Let A = beginbmatrix alpha & -1 \\ 6 & beta endbmatrix$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$, alpha > 0$\alpha > 0$, such that det(A) = 0$\det(A) = 0$ and alpha + beta = 1$\alpha + \beta = 1$. If I$I$ denotes the 2 times 2$2 \times 2$ identity matrix, then the matrix (I + A)^8$(I + A)^8$ is:

### 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

### Related Formula sum_k a_ik A_jk = delta_ij |A| implies C = beginbmatrix |A| & 0 \\ 0 & |A| endbmatrix implies |C| = |A|^2$$\sum_{k} a_{ik} A_{jk} = \delta_{ij} |A| \implies C = \begin{bmatrix} |A| & 0 \\ 0 & |A| \end{bmatrix} \implies |C| = |A|^2$$ ### Core Logic Evaluate the determinant of matrix A$A$: |A| = (log_5 128)(log_4 25) - (log_4 5)(log_5 8)$$|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)$$ Using change of base rules: |A| = left(7log_5 2right)left(2log_4 5right) - left(frac12log_2 5right)left(3log_5 2right)$$|A| = \left(7\log_5 2\right)\left(2\log_4 5\right) - \left(\frac{1}{2}\log_2 5\right)\left(3\log_5 2\right)$$ |A| = 14left(log_5 2 cdot frac12log_2 5right) - frac32 = 7 - 1.5 = 5.5 = frac112$$|A| = 14\left(\log_5 2 \cdot \frac{1}{2}\log_2 5\right) - \frac{3}{2} = 7 - 1.5 = 5.5 = \frac{11}{2}$$ ### Step 1: Compute |C| and evaluate response target Since matrix properties dictate |C| = |A|^2$|C| = |A|^2$: |C| = left(frac112right)^2 = frac1214$$|C| = \left(\frac{11}{2}\right)^2 = \frac{121}{4}$$ Evaluate targeted multiplier: 8|C| = 8 times frac1214 = 2 times 121 = 242$$8|C| = 8 \times \frac{121}{4} = 2 \times 121 = 242$$ ### Pattern Recognition Recognize the core cofactor theorem identity instantly: multiplying rows by cofactors of other rows creates zero elements, yielding basic diagonal scalar structures matching matrix attributes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula textInductive formulation for exponent powers of a pattern matrix$$\text{Inductive formulation for exponent powers of a pattern matrix}$$ ### Core Logic Evaluate lower power forms of A$A$ to establish inductive sequence patterns: A = beginbmatrix 2 & -1 \\ 1 & 0 endbmatrix, quad A^2 = beginbmatrix 3 & -2 \\ 2 & -1 endbmatrix, quad A^3 = beginbmatrix 4 & -3 \\ 3 & -2 endbmatrix$$A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}, \quad A^2 = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}$$ This cleanly establishes general power state rule configuration expression: A^m = beginbmatrix m+1 & -m \\ m & -m+1 endbmatrix$$A^m = \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix}$$ ### Step 1: Setup Matrix Power Equation Using the pattern, find expressions for terms: A^6 = beginbmatrix 7 & -6 \\ 6 & -5 endbmatrix, quad A^-6 = (A^6)^-1 = beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix$$A^6 = \begin{bmatrix} 7 & -6 \\ 6 & -5 \end{bmatrix}, \quad A^{-6} = (A^6)^{-1} = \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix}$$ Substitute into the targeted equation block: A^m^2 + A^m = 3beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix - beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix = beginbmatrix 8 & -6 \\ 6 & -4 endbmatrix$$A^{m^2} + A^m = 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix} = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix}$$ ### Step 2: Equate Corresponding Elements From the bottom-left entry [2,1]$[2,1]$: m^2 + m = 6 implies m^2 + m - 6 = 0$$m^2 + m = 6 \implies m^2 + m - 6 = 0$$ (m+3)(m-2) = 0 implies m = -3, 2$$(m+3)(m-2) = 0 \implies m = -3, 2$$ Both integer states satisfy all matrix components seamlessly. Therefore, the number of elements n(S) = 2$n(S) = 2$. ### Pattern Recognition Always calculate first 2–3 matrix powers to spot linear arithmetic trends across specific cell values (m+1, -m$m+1, -m$), avoiding tedious full Cayley-Hamilton character expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices

### Related Formula Properties of Determinants: 1. det(AB) = det(A) cdot det(B)$\det(AB) = \det(A) \cdot \det(B)$ 2. det(A^T) = det(A)$\det(A^T) = \det(A)$ 3. det(A^n) = (det(A))^n$\det(A^n) = (\det(A))^n$ ### Core Logic First, find the determinant of matrix A$A$ and matrix B$B$: det(A) = |A| = beginvmatrixsqrt2 & 1 \\ -1 & sqrt2endvmatrix = (sqrt2)(sqrt2) - (1)(-1) = 2 + 1 = 3 $$\det(A) = |A| = \begin{vmatrix}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{vmatrix} = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3 $$ det(B) = |B| = beginvmatrix1 & 0 \\ 1 & 1endvmatrix = (1)(1) - (0)(1) = 1 $$\det(B) = |B| = \begin{vmatrix}1 & 0 \\ 1 & 1\end{vmatrix} = (1)(1) - (0)(1) = 1 $$ ### Step 1: Calculate determinant of C Given C = ABA^T$C = ABA^{T}$, we calculate its determinant: det(C) = det(ABA^T) = det(A) cdot det(B) cdot det(A^T) $$\det(C) = \det(ABA^{T}) = \det(A) \cdot \det(B) \cdot \det(A^{T}) $$ Using det(A^T) = det(A) = 3$\det(A^T) = \det(A) = 3$: det(C) = 3 cdot 1 cdot 3 = 9 $$\det(C) = 3 \cdot 1 \cdot 3 = 9 $$ ### Step 2: Calculate determinant of X Given X = A^TC^2A$X = A^{T}C^{2}A$, we determine |X|$|X|$: det(X) = |A^TC^2A| = |A^T| cdot |C^2| cdot |A| $$\det(X) = |A^{T}C^{2}A| = |A^{T}| \cdot |C^{2}| \cdot |A| $$ det(X) = |A| cdot |C|^2 cdot |A| = |A|^2 cdot |C|^2 $$\det(X) = |A| \cdot |C|^{2} \cdot |A| = |A|^{2} \cdot |C|^{2} $$ Substitute the values |A| = 3$|A| = 3$ and |C| = 9$|C| = 9$: det(X) = (3)^2 cdot (9)^2 = 9 cdot 81 = 729 $$\det(X) = (3)^{2} \cdot (9)^{2} = 9 \cdot 81 = 729 $$ ### Pattern Recognition Sees: Transpose and multiplication operations nested inside a determinant statement. Shortcut: Never explicitly compute product matrices like ABA^T$ABA^T$ or A^TC^2A$A^TC^2A$. Apply the distributive identity of determinants |XYZ| = |X||Y||Z|$|XYZ| = |X||Y||Z|$ entirely to work strictly with scalar multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For a system of linear equations representing planes to have infinitely many solutions, the planes must belong to a single family sharing a common line of intersection: P_1 = k_1 P_2 + k_2 P_3$$P_1 = k_1 P_2 + k_2 P_3$$ ### Core Logic Let the planes be defined as: - P_1: 2x + 3y - z - 5 = 0$P_1: 2x + 3y - z - 5 = 0$ - P_2: x + alpha y + 3z + 4 = 0$P_2: x + \alpha y + 3z + 4 = 0$ - P_3: 3x - y + beta z - 7 = 0$P_3: 3x - y + \beta z - 7 = 0$ Expressing P_1$P_1$ as a linear combination of P_2$P_2$ and P_3$P_3$: 2x + 3y - z - 5 = k_1(x + alpha y + 3z + 4) + k_2(3x - y + beta z - 7)$$2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$$ ### Step 1: Evaluate parameters k1 and k2 Comparing the coefficients of x$x$ and the constant terms on both sides: - For x$x$: k_1 + 3k_2 = 2 quad implies (1)$$k_1 + 3k_2 = 2 \quad \implies (1)$$ - For the constants: 4k_1 - 7k_2 = -5 quad implies (2)$$4k_1 - 7k_2 = -5 \quad \implies (2)$$ Multiplying equation (1) by 4 gives 4k_1 + 12k_2 = 8$4k_1 + 12k_2 = 8$. Subtracting equation (2) from this result: (4k_1 + 12k_2) - (4k_1 - 7k_2) = 8 - (-5)$$(4k_1 + 12k_2) - (4k_1 - 7k_2) = 8 - (-5)$$ 19k_2 = 13 implies k_2 = frac1319$$19k_2 = 13 \implies k_2 = \frac{13}{19}$$ Substituting k_2$k_2$ back into equation (1): k_1 + 3left(frac1319right) = 2 implies k_1 = 2 - frac3919 = -frac119$$k_1 + 3\left(\frac{13}{19}\right) = 2 \implies k_1 = 2 - \frac{39}{19} = -\frac{1}{19}$$ ### Step 2: Calculate alpha, beta and final product Comparing coefficients for y$y$ and z$z$: - For y$y$: k_1alpha - k_2 = 3 implies -frac119alpha - frac1319 = 3$$k_1\alpha - k_2 = 3 \implies -\frac{1}{19}\alpha - \frac{13}{19} = 3$$ -alpha - 13 = 57 implies alpha = -70$$-\alpha - 13 = 57 \implies \alpha = -70$$ - For z$z$: 3k_1 + k_2beta = -1 implies 3left(-frac119right) + frac1319beta = -1$$3k_1 + k_2\beta = -1 \implies 3\left(-\frac{1}{19}\right) + \frac{13}{19}\beta = -1$$ -3 + 13beta = -19 implies 13beta = -16 implies beta = -frac1613$$-3 + 13\beta = -19 \implies 13\beta = -16 \implies \beta = -\frac{16}{13}$$ Now, compute 13 alpha beta$13 \alpha \beta$: 13 alpha beta = 13 times (-70) times left(-frac1613right) = 70 times 16 = 1120$$13 \alpha \beta = 13 \times (-70) \times \left(-\frac{16}{13}\right) = 70 \times 16 = 1120$$ ### Pattern Recognition Sees: Infinite solution framework for three linear planes. Shortcut: Using the family of planes equation is significantly less prone to fractional algebraic mistakes compared to establishing Cramer's rule determinants (D = D_x = D_y = D_z = 0$D = D_x = D_y = D_z = 0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants