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

Core Logic

To find A¹⁰⁰$A^{100}$, express A$A$ as I + M$I + M$: A = I + bmatrix 2 & -4 1 & -2 bmatrix$A = I + \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$. Let M = bmatrix 2 & -4 1 & -2 bmatrix$M = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$.

Calculate M²$M^2$ to check for nilpotency:

Thus, M² = 0$M^2 = 0$ (nilpotent matrix of index 2).

Execution

Expand A¹⁰⁰ = (I + M)¹⁰⁰$A^{100} = (I + M)^{100}$ using Binomial Theorem (valid since I$I$ and M$M$ commute):

Since M^k = 0$M^k = 0$ for k ≥ 2$k \geq 2$, all higher terms vanish.

Given the condition A¹⁰⁰ = 100B + I$A^{100} = 100B + I$, we compare the equations: I + 100M = 100B + I ⇒ B = M$I + 100M = 100B + I \Rightarrow B = M$.

We need the sum of all elements of B¹⁰⁰$B^{100}$. Since B = M$B = M$, B² = M² = 0$B^2 = M^2 = 0$, which implies B¹⁰⁰ = 0$B^{100} = 0$. The sum of all elements of a null matrix is 0$0$.

Pattern Recognition

When asked for extreme powers of a non-diagonal matrix, extract the identity matrix I$I$. The residual matrix M$M$ will almost inevitably be nilpotent (M²=0$M^2=0$ or M³=0$M^3=0$), collapsing the binomial expansion.

Chapter Mix

Class 12 Maths: Matrices and Determinants

Related Formula

Core Logic

For a system of 3 linear equations to have infinitely many solutions, the determinant of coefficients and all Cramer determinants must equal zero.

Step 1: Set up determinant equations

The determinant of coefficients is:

Now, set Δ₃ = 0$\Delta_3 = 0$:

Step 2: Solve for mu and compute the sum of squares

Substitute λ = -1$\lambda = -1$ into equation (1):

Now calculate the sum of squares:

Pattern Recognition

Whenever you need to solve for two variables in Cramer's theorem, identifying which determinant lacks the complex variable (like Δ₃$\Delta_3$ which lacks μ$\mu$) is the fastest way to solve for one variable independently.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

Core Logic

We use the given cubic matrix equation recursively to express the fifth power of matrix A$A$ solely in terms of quadratic and linear terms.

Step 1: Simplify the cubic matrix equation

The given equation is:

Multiply by matrix A$A$ to find the fourth power:

Step 2: Reduce the fourth power term

Substitute the expression for A³$\mathrm{A}^3$ into our formula for A⁴$\mathrm{A}^4$:

Multiply by matrix A$A$ to find the fifth power:

Step 3: Reduce the fifth power term and solve

Substitute the expression for A³$\mathrm{A}^3$ again:

Comparing this with A⁵ = α A² + β A + γ I$\mathrm{A}^5 = \alpha \mathrm{A}^2 + \beta \mathrm{A} + \gamma \mathrm{I}$, we find:

• α = 20$\alpha = 20$
• β = 24$\beta = 24$
• γ = -32$\gamma = -32$

Sum the coefficients:

Pattern Recognition

Cayley-Hamilton reduction: For any polynomial equation of a matrix, higher powers A^k$A^k$ can always be reduced down to polynomials of order less than the degree of the characteristic equation by recursive substitution.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

For a matrix satisfying A² = A$A^2 = A$ (Idempotent Matrix):

Core Logic

Given (A) = αβ + 6 = 0 αβ = -6$\det(A) = \alpha\beta + 6 = 0 \implies \alpha\beta = -6$ and α + β = 1$\alpha + \beta = 1$. Solving these gives α = 3, β = -2$\alpha = 3, \beta = -2$ (since α > 0$\alpha > 0$).

Step 1: Check Powers of A

Substitute values into A$A$:

Compute A²$A^2$:

Step 2: Expand Matrix Expression

Since A² = A$A^2 = A$, it follows that Aⁿ = A$A^n = A$ for all integers n ≥ 1$n \ge 1$.

Step 3: Construct the Final Matrix

Pattern Recognition

Whenever tr(A) = 1$\text{tr}(A) = 1$ and (A) = 0$\det(A) = 0$ for a 2 × 2$2 \times 2$ matrix, Cayley-Hamilton theorem gives A² - tr(A)A + (A)I = 0 A² = A$A^2 - \text{tr}(A)A + \det(A)I = 0 \implies A^2 = A$. Thus A$A$ is idempotent, simplifying polynomial expansions exponentially.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

Cramer's Rule for infinite solutions specifies that the main determinant and all component determinants must vanish:

Core Logic

Set the key system determinants to zero to form equations linking α$\alpha$ and β$\beta$, then isolate the constants.

Step 1: Set Main Determinant to Zero

Expand along the first row:

Step 2: Set Subsidiary Determinant to Zero

Using Δ₃ = 0$\Delta_3 = 0$ by substituting the constants vector into the third column:

Expand along the first row:

Step 3: Solve for Beta and Final Expression

Substitute α = -(19)/(9)$\alpha = -\frac{19}{9}$ into equation (1):

Now compute 22β - 9α$22\beta - 9\alpha$:

Pattern Recognition

Choosing Δ₃$\Delta_3$ over Δ₁$\Delta_1$ or Δ₂$\Delta_2$ eliminates β$\beta$ entirely because the variable parameters are localized in specific positions. This yields α$\alpha$ directly without requiring a coupled system solution.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants