Among the statements : I: If |1 cos α cos β; cos α 1 cos γ; cos β cos γ 1| = |0 cos α cos β; cos α 0

Q: Why can't I just expand the determinant in Statement II? It looks like standard algebra.

You *could*, but it is a trap! Expanding a $3 \times 3$ determinant with polynomial entries leads to a cubic or quadratic expression that is incredibly easy to mess up. In JEE Advanced, time is your currency. Using the identity property by substituting $x=0$ and $x=1$ is the 'pro' strategy that saves time and minimizes calculation errors.

Q: In Statement I, I got stuck on the expansion. Is there a shortcut?

While there are properties of determinants (like row operations) that can simplify things, sometimes direct expansion is the most reliable path. Don't fear the algebra! Just be methodical. The key was realizing that the term $2 \cos \alpha \cos \beta \cos \gamma$ was the 'bridge' between the two sides.

Q: How do I know which values of $x$ to pick for Statement II?

Great question! You want to pick values that make the determinant easy to calculate. $x=0$ is almost always the best first choice because it kills off the $x^2$ and $x$ terms. $x=1$ is usually the second best because it simplifies the coefficients to small integers. Always look for the 'path of least resistance'.

Question

Among the statements :
I: If $\begin{vmatrix} 1 & \cos \alpha & \cos \beta \ \cos \alpha & 1 & \cos \gamma \ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \ \cos \alpha & 0 & \cos \gamma \ \cos \beta & \cos \gamma & 0 \end{vmatrix}$, then $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{3}{2}$, and
II : If $\begin{vmatrix} x^2 + x & x + 1 & x - 2 \ 2x^2 + 3x - 1 & 3x & 3x - 3 \ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q$, then $p^2 = 196q^2$,

Options:

(A) only II is true

(B) both are false

(C) both are true

(D) only I is true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are engaging in a duel with two distinct mathematical entities. In the world of JEE Advanced, a determinant is not merely a grid of numbers—it is a structure waiting to be unraveled.

We have two statements before us, and our mission is to determine their truth. Let us embark on this journey with precision and patience.

### The Trigonometric Dance

We begin with Statement I. We are presented with two determinants, $D_1$ and $D_2$. At first glance, they look intimidating, filled with trigonometric functions.

We define $D_1$ as the determinant with ones on the diagonal and cosines elsewhere. When we expand $D_1$ along the first row, we perform a systematic decomposition. As we expand this, we see a beautiful pattern emerge:

$$D_1 = 1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma$$

Now, look at $D_2$. It is a skew-symmetric-like structure with zeros on the diagonal. Expanding this is even more satisfying because the zeros simplify our work significantly:

$$D_2 = 2 \cos \alpha \cos \beta \cos \gamma$$

Here is the moment of truth. The problem states $D_1 = D_2$. When we equate them, the term $2 \cos \alpha \cos \beta \cos \gamma$ appears on both sides and cancels out.

We are left with the elegant result: $1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma = 0$, which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Statement I claims this sum is $\frac{3}{2}$. Since $1 
eq \frac{3}{2}$, we can confidently declare: **Statement I is false.**

### The Polynomial Trap

Now, we turn our attention to Statement II. We have a determinant $f(x)$ filled with polynomials in $x$, and we are told it equals $px + q$. Many students would immediately try to expand this determinant in terms of $x$.

Do not fall into this trap! Expanding a $3 	imes 3$ determinant with polynomial entries is a recipe for a calculation error. Instead, we use the power of the 'Identity'. Since the equation holds for all $x$, we can choose values that make our lives easier.

First, let $x = 0$. The term $px$ vanishes, leaving us with $q = f(0)$. Plugging $x=0$ into the matrix, we get a simple numerical determinant:

$$q = \begin{vmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 3 & -1 & -1 \end{vmatrix} = -12$$

Next, we need $p$. Let $x = 1$. This gives us $p + q = f(1)$. Substituting $x=1$ into the matrix yields:

$$p + q = \begin{vmatrix} 2 & 2 & -1 \ 4 & 3 & 0 \ 6 & 1 & 1 \end{vmatrix} = 12$$

Since we already know $q = -12$, we solve for $p$: $p - 12 = 12$, which implies $p = 24$.

### The Final Verdict

We have our values: $p = 24$ and $q = -12$. Statement II asks us to verify if $p^2 = 196q^2$.

Let us calculate the left side: $p^2 = (24)^2 = 576$. Now the right side: $196q^2 = 196(-12)^2 = 196 	imes 144 = 28224$.

Clearly, $576 
eq 28224$. Thus, **Statement II is also false.**

In this problem, we learned that the most complex-looking expressions often yield to simple, strategic choices. Both statements were false, but the journey to that truth was a masterclass in algebraic strategy.

Select Answer:

Visualized Solution

Analyzing the Statements

Setting up Statement I

Expansion of $D_{1}$

Expansion of $D_{2}$

Comparing $D_{1}$ and $D_{2}$

Verifying Statement I

Setting up Statement II

Substituting $x = 0$

Evaluating the Determinant for $q$

Substituting $x = 1$

Evaluating the Determinant for $p + q$

Solving for $p$

Checking the Condition $p^{2} = 196 q^{2}$

Final Answer

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Trigonometric Dance

The Polynomial Trap

The Final Verdict

Similar Questions

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Let $m$ and $M$ be respectively the minimum and maximum values of $cos^{2} x 1 + cos^{2} x cos^{2} x 1 + sin^{2} x sin^{2} x sin^{2} x sin 2 x sin 2 x 1 + sin 2 x$ . Then the ordered pair $(m, M)$ is equal to :

If $α \neq = a, β \neq = b, γ \neq = c$ and $α a a b β b c c γ = 0$ , then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to:

If $a - b - c 2 b 2 c 2 a b - c - a 2 c 2 a 2 b c - a - b = (a + b + c) (x + a + b + c)^{2}$ , $x \neq = 0$ and $a + b + c \neq = 0$ , then $x$ is equal to:

Select Answer:

Visualized Solution

Analyzing the Statements

Setting up Statement I

Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​

Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2​

Comparing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2​

Verifying Statement I

Setting up Statement II

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Evaluating the Determinant for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Evaluating the Determinant for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p+q

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Checking the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2=196q2