variable Constraints Evaluate Delta_x = 0$\Delta_x = 0$: Delta_x = beginvmatrix 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 endvmatrix = 0$$\Delta_x = \begin{vmatrix} 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 \end{vmatrix} = 0$$ 1(20 - 16) - 1(10m - 4m^2) + 1(4m - 2m^2) = 0$$1(20 - 16) - 1(10m - 4m^2) + 1(4m - 2m^2) = 0$$ 4 - 10m + 4m^2 + 4m - 2m^2 = 0$$4 - 10m + 4m^2 + 4m - 2m^2 = 0$$ 2m^2 - 6m + 4 = 0 implies m^2 - 3m + 2 = 0$$2m^2 - 6m + 4 = 0 \implies m^2 - 3m + 2 = 0$$ Thus, m = 1, 2$m = 1, 2$, which gives alpha = 1, beta = 2$\alpha = 1, \beta = 2$. ### Step 2: Calculate Sigma Expression sum_n=1^10 (n^1 + n^2) = sum_n=1^10 n + sum_n=1^10 n^2$$\sum_{n=1}^{10} (n^1 + n^2) = \sum_{n=1}^{10} n + \sum_{n=1}^{10} n^2$$ = frac10(11)2 + frac10(11)(21)6 = 55 + 385 = 440$$= \frac{10(11)}{2} + \frac{10(11)(21)}{6} = 55 + 385 = 440$$ ### Pattern Recognition When infinitely many solutions are required, solving the determinant created by replacing one column with the constant vector provides parameter roots directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula Element formula entry rule: a_ij = (sqrt2)^i+j$$a_{ij} = (\sqrt{2})^{i+j}$$ ### Core Logic Constructing the initial matrix structure from the formula entries: A = beginbmatrix 2 & 2sqrt2 & 4 \\ 2sqrt2 & 4 & 4sqrt2 \\ 4 & 4sqrt2 & 8 endbmatrix$$A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix}$$ Factoring scalar factor 4 out to ease squaring multiplication lines: A = 2beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix$$A = 2\begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix}$$ ### Step 1: Calculate Rows of Power Matrix Squaring matrix A^2$A^2$ matches scalar multipliers: A^2 = 4 beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix$$A^2 = 4 \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix} \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix}$$ Extract third row entries explicitly: textRow 3 = 4 beginbmatrix (2+4+8) & (2sqrt2+4sqrt2+8sqrt2) & (4+8+16) endbmatrix$$\text{Row 3} = 4 \begin{bmatrix} (2+4+8) & (2\sqrt{2}+4\sqrt{2}+8\sqrt{2}) & (4+8+16) \end{bmatrix}$$ = 4 beginbmatrix 14 & 14sqrt2 & 28 endbmatrix$$= 4 \begin{bmatrix} 14 & 14\sqrt{2} & 28 \end{bmatrix}$$ ### Step 2: Aggregate Entries textSum of row elements = 4(14 + 14sqrt2 + 28)$$\text{Sum of row elements} = 4(14 + 14\sqrt{2} + 28)$$ = 4(42 + 14sqrt2) = 168 + 56sqrt2$$= 4(42 + 14\sqrt{2}) = 168 + 56\sqrt{2}$$ Matching structural parameters: alpha = 168, quad beta = 56$$\alpha = 168, \quad \beta = 56$$ alpha + beta = 168 + 56 = 224$$\alpha + \beta = 168 + 56 = 224$$ ### Pattern Recognition Pull common scaling scalar integers out of matrices before running large multiplications. It eliminates algebraic tracking errors across geometric indices. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula Set Principle of Inclusion-Exclusion: n(S_1 cup S_2 cup S_3) = n(S_1) + n(S_2) + n(S_3) - n(S_1 cap S_2) - n(S_2 cap S_3) - n(S_1 cap S_3) + n(S_1 cap S_2 cap S_3)$$n(S_1 \cup S_2 \cup S_3) = n(S_1) + n(S_2) + n(S_3) - n(S_1 \cap S_2) - n(S_2 \cap S_3) - n(S_1 \cap S_3) + n(S_1 \cap S_2 \cap S_3)$$ ### Core Logic Let's count each subset based on the 5 elements available in S$S$: 1. For S_1$S_1$ (Symmetric matrices): 6 independent element choices implies n(S_1) = 5^6$\implies n(S_1) = 5^6$. 2. For S_2$S_2$ (Skew-symmetric matrices): Diagonal elements must be 0, but 0 notin S$0 \notin S$, so n(S_2) = 0$n(S_2) = 0$. Since n(S_2) = 0$n(S_2) = 0$, any intersection term involving S_2$S_2$ also becomes 0. ### Step 1: Calculating Trace Matrix Variations For S_3$S_3$ (Trace equal to zero conditions): The condition a_11 + a_22 + a_33 = 0$a_{11} + a_{22} + a_{33} = 0$ over S = \-3, -2, -1, 1, 2\$S = \{-3, -2, -1, 1, 2\}$ has exactly 12 valid tuple combinations. The remaining 6 elements can be chosen freely. n(S_3) = 12 times 5^6$$n(S_3) = 12 \times 5^6$$ For the intersection n(S_1 cap S_3)$n(S_1 \cap S_3)$: n(S_1 cap S_3) = 12 times 5^3$$n(S_1 \cap S_3) = 12 \times 5^3$$ ### Step 2: Final Inclusion-Exclusion Assembly n(S_1 cup S_2 cup S_3) = 5^6 + 12 times 5^6 - 12 times 5^3$$n(S_1 \cup S_2 \cup S_3) = 5^6 + 12 \times 5^6 - 12 \times 5^3$$ = 5^3 times [13 times 5^3 - 12] = 125 times 1613$$= 5^3 \times [13 \times 5^3 - 12] = 125 \times 1613$$ Thus, alpha = 1613$\alpha = 1613$. ### Pattern Recognition Always check if the set contains 0$0$. Missing zero elements in skew-symmetric matrix setups instantly zeros out large blocks of permutations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants