Let for any three distinct consecutive terms a, b, c$a, b, c$ of an A.P, the lines ax + by + c = 0$ax + by + c = 0$ be concurrent at the point P$P$ and Q (α, β)$Q (\alpha, \beta)$ be a point such that the system of equations x + y + z = 6$x + y + z = 6$, 2 x + 5 y + α z = β$2 x + 5 y + \alpha z = \beta$ and x + 2y + 3z = 4$x + 2y + 3z = 4$ has infinitely many solutions. Then (PQ)²$(PQ)^2$ is equal to

Related Formula

For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero: Δ ≠ 0$\Delta \neq 0$

Core Logic

Construct the determinant of coefficients:

Expanding the determinant yields:

Hence, n ≠ 1, 2$n \neq 1, 2$.

Step 1: Probability Calculation

The possible outcomes on rolling a fair die are n in 1, 2, 3, 4, 5, 6$n \in \{1, 2, 3, 4, 5, 6\}$. The favorable values of n$n$ for a unique solution are n = 3, 4, 5, 6$n = 3, 4, 5, 6$ (4 values). Therefore, the probability is:

Step 2: Required Sum

Sum = k + (3 + 4 + 5 + 6) = 4 + 18 = 22$= k + (3 + 4 + 5 + 6) = 4 + 18 = 22$.

Pattern Recognition

Sees unique solution condition for linear system Δ ≠ 0$\Delta \neq 0$. Exclude roots of coefficient determinant from sample space of die rolls.

Chapter Mix

Class 12 Maths: Matrices and Determinants Class 11 Maths: Probability

Related Formula

Solution of matrix system AX = B$AX = B$ is:

Core Logic

Multiply adj A$\operatorname{adj} A$ by B$B$:

Since X = ± (1)/(10) bmatrix 20 -10 10 bmatrix = ± bmatrix 2 -1 1 bmatrix$X = \pm \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \pm \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$.

Step 1: Evaluation of |x + y + z|

Therefore, |x + y + z| = 2$|x + y + z| = 2$.

Pattern Recognition

Direct matrix multiplication (adj A) B$(\operatorname{adj} A) B$ gives component ratios directly without computing |A|$|A|$ explicitly.

Chapter Mix

Class 12 Maths: Matrices and Determinants

Core Logic

Evaluate Statement I: Let x = α$x = \cos\alpha$, y = β$y = \cos\beta$, z = γ$z = \cos\gamma$. The given relation is:

Expanding the left-hand side (LHS): LHS = 1(1 - z²) - x(x - yz) + y(xz - y) = 1 - z² - x² + xyz + xyz - y² = 1 - (x² + y² + z²) + 2xyz$= 1(1 - z^2) - x(x - yz) + y(xz - y) = 1 - z^2 - x^2 + xyz + xyz - y^2 = 1 - (x^2 + y^2 + z^2) + 2xyz$ Expanding the right-hand side (RHS): RHS = 0 - x(0 - yz) + y(xz - 0) = xyz + xyz = 2xyz$= 0 - x(0 - yz) + y(xz - 0) = xyz + xyz = 2xyz$ Equating LHS and RHS: 1 - x² - y² - z² + 2xyz = 2xyz$1 - x^2 - y^2 - z^2 + 2xyz = 2xyz$ x² + y² + z² = 1$x^2 + y^2 + z^2 = 1$ This implies ²α + ²β + ²γ = 1$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, not (3)/(2)$\frac{3}{2}$. Thus, Statement I is false.

Step 1: Evaluate Statement II

We have:

Put x = 0$x = 0$ to find q$q$:

Now put x = 1$x = 1$ to find p + q$p + q$:

Since q = -12$q = -12$, p - 12 = 12 ⇒ p = 24$p - 12 = 12 \Rightarrow p = 24$. Check the relation p² = 196q²$p^2 = 196q^2$: p² = 24² = 576$p^2 = 24^2 = 576$ 196q² = 196(-12)² = 196(144) = 28224$196q^2 = 196(-12)^2 = 196(144) = 28224$ Clearly, 576 ≠ 28224$576 \neq 28224$. Thus, Statement II is false.

Step 2: Final Verdict

Both statements I and II are false.

Pattern Recognition

Instead of expanding 3 × 3$3 \times 3$ polynomial determinants algebraically, substituting boundary values (like x=0, x=1$x=0, x=1$) isolates q$q$ and p$p$ almost instantaneously.

Chapter Mix

Class 12 Maths: Determinants

Related Formula

Core Logic

Analyze the innermost term: adj(2A) = 2³⁻¹ adj(A) = 4adj(A)$\text{adj}(2A) = 2^{3-1} \text{adj}(A) = 4\text{adj}(A)$ So, A² · adj(2A) = 4A² · adj(A) = 4A(A · adj(A))$A^2 \cdot \text{adj}(2A) = 4A^2 \cdot \text{adj}(A) = 4A(A \cdot \text{adj}(A))$ Using A · adj(A) = |A|I₃$A \cdot \text{adj}(A) = |A|I_3$:

Step 1: Expand the Adjoint Layers

Substitute this back into the next layer: 3adj(24A) = 3 · (24)³⁻¹adj(A) = 3(24)²adj(A)$3\text{adj}(24A) = 3 \cdot (24)^{3-1}\text{adj}(A) = 3(24)^2\text{adj}(A)$

Step 2: Evaluate the Determinant

Now we need |adj(3(24)²adj(A))|$|\text{adj}(3(24)^2\text{adj}(A))|$. Let k = 3(24)²$k = 3(24)^2$. The expression is |adj(k adj(A))|$|\text{adj}(k \text{adj}(A))|$. Since adj(kX) = k²adj(X)$\text{adj}(kX) = k^2\text{adj}(X)$ for a 3 × 3$3 \times 3$ matrix, the matrix inside the determinant is k² adj(adj(A))$k^2 \text{adj}(\text{adj}(A))$. Wait, taking the determinant directly is simpler: |adj(X)| = |X|²$|\text{adj}(X)| = |X|^2$. Let X = k adj(A)$X = k \text{adj}(A)$. Then |X| = |k adj(A)| = k³ |adj(A)| = k³ |A|²$|X| = |k \text{adj}(A)| = k^3 |\text{adj}(A)| = k^3 |A|^2$. Thus, |adj(X)| = |X|² = (k³ |A|²)² = k⁶ |A|⁴$|\text{adj}(X)| = |X|^2 = (k^3 |A|^2)^2 = k^6 |A|^4$.

Step 3: Calculate Powers of 2 and 3

Substitute k = 3 · 24² = 3 · (2³ · 3)² = 3 · 2⁶ · 3² = 2⁶ · 3³$k = 3 \cdot 24^2 = 3 \cdot (2^3 \cdot 3)^2 = 3 \cdot 2^6 \cdot 3^2 = 2^6 \cdot 3^3$.

We know |A| = 6 = 2 · 3$|A| = 6 = 2 \cdot 3$. So |A|⁴ = 2⁴ · 3⁴$|A|^4 = 2^4 \cdot 3^4$. Total determinant = (2³⁶ · 3¹⁸) × (2⁴ · 3⁴) = 2⁴⁰ · 3²²$= (2^{36} \cdot 3^{18}) \times (2^4 \cdot 3^4) = 2^{40} \cdot 3^{22}$. Equating to 2^m · 3ⁿ$2^m \cdot 3^n$, we get m = 40$m = 40$ and n = 22$n = 22$.

Step 4: Final Sum

Pattern Recognition

Telescoping determinant operations demand rigorous sequential application of |kA| = kⁿ|A|$|kA| = k^n|A|$ and adj(kA) = kⁿ⁻¹adj(A)$\text{adj}(kA) = k^{n-1}\text{adj}(A)$. Collapsing A · adj(A)$A \cdot \text{adj}(A)$ into |A|I$|A|I$ early prevents chaotic polynomial matrix expressions.

Chapter Mix

Class 12 Maths: Matrices Class 12 Maths: Determinants

Related Formula

Using Cramer's Rule: If D = 0$D = 0$ and at least one of D₁, D₂, D₃$D_1, D_2, D_3$ is non-zero, the system has no solution. If D = 0$D = 0$ and D₁ = D₂ = D₃ = 0$D_1 = D_2 = D_3 = 0$, the system generally has infinitely many solutions.

Core Logic

First, evaluate the main determinant D$D$:

Setting D = 0 a = 8$D = 0 \implies a = 8$.

Now, evaluate D₃$D_3$ to check the consistency condition:

Setting D₃ = 0 3b - 42 = 0 b = 14$D_3 = 0 \implies 3b - 42 = 0 \implies b = 14$.

Step 1: Confirming Infinitely Many Solutions

For a=8$a=8$ and b=14$b=14$, evaluate D₁$D_1$ and D₂$D_2$ to ensure they are also zero.

Since D = D₁ = D₂ = D₃ = 0$D = D_1 = D_2 = D_3 = 0$, the system possesses infinitely many solutions.

Pattern Recognition

For a 3 × 3$3 \times 3$ system heavily relying on constants, D=0$D=0$ determines the parameter 'a' acting on the variable array, and D₃=0$D_3=0$ immediately forces the final constant 'b'.

Chapter Mix

Class 12 Maths: Matrices and Determinants