Let A= bmatrix 2 & 0 & 1 1 & 1 & 0 1 & 0 & 1 bmatrix$A=\begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$, B=[B₁, B₂, B₃]$B=[B_{1}, B_{2}, B_{3}]$, where B₁, B₂, B₃$B_{1}, B_{2}, B_{3}$ are column matrices, and AB₁= bmatrix 1 0 0 bmatrix$AB_{1}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, AB₂= bmatrix 2 3 0 bmatrix$AB_{2}=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$, AB₃= bmatrix 3 2 1 bmatrix$AB_{3}=\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$. If α=|B|$\alpha=|B|$ and β$\beta$ is the sum of all the diagonal elements of B$B$, then α³+β³$\alpha^{3}+\beta^{3}$ is equal to:

Related Formula

For a 2 × 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states:

Also, |adj(M)| = |M|ⁿ⁻¹$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 × 2$2 \times 2$ matrix, |adj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.

Core Logic

Using the characteristic equation A² - Tr(A)A + (A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$: Comparing with A² - 4A + 2I = O$A^2 - 4A + 2I = O$: Tr(A) = 4 ⇒ α + 2 = 4 ⇒ α = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$

Comparing with B² - 3B + I = O$B^2 - 3B + I = O$: Tr(B) = 3 ⇒ 1 + β = 3 ⇒ β = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$

Step 1: Compute A³$A^3$ and B³$B^3$ via reduction

A² = 4A - 2I$A^2 = 4A - 2I$

Similarly, B = bmatrix 1 & 1 1 & 2 bmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ B² = 3B - I$B^2 = 3B - I$

Step 2: Difference and Determinant

Step 3: Final Answer

For a 2 × 2$2 \times 2$ matrix, |adj(M)| = |M|$|\operatorname{adj}(M)| = |M|$. Thus, (adj(A³ - B³)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$.

Pattern Recognition

Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M³$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic.

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Class 12 Maths: Matrices

Related Formula

(provided N² = 0$N^2 = 0$, meaning N$N$ is nilpotent of order 2).

Core Logic

Evaluate powers of matrix A$A$ to detect a pattern.

Let N = bmatrix2 & -4 1 & -2 bmatrix$N = \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}$. Check N²$N^2$:

Step 1: Compute A^15

Since N² = 0$N^2 = 0$, Aⁿ = I + nN$A^n = I + nN$.

Step 2: Evaluate Matrix Operation

Given (A¹⁵ + B) bmatrixx y bmatrix = bmatrix0 0 bmatrix$(A^{15} + B)\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$:

This yields a single independent equation:

Checking the options, for x=11, y=2$x=11, y=2$, 2(11) = 11(2) 22 = 22$2(11) = 11(2) \implies 22 = 22$. Thus, Option 3 satisfies the condition.

Pattern Recognition

Whenever you need a high power of a 2× 2$2\times 2$ matrix, check its trace and determinant. If |A-λ I|=0$|A-\lambda I|=0$ yields λ=1,1$\lambda=1,1$, then A$A$ can be written as I+N$I+N$ where N$N$ is nilpotent. This bypasses lengthy matrix multiplication.

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Class 12 Maths: Matrices and Determinants

Related Formula

Core Logic

Evaluate the main determinant Δ$\Delta$ to 0$0$ to find α$\alpha$. Then quickly verify Δₓ ≠ 0$\Delta_x \neq 0$ to confirm it produces no solution.

Step 1: Calculate Delta

Expand along the first row:

Step 2: Verification of Delta_x

For α = 19$\alpha = 19$:

Since Δ = 0$\Delta = 0$ and Δₓ ≠ 0$\Delta_x \neq 0$, the system has no solution for α = 19$\alpha = 19$.

Pattern Recognition

Unless the problem specifically asks to check between infinite and no solution, setting the coefficient determinant Δ = 0$\Delta = 0$ directly yields the unique required parameter value.

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Class 12 Maths: Matrices and Determinants

Related Formula

|AB| = |A||B|$|AB| = |A||B|$ |Aⁿ| = |A|ⁿ$|A^n| = |A|^n$

Core Logic

Given A = bmatrix2 & 3 3 & 5 bmatrix$A = \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix}$.

The determinant |A| = (2)(5) - (3)(3) = 10 - 9 = 1$|A| = (2)(5) - (3)(3) = 10 - 9 = 1$.

We need to find the determinant of X = A²⁰²⁵ - 3A²⁰²⁴ + A²⁰²³$X = A^{2025} - 3A^{2024} + A^{2023}$. Factor out the lowest power of A:

Step 1: Determinant Factoring

Using properties of determinants:

Since |A| = 1$|A| = 1$, |A|²⁰²³ = 1$|A|^{2023} = 1$.

Therefore, |X| = |A² - 3A + I|$|X| = |A^2 - 3A + I|$.

Step 2: Matrix Computation

Calculate A²$A^2$:

Calculate -3A$-3A$:

Assemble A² - 3A + I$A^2 - 3A + I$:

Step 3: Final Determinant

Pattern Recognition

High-powered matrix polynomials simplify radically by factoring out the lowest power. Since |A| = 1$|A| = 1$, the leading coefficient collapses completely, leaving only a basic A²$A^2$ evaluation.

Chapter Mix

Class 12 Maths: Matrices and Determinants

Related Formula

Core Logic

Since A$A$ is a 3 × 3$3 \times 3$ skew-symmetric matrix, let A = bmatrix 0 & a & b -a & 0 & c -b & -c & 0 bmatrix$A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.

Given A bmatrix 1 -1 0 bmatrix = bmatrix 3 3 2 bmatrix$A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$:

Wait, (-a)(1) + (0)(-1) + (c)(0) = -a$(-a)(1) + (0)(-1) + (c)(0) = -a$. But wait, -b(1) - c(-1) + 0 = -b+c$-b(1) - c(-1) + 0 = -b+c$.

Equating the result: -a = 3 a = -3$-a = 3 \implies a = -3$ -b + c = 2 (1)$-b + c = 2 \quad \dots(1)$

Step 1: Finding Matrix A Using Second Given Condition

Given A² bmatrix 1 -1 0 bmatrix = bmatrix -3 19 -24 bmatrix$A^2 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$.

We know A bmatrix 1 -1 0 bmatrix = bmatrix 3 3 2 bmatrix$A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$. Therefore, A bmatrix 3 3 2 bmatrix = bmatrix -3 19 -24 bmatrix$A \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$.

Substitute a = -3$a = -3$: 3(-3) + 2b = -3 -9 + 2b = -3 2b = 6 b = 3$3(-3) + 2b = -3 \implies -9 + 2b = -3 \implies 2b = 6 \implies b = 3$

Use equation (1): -b + c = 2 -3 + c = 2 c = 5$-b + c = 2 \implies -3 + c = 2 \implies c = 5$.

So, a = -3, b = 3, c = 5$a = -3, b = 3, c = 5$. Matrix A = bmatrix 0 & -3 & 3 3 & 0 & 5 -3 & -5 & 0 bmatrix$A = \begin{bmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{bmatrix}$.

Step 2: Calculate Determinant of A + I

Step 3: Calculating Adjoint Target Expression

Target: (adj(2adj(A + I)))$\det(\text{adj}(2\text{adj}(A + I)))$. Let M = A+I$M = A+I$. For a 3 × 3$3 \times 3$ matrix, |adj(X)| = |X|²$|\text{adj}(X)| = |X|^2$.

We know |kY| = k³|Y|$|kY| = k^3|Y|$ for 3 × 3$3 \times 3$. So |2adj M| = 2³ |adj M| = 8|M|²$|2\text{adj} M| = 2^3 |\text{adj} M| = 8|M|^2$.

Substitute |M| = 44$|M| = 44$:

Comparing to 2^α · 3^β · 11^γ$2^\alpha \cdot 3^\beta \cdot 11^\gamma$: α = 14, β = 0, γ = 4$\alpha = 14, \beta = 0, \gamma = 4$.

α + β + γ = 14 + 0 + 4 = 18$\alpha + \beta + \gamma = 14 + 0 + 4 = 18$.

Pattern Recognition

Skew-symmetric matrices only have 3 unknowns. Matrix multiplication acts linearly; A² x = A(Ax)$A^2 x = A(Ax)$ provides immediate simultaneous equations without having to compute the heavy A²$A^2$ explicitly. Determinant adjoint loops always reduce to |A|(n-1)^k$|A|^{(n-1)^k}$ modulated by constant pullouts.

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Class 12 Maths: Matrices and Determinants