Let A$A$ be a 3 × 3$3 \times 3$ matrix such that |adj(adj(adj A))| = 81.$$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 81.$$ If S = n in Z : (|adj(adj A)|)((n - 1)²)/(2) = |A|3n² - 5n - 4$S = \left\{n \in \mathbb{Z} : \left(|\operatorname{adj}(\operatorname{adj} A)|\right)^{\frac{(n - 1)^2}{2}} = |A|^{3n^2 - 5n - 4}\right\}$, then Σn in S |An² + n|$\sum_{n \in S} |A^{n^2 + n}|$ is equal to

Related Formula

For a skew-symmetric matrix A$A$, A^T = -A$A^T = -A$. (XY)^T = Y^T X^T$(XY)^T = Y^T X^T$, and (X⁻¹)^T = (X^T)⁻¹$(X^{-1})^T = (X^T)^{-1}$.

Core Logic

Observe the matrix A$A$. aᵢⱼ = -aⱼᵢ$a_{ij} = -a_{ji}$ and diagonals are zero. So, A$A$ is a skew-symmetric matrix: A^T = -A$A^T = -A$.

We have B(I - A) = I + A B = (I + A)(I - A)⁻¹$B(I - A) = I + A \implies B = (I + A)(I - A)^{-1}$. We need to compute B^T B$B^T B$.

Step 1: Transpose Manipulation

Since A^T = -A$A^T = -A$:

Therefore, B^T = (I + A)⁻¹ (I - A)$B^T = (I + A)^{-1} (I - A)$.

Step 2: Trace Calculation

Now compute B^T B$B^T B$:

Since (I-A)$(I-A)$ and (I+A)$(I+A)$ commute, (I-A)(I+A) = I² - A² = (I+A)(I-A)$(I-A)(I+A) = I^2 - A^2 = (I+A)(I-A)$:

Thus, B^T B$B^T B$ is simply the 3 × 3$3 \times 3$ identity matrix I₃$I_3$. The sum of the diagonal elements is the trace of the identity matrix:

Pattern Recognition

The expression B = (I+A)(I-A)⁻¹$B = (I+A)(I-A)^{-1}$ is the classic Cayley transform, which maps any skew-symmetric matrix A$A$ exactly onto an orthogonal matrix B$B$. Orthogonal matrices intrinsically satisfy B^T B = I$B^T B = I$.

Chapter Mix

Class 12 Maths: Matrices and Determinants

Related Formula

Core Logic

Let A = pmatrix a₁ & b₁ a₂ & b₂ a₃ & b₃ pmatrix$A = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}$. A^T A = pmatrix a₁ & a₂ & a₃ b₁ & b₂ & b₃ pmatrix pmatrix a₁ & b₁ a₂ & b₂ a₃ & b₃ pmatrix$A^T A = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}$ A^T A = pmatrix a₁² + a₂² + a₃² & & b₁² + b₂² + b₃² pmatrix$A^T A = \begin{pmatrix} a_1^2 + a_2^2 + a_3^2 & \dots \\ \dots & b_1^2 + b_2^2 + b_3^2 \end{pmatrix}$ Trace = a₁² + a₂² + a₃² + b₁² + b₂² + b₃² = 5$a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 = 5$. We need to find the number of ways to pick 6$6$ elements from -2, -1, 0, 1, 2$\{-2, -1, 0, 1, 2\}$ such that the sum of their squares is 5$5$.

Step 1: Identifying Valid Squares

Available squares from set: 0²=0, (± 1)²=1, (± 2)²=4$0^2=0, (\pm 1)^2=1, (\pm 2)^2=4$. Possible sets of 6 squares that sum to 5: Case 1: 4, 1, 0, 0, 0, 0$4, 1, 0, 0, 0, 0$ Case 2: 1, 1, 1, 1, 1, 0$1, 1, 1, 1, 1, 0$

Step 2: Calculating Permutations for Case 1

Elements for 4$4$ can be ± 2$\pm 2$ (2 options). Elements for 1$1$ can be ± 1$\pm 1$ (2 options). Zeroes are fixed (1 option). Arrangements of 4, 1, 0, 0, 0, 0$\{4, 1, 0, 0, 0, 0\}$: No. of ways = (6!)/(1!1!4!) = 30$\frac{6!}{1!1!4!} = 30$. For each arrangement, element choices = 2 × 2 = 4$2 \times 2 = 4$. Total for Case 1 = 30 × 4 = 120$30 \times 4 = 120$ ways.

Step 3: Calculating Permutations for Case 2

Elements for 1$1$ can be ± 1$\pm 1$ (2 options each, so 2⁵$2^5$ choices). Zero is fixed. Arrangements of 1, 1, 1, 1, 1, 0$\{1, 1, 1, 1, 1, 0\}$: No. of ways = (6!)/(5!1!) = 6$\frac{6!}{5!1!} = 6$. For each arrangement, element choices = 2⁵ = 32$2^5 = 32$. Total for Case 2 = 6 × 32 = 192$6 \times 32 = 192$ ways. (Wait, following the source logic exactly: source states: "No of ways = (6!)/(4!) × 4 + 2 × (6!)/(5!) + 2 × (6!)/(4!) + 2 × (6!)/(3!2!)$= \frac{6!}{4!} \times 4 + 2 \times \frac{6!}{5!} + 2 \times \frac{6!}{4!} + 2 \times \frac{6!}{3!2!} $" Wait, the pdf says = 120 + 120 + 12 + 60 = 312$= 120 + 120 + 12 + 60 = 312$. Ah, expanding the 2⁵ = 32$2^5 = 32$ manually: they treated subsets of sign assignments differently, but 6 × 32 = 192$6 \times 32 = 192$. 192 + 120 = 312$192 + 120 = 312$. The final numerical answer matches perfectly).

Pattern Recognition

The trace of A^T A$A^T A$ is universally the sum of squares of all elements in matrix A (Frobenius norm squared). Transitioning immediately to an integer-sum-of-squares combinatorial problem sidesteps matrix multiplication.

Chapter Mix

Class 12 Maths: Matrices and Determinants Class 11 Maths: Permutations and Combinations

Related Formula

Core Logic

First, evaluate the integral for f(x)$f(x)$ by factoring out the highest power of x$x$ from the denominator.

Let t = 1x⁹ + 1x⁷ + 2 (dt)/(dx) = - 9x¹⁰ - 7x⁸$t = \frac{1}{x^{9}} + \frac{1}{x^{7}} + 2 \implies \frac{dt}{dx} = -\frac{9}{x^{10}} - \frac{7}{x^{8}}$

Step 1: Finding Constant and Derivative

Given f(1) = (1)/(4) (1)/(4) + C = (1)/(4) C = 0$f(1) = \frac{1}{4} \implies \frac{1}{4} + C = \frac{1}{4} \implies C = 0$

Now, differentiate using the quotient rule to find f'(x)$f'(x)$:

At x = 1$x = 1$:

Step 2: Evaluating Determinants

Matrix A$A$ becomes:

Given B = adj(adj A)$B = \text{adj}(\text{adj } A)$ and |B| = 81$|B| = 81$. Since A$A$ is a 3 × 3$3 \times 3$ matrix (n=3$n=3$):

Step 3: Finding Alpha

Thus, α² = 4$\alpha^{2} = 4$.

Pattern Recognition

When an integrand is a rational function with large powers of x$x$, dividing numerator and denominator by the highest power of x$x$ inside the bracket transforms the numerator into the exact derivative of the modified denominator.

Chapter Mix

Class 12 Maths: Matrices Class 12 Maths: Indefinite Integration

Related Formula

Where n$n$ is the order of the matrix.

Core Logic

Expand the matrix Q$Q$ given the condition qᵢⱼ = 2(i + j - 1) pᵢⱼ$q_{ij} = 2^{(i + j - 1)} p_{ij}$:

Step 1: Factoring from Rows

Take out common factors from each row: Row 1: 2¹$2^1$ Row 2: 2²$2^2$ Row 3: 2³$2^3$

Step 2: Factoring from Columns

Now take out common factors from each column: Col 1: 2⁰ = 1$2^0 = 1$ Col 2: 2¹ = 2$2^1 = 2$ Col 3: 2² = 4$2^2 = 4$

Step 3: Finding |P| and Target Value

Given |Q| = 2¹⁰$|Q| = 2^{10}$:

We need to find |adj(adj P)|$|\text{adj}(\text{adj } P)|$ for n=3$n=3$:

Pattern Recognition

When matrix elements are scaled by ki+j-c$k^{i+j-c}$, factoring rows and columns systematically pulls out kΣ row powers + Σ col powers$k^{\sum \text{row powers} + \sum \text{col powers}}$ cleanly. Always split the exponent index i+j$i+j$ visually into row operations and column operations.

Chapter Mix

Class 12 Maths: Determinants

Core Logic

Given:

• B = (I + A)⁻¹$B = (I + A)^{-1}$
• A + C = I A = I - C$A + C = I \implies A = I - C$

Substitute A$A$ into the first equation:

This implies B(2I - C) = I 2B - BC = I$B(2I - C) = I \implies 2B - BC = I$. Also, (2I - C)B = I 2B - CB = I$(2I - C)B = I \implies 2B - CB = I$. Therefore, 2B - BC = 2B - CB BC = CB$2B - BC = 2B - CB \implies BC = CB$.

Step 1: Matrix Evaluation

Since BC = CB$BC = CB$, we can substitute BC$BC$ into the given linear equation:

Step 2: Solve the Matrix Equation

Calculate the inverse: Determinant = (1)(2) - (-5)(-1) = 2 - 5 = -3$(1)(2) - (-5)(-1) = 2 - 5 = -3$.

Step 3: Final Answer

x₁ = 2$x_1 = 2$ and x₂ = -2$x_2 = -2$. Therefore, x₁ + x₂ = 2 - 2 = 0$x_1 + x_2 = 2 - 2 = 0$.

Pattern Recognition

If a matrix B$B$ is the inverse of a polynomial in C$C$ (like 2I - C$2I - C$), then B$B$ naturally commutes with C$C$. Therefore, BC = CB$BC = CB$ without having to know the elements of B$B$ or C$C$ independently.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants