### 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

### Related Formula For a matrix M$M$ of order ntimes n$n\times n$: 1. |kM| = k^n|M|$|kM| = k^n|M|$ 2. |operatornameadj(M)| = |M|^n-1$|operatorname{adj}(M)| = |M|^{n-1}$ ### Core Logic Given n=3$n=3$ and |A|=5$|A|=5$[cite: 496]. Let us simplify the expression stepwise[cite: 1188]: |2operatornameadj(3Aoperatornameadj(2A))| = 2^3 cdot |operatornameadj(3Aoperatornameadj(2A))|$|2operatorname{adj}(3Aoperatorname{adj}(2A))| = 2^3 cdot |operatorname{adj}(3Aoperatorname{adj}(2A))|$ [cite: 1188] Using the adjoint determinant rule [cite: 1188]: = 2^3 cdot |3Aoperatornameadj(2A)|^3-1 = 2^3 cdot |3Aoperatornameadj(2A)|^2$= 2^3 cdot |3Aoperatorname{adj}(2A)|^{3-1} = 2^3 cdot |3Aoperatorname{adj}(2A)|^2$ [cite: 1188] Now, apply the constant multiple property inside the determinant [cite: 1189]: = 2^3 cdot (3^3)^2 cdot |A|^2 cdot |operatornameadj(2A)|^2$= 2^3 cdot (3^3)^2 cdot |A|^2 cdot |operatorname{adj}(2A)|^2$ [cite: 1189] = 2^3 cdot 3^6 cdot |A|^2 cdot (|2A|^3-1)^2 = 2^3 cdot 3^6 cdot |A|^2 cdot |2A|^4$= 2^3 cdot 3^6 cdot |A|^2 cdot (|2A|^{3-1})^2 = 2^3 cdot 3^6 cdot |A|^2 cdot |2A|^4$ [cite: 1190] Substitute |2A| = 2^3|A|$|2A| = 2^3|A|$ [cite: 1191]: = 2^3 cdot 3^6 cdot |A|^2 cdot (2^3|A|)^4 = 2^3 cdot 3^6 cdot |A|^2 cdot 2^12 cdot |A|^4$= 2^3 cdot 3^6 cdot |A|^2 cdot (2^3|A|)^4 = 2^3 cdot 3^6 cdot |A|^2 cdot 2^{12} cdot |A|^4$ [cite: 1191] = 2^15 cdot 3^6 cdot |A|^6$= 2^{15} cdot 3^6 cdot |A|^6$ [cite: 1192] ### Step 1: Substituting the value of |A| Substitute |A|=5$|A|=5$ into the simplified form [cite: 1193]: 2^15 cdot 3^6 cdot 5^6 = 2^alpha cdot 3^beta cdot 5^gamma$2^{15} cdot 3^6 cdot 5^6 = 2^{alpha} cdot 3^{\beta} cdot 5^{gamma}$ [cite: 1193] Comparing exponents [cite: 1194]: alpha = 15, quad beta = 6, quad gamma = 6$alpha = 15, \quad \beta = 6, \quad \gamma = 6$ [cite: 1194] Therefore, the sum is [cite: 1195]: alpha + beta + gamma = 15 + 6 + 6 = 27$\alpha + \beta + \gamma = 15 + 6 + 6 = 27$ [cite: 1195] ### Pattern Recognition Always evaluate scaling factor transformations from the outermost function inward. Keep absolute track of matrix dimensions as powers amplify rapidly with each step layer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants