Let pλ⁴ + qλ³ + rλ² + sλ + t = |λ²+3λ λ-1 λ+3; λ+1 -2λ λ-4; λ-3 λ+4 3λ| be an identity in λ, where p

Q: Why can't I just expand the determinant as a polynomial in $\lambda$ and compare coefficients?

You absolutely could, and it would be mathematically correct! However, in a JEE Advanced exam, time is your most precious currency. Expanding a $3 \times 3$ determinant with variable entries is a breeding ground for sign errors and takes minutes. The substitution method is a 'surgical strike'—it gets you the answer in seconds with minimal risk. Always look for the shortcut before you commit to the grind.

Q: Is it always valid to substitute $\lambda = 0$?

Yes, because the problem explicitly states this is an *identity*. An identity is not just a conditional equation; it is a statement that the two expressions are identical for all values of the variable. If it holds for all $\lambda$, it must hold for $\lambda = 0$. This is a powerful property of polynomials that you should keep in your toolkit.

Q: What if the constant term was not $t$ but something like $t+1$?

The logic remains identical! You would set $\lambda = 0$, solve the determinant to get a numerical value (let's say $K$), and then equate $t+1 = K$. The strategy of 'killing' the variable terms is universal, regardless of how the constant term is presented. Never let the algebraic packaging distract you from the core principle.

Question

Let $p\lambda^4 + q\lambda^3 + r\lambda^2 + s\lambda + t = \begin{vmatrix} \lambda^2 + 3\lambda & \lambda - 1 & \lambda + 3 \ \lambda + 1 & -2\lambda & \lambda - 4 \ \lambda - 3 & \lambda + 4 & 3\lambda \end{vmatrix}$ be an identity in $\lambda$, where $p, q, r, s$ and $t$ are constants. Then, the value of $t$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents a polynomial identity involving a determinant:

$$p\lambda^4 + q\lambda^3 + r\lambda^2 + s\lambda + t = \begin{vmatrix} \lambda^2 + 3\lambda & \lambda - 1 & \lambda + 3 \ \lambda + 1 & -2\lambda & \lambda - 4 \ \lambda - 3 & \lambda + 4 & 3\lambda \end{vmatrix}$$

Many students immediately attempt a full expansion, which is a common "trap of brute force." In JEE Advanced, the ability to identify an identity is your greatest weapon.

### The Anatomy of an Identity

An identity is a statement of equality that holds true for all values of $\lambda$. Because it is true for any $\lambda$, we can choose values that simplify the expression significantly.

The most powerful choice is $\lambda = 0$. By substituting $\lambda = 0$, we eliminate all terms containing $\lambda$ on the left side, leaving only the constant term $t$.

### The Determinant as a Function

When we set $\lambda = 0$, the determinant transforms from a complex variable expression into a simple numerical matrix. Substituting $\lambda = 0$ into the right-hand side yields:

$$	ext{RHS} = \begin{vmatrix} 0^2 + 3(0) & 0 - 1 & 0 + 3 \ 0 + 1 & -2(0) & 0 - 4 \ 0 - 3 & 0 + 4 & 3(0) \end{vmatrix}$$

Simplifying the entries, we obtain:

$$	ext{RHS} = \begin{vmatrix} 0 & -1 & 3 \ 1 & 0 & -4 \ -3 & 4 & 0 \end{vmatrix}$$

### The Execution

We now expand this determinant along the first row using the standard sign convention (plus, minus, plus):

$$t = 0 \cdot \begin{vmatrix} 0 & -4 \ 4 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -4 \ -3 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 0 \ -3 & 4 \end{vmatrix}$$

Calculating the minors:

1. The first term is $0$.
2. The second term is $1 \cdot ((1)(0) - (-3)(-4)) = 1 \cdot (0 - 12) = -12$.
3. The third term is $3 \cdot ((1)(4) - (-3)(0)) = 3 \cdot (4 - 0) = 12$.

Summing these values, we find:

$$t = 0 - 12 + 12 = 0$$

### Conclusion: The Topper's Mindset

The final result is **$t = 0$**. By leveraging the properties of an identity, we bypassed the need to calculate the coefficients $p, q, r,$ and $s$ entirely.

Always look for symmetry or special cases before committing to lengthy algebraic expansions. This strategic approach is the hallmark of a champion and will save you significant time during the examination.

Animated Solution for Mathematics - Matrices and Determinants: Let $p λ^{4} + q λ^{3} + r λ^{2} + s λ + t = λ^{2} + 3 λ λ + 1 λ - 3 λ - 1 - 2 λ λ + 4 λ + 3 λ - 4 3 λ$ be an identity in $λ$ , where $p, q, r, s$ and $t$ are constants. Then, the value of $t$ is .........

Enter Numerical Value:

Visualized Solution

Understanding the Identity

Strategy to Isolate $t$

Substituting $λ = 0$ in LHS

Substituting $λ = 0$ in RHS

Simplifying the Determinant

Expansion Strategy along Row 1

Evaluating the First and Second Terms

Evaluating the Third Term

Computing the $2 \times 2$ Minors

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Anatomy of an Identity

The Determinant as a Function

The Execution

Conclusion

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