Let M$M$ denote the set of all real matrices of order 3 × 3$3 \times 3$ and let S = -3, -2, -1, 1, 2$S = \{-3, -2, -1, 1, 2\}$. Let S₁ = A = [ aᵢⱼ ] in M: A = A^T and aᵢⱼ in S, i, j$$S_1 = \left\{A = \left[ a_{ij} \right] \in M: A = A^T \text{ and } a_{ij} \in S, \forall i, j \right\}$$ S₂ = A = [ aᵢⱼ ] in M: A = -A^T and aᵢⱼ in S, i, j$$S_2 = \left\{A = \left[ a_{ij} \right] \in M: A = -A^T \text{ and } a_{ij} \in S, \forall i, j \right\}$$ S₃ = A = [ aᵢⱼ ] in M: a₁₁ + a₂₂ + a₃₃ = 0 and aᵢⱼ in S, i, j$$S_3 = \left\{A = \left[ a_{ij} \right] \in M: a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j \right\}$$ If n(S₁ S₂ S₃) = 125α$n(S_1 \cup S_2 \cup S_3) = 125\alpha$, then α$\alpha$ equals.

Related Formula

Core Logic

Since |P⁻¹| · |P| = 1$|P^{-1}| \cdot |P| = 1$, the expression simplifies completely to the determinant of A - 2I$A - 2I$.

First, let us construct the matrix A - 2I$A - 2I$:

Now, evaluating the determinant:

Step 1: Finding Prime Factors

The number obtained is 69$69$. Let us find its prime factorization:

Both 3$3$ and 23$23$ are prime numbers. Their sum is:

Pattern Recognition

Whenever you encounter a matrix expression of the form P⁻¹AP - kI$P^{-1}AP - kI$, always factor out P⁻¹$P^{-1}$ and P$P$ to simplify it to |A - kI|$|A - kI|$. This saves tremendous time over computing matrix multiplications.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Related Formula

For infinite solutions of a system of equations, the main determinant D$D$ and auxiliary determinants D₁, D₂, D₃$D_1, D_2, D_3$ must all equal 0.

Core Logic

Since a, b, c$a, b, c$ are in A.P., we have 2b = a + c a - 2b + c = 0$2b = a + c \implies a - 2b + c = 0$. Comparing this identity with the line equation ax + by + c = 0$ax + by + c = 0$, we immediately find that the lines always pass through the fixed point configuration (1, -2)$(1, -2)$. Thus, P = (1, -2)$P = (1, -2)$.

Step 1: Evaluation of the Matrix for Infinite Solutions

Let us set the system determinant D = 0$D = 0$:

Now set D₁ = 0$D_1 = 0$ by substituting columns:

Thus, Q = (8, 6)$Q = (8, 6)$.

Step 2: Distance Metric Calculation

Using the distance formula between P(1, -2)$P(1, -2)$ and Q(8, 6)$Q(8, 6)$:

Pattern Recognition

A.P. coefficients inside a standard linear equation reveal a fixed point of concurrency by mapping matching coefficient components (1, -2, 1$1, -2, 1$).

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines

Related Formula

Core Logic

Evaluate f(-x)$f(-x)$:

Checking if f(-x)$f(-x)$ is the inverse by evaluating f(x) · f(-x)$f(x) \cdot f(-x)$:

Thus, Statement I is true.

Step 1: Checking Statement II

Evaluate the matrix multiplication f(x) · f(y)$f(x) \cdot f(y)$:

Apply standard trigonometric compound angle formulas:

Thus, Statement II is also true.

Step 2: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

This specific matrix represents a standard 2D rotation matrix embedded in 3D space. Rotation matrices naturally follow R(x)R(y) = R(x+y)$R(x)R(y) = R(x+y)$ (additive property of angles) and their inverse is always obtained by negating the angle R(-x) = R(x)⁻¹$R(-x) = R(x)^{-1}$.

Chapter Mix

Class 12 Maths: Matrices Class 11 Maths: Trigonometric Functions

Related Formula

|AB| = |A||B|$|AB| = |A||B|$ Trace(B$B$) = Σ bᵢᵢ$\sum b_{ii}$

Core Logic

Since B = [B₁, B₂, B₃]$B = [B_1, B_2, B_3]$, the matrix multiplication AB$AB$ effectively applies A$A$ to each column of B$B$:

Let this target matrix be C$C$. We know AB = C$AB = C$. Thus, taking determinants on both sides: |A| |B| = |C|$|A| |B| = |C|$

Step 1: Finding Determinants

Calculate determinant of A$A$:

Calculate determinant of C$C$. Since C$C$ is an upper triangular matrix, its determinant is simply the product of its main diagonal:

Therefore, 1 × |B| = 3 ⇒ |B| = 3$1 \times |B| = 3 \Rightarrow |B| = 3$. So α = 3$\alpha = 3$.

Step 2: Finding Matrix B

To find β$\beta$ (the trace of B$B$), we must explicitly find B = A⁻¹C$B = A^{-1}C$. Rather than finding the full inverse, manually solve A Bᵢ = Cᵢ$A B_i = C_i$: For B₁ = [x, y, z]^T$B_1 = [x, y, z]^T$: 2x+z=1$2x+z=1$, x+y=0$x+y=0$, x+z=0 ⇒ x=1, z=-1, y=-1$x+z=0 \Rightarrow x=1, z=-1, y=-1$. Thus B₁ = [1, -1, -1]^T$B_1 = [1, -1, -1]^T$. For B₂$B_2$: 2x+z=2$2x+z=2$, x+y=3$x+y=3$, x+z=0 ⇒ x=2, z=-2, y=1$x+z=0 \Rightarrow x=2, z=-2, y=1$. Thus B₂ = [2, 1, -2]^T$B_2 = [2, 1, -2]^T$. For B₃$B_3$: 2x+z=3$2x+z=3$, x+y=2$x+y=2$, x+z=1 ⇒ x=2, z=-1, y=0$x+z=1 \Rightarrow x=2, z=-1, y=0$. Thus B₃ = [3, 0, -1]^T$B_3 = [3, 0, -1]^T$.

Step 3: Calculating Trace and Final Output

Constructing Matrix B:

The diagonal elements are 1, 1, -1$1, 1, -1$. Trace β = 1 + 1 - 1 = 1$\beta = 1 + 1 - 1 = 1$.

Finally compute α³ + β³$\alpha^3 + \beta^3$:

Pattern Recognition

When given AXᵢ = Yᵢ$AX_i = Y_i$ for multiple columns, they collectively form A X = Y$A X = Y$. Using |A||X| = |Y|$|A||X| = |Y|$ bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix.

Chapter Mix

Class 12 Maths: Matrices

Related Formula

|kA| = kⁿ |A|$|kA| = k^n |A|$ Where A$A$ is an n × n$n \times n$ matrix, and k$k$ is a scalar.

Core Logic

Find the determinant of the 3 × 3$3 \times 3$ matrix A$A$:

Given the condition |2A|³ = 2²¹$|2A|^3 = 2^{21}$. Since A$A$ is a 3 × 3$3 \times 3$ matrix, applying the scalar property |kA| = k³|A|$|kA| = k^3|A|$:

Substitute this back into the original condition:

Taking the cube root of both sides: |A| = 2⁴ = 16$|A| = 2^4 = 16$

Step 1: Solve the Diophantine Equation

We have:

Since α$\alpha$ and β$\beta$ are integers, their sum and difference must also be integers. Also, (α + β)$(\alpha + \beta)$ and (α - β)$(\alpha - \beta)$ must share the same parity (both even or both odd) because their sum is 2α$2\alpha$ (an even number).

Since their product is 16, the only valid integer factor pairs of 16 that share the same parity are (8, 2)$(8, 2)$ and (-8, -2)$(-8, -2)$ and (4, 4)$(4, 4)$ and (-4, -4)$(-4, -4)$.

Case 1: (α + β) = 8$(\alpha + \beta) = 8$ and (α - β) = 2$(\alpha - \beta) = 2$ Adding them gives 2α = 10 ⇒ α = 5$2\alpha = 10 \Rightarrow \alpha = 5$. Thus β = 3$\beta = 3$.

Case 2: (α + β) = 4$(\alpha + \beta) = 4$ and (α - β) = 4$(\alpha - \beta) = 4$ Adding them gives 2α = 8 ⇒ α = 4$2\alpha = 8 \Rightarrow \alpha = 4$. Thus β = 0$\beta = 0$.

Looking at the options provided (3, 5, 17, 9), the value α = 5$\alpha = 5$ is listed.

Pattern Recognition

Extracting scalar multipliers from determinants always depends on the dimension n$n$ of the matrix. For Diophantine equations like x² - y² = k$x^2 - y^2 = k$, factoring into (x-y)(x+y)$(x-y)(x+y)$ and analyzing parity constraints restricts the solution space instantly.

Chapter Mix

Class 12 Mathematics: Determinants