Question ID: #791
Among the statements:
I: If $\left|\begin{matrix}1&\cos \alpha&\cos \beta\\ \cos \alpha&1&\cos \gamma\\ \cos \beta&\cos \gamma&1\end{matrix}\right|=\left|\begin{matrix}0&\cos \alpha&\cos \beta\\ \cos \alpha&0&\cos \gamma\\ \cos \beta&\cos \gamma&0\end{matrix}\right|$, then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$.
II: If $\left|\begin{matrix}x^{2}+x&x+1&x-2\\ 2x^{2}+3x-1&3x&3x-3\\ x^{2}+2x+3&2x-1&2x-1\end{matrix}\right|=px+q$, then $p^{2}=196q^{2}$.
is:
- (1) both are false
- (2) only II is true
- (3) both are true
- (4) only I is true
Solution:
**For Statement I:**
Let $x = \cos \alpha, y = \cos \beta, z = \cos \gamma$.
Given equation:
$$ \left|\begin{matrix}1&x&y\\ x&1&z\\ y&z&1\end{matrix}\right|=\left|\begin{matrix}0&x&y\\ x&0&z\\ y&z&0\end{matrix}\right| $$
Expanding LHS:
$$ 1(1-z^2) – x(x-yz) + y(xz-y) $$
$$ = 1 – z^2 – x^2 + xyz + xyz – y^2 $$
$$ = 1 – (x^2 + y^2 + z^2) + 2xyz $$
Expanding RHS:
$$ 0 – x(0-yz) + y(xz-0) $$
$$ = xyz + xyz = 2xyz $$
Equating LHS and RHS:
$$ 1 – (x^2 + y^2 + z^2) + 2xyz = 2xyz $$
$$ 1 – (x^2 + y^2 + z^2) = 0 $$
$$ x^2 + y^2 + z^2 = 1 $$
Substituting back:
$$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $$
The statement claims the sum is $\frac{3}{2}$. Thus, **Statement I is False**.
**For Statement II:**
Let $\Delta(x) = px + q$.
$$ \left|\begin{matrix}x^{2}+x&x+1&x-2\\ 2x^{2}+3x-1&3x&3x-3\\ x^{2}+2x+3&2x-1&2x-1\end{matrix}\right| = px + q $$
Put $x = 0$ to find $q$:
$$ q = \left|\begin{matrix}0&1&-2\\ -1&0&-3\\ 3&-1&-1\end{matrix}\right| $$
$$ q = 0(0-3) – 1(1+9) – 2(1-0) $$
$$ q = -10 – 2 = -12 $$
Put $x = 1$ to find $p+q$:
$$ p+q = \left|\begin{matrix}2&2&-1\\ 4&3&0\\ 6&1&1\end{matrix}\right| $$
$$ p+q = 2(3-0) – 2(4-0) – 1(4-18) $$
$$ p+q = 6 – 8 + 14 = 12 $$
Since $q = -12$:
$$ p – 12 = 12 \Rightarrow p = 24 $$
Check the condition $p^2 = 196q^2$:
$$ LHS = p^2 = 24^2 = 576 $$
$$ RHS = 196(-12)^2 = 196(144) $$
Clearly $576 \ne 196 \times 144$.
Thus, **Statement II is False**.
Ans. (1)
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