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:
vmatrix 1 & x & y x & 1 & z y & z & 1 vmatrix = vmatrix 0 & x & y x & 0 & z y & z & 0 vmatrix$$ \begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix} = \begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix} $$
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:
vmatrix x² + x & x + 1 & x - 2 2x² + 3x - 1 & 3x & 3x - 3 x² + 2x + 3 & 2x - 1 & 2x - 1 vmatrix = px + q$$ \begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q $$
Put x = 0$x = 0$ to find q$q$:
q = vmatrix 0 & 1 & -2 -1 & 0 & -3 3 & -1 & -1 vmatrix = 0 - 1(1 - (-9)) - 2(1 - 0) = -10 - 2 = -12$$ q = \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} = 0 - 1(1 - (-9)) - 2(1 - 0) = -10 - 2 = -12 $$
Now put x = 1$x = 1$ to find p + q$p + q$:
p + q = vmatrix 2 & 2 & -1 4 & 3 & 0 6 & 1 & 1 vmatrix = 2(3 - 0) - 2(4 - 0) - 1(4 - 18) = 6 - 8 + 14 = 12$$ p + q = \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 0 \\ 6 & 1 & 1 \end{vmatrix} = 2(3 - 0) - 2(4 - 0) - 1(4 - 18) = 6 - 8 + 14 = 12 $$
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