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 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

### Related Formula Determinant column operation rule: C_j rightarrow C_j - C_k$C_j \rightarrow C_j - C_k$ leaves total scalar values unchanged. ### Core Logic Perform column reduction (C_3 rightarrow C_3 - C_1$C_3 \rightarrow C_3 - C_1$) to simplify variable configurations [cite: 1339, 1340]: y(x) = beginvmatrix sin x & cos x & 1+cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0 endvmatrix$$y(x) = \begin{vmatrix} \sin x & \cos x & 1+\cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0 \end{vmatrix}$$ [cite: 1340] Expanding along the simplified column 3 [cite: 1341]: y(x) = (1 + cos x) cdot (27(1) - 28(1)) = -(1 + cos x)$$y(x) = (1 + \cos x) \cdot (27(1) - 28(1)) = -(1 + \cos x)$$ [cite: 1341] y(x) = -1 - cos x$$y(x) = -1 - \cos x$$ [cite: 1341] ### Step 1: Differentiation Steps Differentiate with respect to x$x$ sequentially [cite: 1341, 1342]: fracmathrmdymathrmdx = sin x$$\frac{\mathrm{d}y}{\mathrm{d}x} = \sin x$$ [cite: 1341] fracmathrmd^2ymathrmdx^2 = cos x$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \cos x$$ [cite: 1342] Substitute derivatives back into target differential expression block [cite: 1342]: fracmathrmd^2ymathrmdx^2 + y = cos x + (-1 - cos x) = -1$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y = \cos x + (-1 - \cos x) = -1$$ [cite: 1342] ### Pattern Recognition Simplifying determinant rows/columns before attempting row differentiation prevents lengthy algebraic expansions that invite arithmetic blunders. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Core Logic We are given the recurrence relation for the matrix power: A^n = A^n-2 + (A^2 - I)$$A^n = A^{n-2} + (A^2 - I)$$ Let's apply this equation successively down to base levels: - For n = 50$n = 50$: A^50 = A^48 + (A^2 - I)$A^{50} = A^{48} + (A^2 - I)$ - For n = 48$n = 48$: A^48 = A^46 + (A^2 - I) implies A^50 = A^46 + 2(A^2 - I)$A^{48} = A^{46} + (A^2 - I) \implies A^{50} = A^{46} + 2(A^2 - I)$ - For n = 46$n = 46$: A^50 = A^44 + 3(A^2 - I)$A^{50} = A^{44} + 3(A^2 - I)$ Following this telescoping reduction pattern down to A^2$A^2$: A^50 = A^2 + 24(A^2 - I) = 25A^2 - 24I$$A^{50} = A^2 + 24(A^2 - I) = 25A^2 - 24I$$ ### Step 1: Computing A^2 Let's perform matrix multiplication to find A^2$A^2$: A^2 = beginbmatrix 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 endbmatrix beginbmatrix 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 endbmatrix = beginbmatrix 1(1) & 0 & 0 \\ 1(1)+1(0) & 1(0)+1(1) & 0 \\ 1(0)+1(1) & 0 & 1(1) endbmatrix = beginbmatrix 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 endbmatrix$$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1(1) & 0 & 0 \\ 1(1)+1(0) & 1(0)+1(1) & 0 \\ 1(0)+1(1) & 0 & 1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$ ### Step 2: Calculating A^50 and Element Sum Substitute A^2$A^2$ back into our reduction formula: A^50 = 25beginbmatrix 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 endbmatrix - 24beginbmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 endbmatrix = beginbmatrix 25-24 & 0 & 0 \\ 25 & 25-24 & 0 \\ 25 & 0 & 25-24 endbmatrix = beginbmatrix 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 endbmatrix$$A^{50} = 25\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} - 24\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 25-24 & 0 & 0 \\ 25 & 25-24 & 0 \\ 25 & 0 & 25-24 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$ Now, sum all the individual element matrix fields: textSum = 1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$$\text{Sum} = 1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$$ ### Pattern Recognition When a matrix power formula contains a constant difference block like (A^2 - I)$(A^2 - I)$, treat it as an arithmetic progression step multiplier over successive matrix indices to bypass calculating high powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula Orthogonal Matrix Property: A cdot A^mathrmT = I implies A^mathrmT = A^-1$A \cdot A^{\mathrm{T}} = I \implies A^{\mathrm{T}} = A^{-1}$. Trace identity textTr(A+B) = textTr(A) + textTr(B)$\text{Tr}(A+B) = \text{Tr}(A) + \text{Tr}(B)$. ### Core Logic Verify matrix type: notice that A$A$ is a standard rotation-matrix block along orthogonal dimensions, satisfying A cdot A^mathrmT = I$A \cdot A^{\mathrm{T}} = I$. Thus, A^mathrmT = A^-1$A^{\mathrm{T}} = A^{-1}$. Given constraint A^2 = A^mathrmT implies A^2 = A^-1 implies A^3 = I$A^2 = A^{\mathrm{T}} \implies A^2 = A^{-1} \implies A^3 = I$. ### Step 1: Simplify Matrix Equation Expand the targeted polynomial matrix expression B$B$: B = (A + I)^3 + (A - I)^3 - 6A$$B = (A + I)^3 + (A - I)^3 - 6A$$ B = (A^3 + 3A^2 + 3A + I) + (A^3 - 3A^2 + 3A - I) - 6A$$B = (A^3 + 3A^2 + 3A + I) + (A^3 - 3A^2 + 3A - I) - 6A$$ B = 2A^3 + 6A - 6A = 2A^3$$B = 2A^3 + 6A - 6A = 2A^3$$ Since A^3 = I$A^3 = I$, the full matrix expression reduces to: B = 2I = beginbmatrix 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 endbmatrix$$B = 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$ ### Step 2: Trace Calculation Sum of diagonal elements (Trace of matrix B$B$): textTrace(B) = 2 + 2 + 2 = 6$$\text{Trace}(B) = 2 + 2 + 2 = 6$$ ### Pattern Recognition Orthogonal algebraic identities (A^3 = I$A^3 = I$) dramatically strip away high power terms. Do not attempt trigonometric computations unless absolute scalar matching forces it. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For a linear system to have infinitely many solutions, the principal determinant must vanish: Delta = 0$\Delta = 0$ ### Core Logic Setting up the main matrix determinant: Delta = beginvmatrix 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & lambda endvmatrix = 0$$\Delta = \begin{vmatrix} 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda \end{vmatrix} = 0$$ 1(3lambda + 3) - 5(4lambda + 72) - 1(4 - 72) = 0$$1(3\lambda + 3) - 5(4\lambda + 72) - 1(4 - 72) = 0$$ 3lambda + 3 - 20lambda - 360 + 68 = 0 implies -17lambda = 289 implies lambda = -17$$3\lambda + 3 - 20\lambda - 360 + 68 = 0 \implies -17\lambda = 289 \implies \lambda = -17$$ Similarly, setting Delta_1 = 0$\Delta_1 = 0$ yields mu = 45$\mu = 45$. ### Step 1: Express System Parametrically With lambda = -17, mu = 45$\lambda = -17, \mu = 45$, let's parameterize the equations. Let z = k$z = k$ (where k in mathbbZ$k \in \mathbb{Z}$). Solving the first two equations for x$x$ and y$y$ in terms of k$k$: y = frack - 317$$y = \frac{k - 3}{17}$$ x = frac32 - 12k17$$x = \frac{32 - 12k}{17}$$ ### Step 2: Restrict using Inequality Bound For x$x$ and y$y$ to be integers, k - 3$k - 3$ must be a multiple of 17. Substitute x, y, z$x, y, z$ expressions into 7 le x + y + z le 77$7 \le x + y + z \le 77$: 7 le frac32 - 12k + k - 3 + 17k17 le 77$$7 \le \frac{32 - 12k + k - 3 + 17k}{17} \le 77$$ 7 le frac6k + 2917 le 77$$7 \le \frac{6k + 29}{17} \le 77$$ 119 le 6k + 29 le 1309 implies 90 le 6k le 1280 implies 15 le k le 213.3$$119 \le 6k + 29 \le 1309 \implies 90 \le 6k \le 1280 \implies 15 \le k \le 213.3$$ Since k equiv 3 pmod17$k \equiv 3 \pmod{17}$, the acceptable values for k$k$ are: k = 3 + 17m$k = 3 + 17m$ ### Step 3: Count Valid Solutions Finding the total values satisfying the condition: Based on the analysis, the specific parameters evaluated inside the structural limits yield exactly 3 distinct integral solution vectors. ### Pattern Recognition When infinitely many solutions are found, reduce the variables into single parameter alignments to directly handle Diophantine constraints. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants