Let α, β$\alpha, \beta$ ( α ≠ β$\alpha \neq \beta$ ) be the values of m$m$, for which the equations x + y + z = 1$x + y + z = 1$ ; x + 2y + 4z = m$x + 2y + 4z = m$ and x + 4y + 10z = m²$x + 4y + 10z = m^2$ have infinitely many solutions. Then the value of Σn=1¹⁰ (n^α + n^β)$\sum_{n=1}^{10} (n^\alpha + n^\beta)$ is equal to:

Related Formula

For a square matrix M$M$ of order n × n$n \times n$:

• |kM| = kⁿ|M|$|kM| = k^n|M|$
• |adj(M)| = |M|ⁿ⁻¹$|\operatorname{adj}(M)| = |M|^{n-1}$
• |XY| = |X||Y|$|XY| = |X||Y|$

Core Logic

Given n = 3$n = 3$ and |A| = 5$|A| = 5$. Let us simplify the determinant expression stepwise:

Using the determinant rule for adjoints, |adj(M)| = |M|ⁿ⁻¹ = |M|²$|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2$:

Factoring the scalar 3$3$ out of the 3 × 3$3 \times 3$ determinant:

Substitute |2A| = 2³|A|$|2A| = 2^3|A|$:

Step 1: Substituting the Value of |A|$|A|$

Substitute |A| = 5$|A| = 5$:

Comparing exponents:

Computing the required sum:

Pattern Recognition

Shortcut: Evaluate scalar properties from the outside in. Each scalar factor k$k$ pulled from an n × n$n \times n$ determinant picks up a power of n$n$, and each adjoint operation elevates the inner determinant to the power of (n-1)$(n-1)$.

Evaluation Rubric / Model Answer

27

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

Determinant column operation rule: Cⱼ arrow Cⱼ - Ck$C_j \rightarrow C_j - C_k$ leaves total scalar values unchanged.

Core Logic

Perform column reduction (C₃ arrow C₃ - C₁$C_3 \rightarrow C_3 - C_1$) to simplify variable configurations [cite: 1339, 1340]:

y(x) = vmatrix x & x & 1+ x 27 & 28 & 0 1 & 1 & 0 vmatrix$$y(x) = \begin{vmatrix} \sin x & \cos x & 1+\cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0 \end{vmatrix}$$ [cite: 1340]

Expanding along the simplified column 3 [cite: 1341]: y(x) = (1 + x) · (27(1) - 28(1)) = -(1 + x)$$y(x) = (1 + \cos x) \cdot (27(1) - 28(1)) = -(1 + \cos x)$$ [cite: 1341]

y(x) = -1 - x$$y(x) = -1 - \cos x$$ [cite: 1341]

Step 1: Differentiation Steps

Differentiate with respect to x$x$ sequentially [cite: 1341, 1342]: dydx = x$$\frac{\mathrm{d}y}{\mathrm{d}x} = \sin x$$ [cite: 1341] d²ydx² = x$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \cos x$$ [cite: 1342]

Substitute derivatives back into target differential expression block [cite: 1342]: d²ydx² + y = x + (-1 - 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.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Core Logic

We are given the recurrence relation for the matrix power:

Let's apply this equation successively down to base levels:

• For n = 50$n = 50$: A⁵⁰ = A⁴⁸ + (A² - I)$A^{50} = A^{48} + (A^2 - I)$
• For n = 48$n = 48$: A⁴⁸ = A⁴⁶ + (A² - I) A⁵⁰ = A⁴⁶ + 2(A² - I)$A^{48} = A^{46} + (A^2 - I) \implies A^{50} = A^{46} + 2(A^2 - I)$
• For n = 46$n = 46$: A⁵⁰ = A⁴⁴ + 3(A² - I)$A^{50} = A^{44} + 3(A^2 - I)$

Following this telescoping reduction pattern down to A²$A^2$:

Step 1: Computing A^2

Let's perform matrix multiplication to find A²$A^2$:

Step 2: Calculating A^50 and Element Sum

Substitute A²$A^2$ back into our reduction formula:

Now, sum all the individual element matrix fields:

Pattern Recognition

When a matrix power formula contains a constant difference block like (A² - I)$(A^2 - I)$, treat it as an arithmetic progression step multiplier over successive matrix indices to bypass calculating high powers.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

Orthogonal Matrix Property: A · AT = I AT = A⁻¹$A \cdot A^{\mathrm{T}} = I \implies A^{\mathrm{T}} = A^{-1}$. Trace identity Tr(A+B) = Tr(A) + Tr(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 · AT = I$A \cdot A^{\mathrm{T}} = I$. Thus, AT = A⁻¹$A^{\mathrm{T}} = A^{-1}$. Given constraint A² = AT A² = A⁻¹ A³ = 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$B$:

Since A³ = I$A^3 = I$, the full matrix expression reduces to:

Step 2: Trace Calculation

Sum of diagonal elements (Trace of matrix B$B$):

Pattern Recognition

Orthogonal algebraic identities (A³ = I$A^3 = I$) dramatically strip away high power terms. Do not attempt trigonometric computations unless absolute scalar matching forces it.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

For a linear system to have infinitely many solutions, the principal determinant must vanish:

Δ = 0$\Delta = 0$

Core Logic

Setting up the main matrix determinant:

Similarly, setting Δ₁ = 0$\Delta_1 = 0$ yields μ = 45$\mu = 45$.

Step 1: Express System Parametrically

With λ = -17, μ = 45$\lambda = -17, \mu = 45$, let's parameterize the equations. Let z = k$z = k$ (where k in Z$k \in \mathbb{Z}$). Solving the first two equations for x$x$ and y$y$ in terms of k$k$:

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 ≤ x + y + z ≤ 77$7 \le x + y + z \le 77$:

Since k ≡ 3 17$k \equiv 3 \pmod{17}$, 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.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants