Q.The maximum value of $\begin{vmatrix} 1 & 1 & 1 \\ 1 & (1 + \sin\theta) & 1 \\ 1 & 1 & (1 + \cos\theta) \end{vmatrix}$ is $\dfrac{1}{2}$.

Concept: Maximum Value Sine Cosine — The determinant simplifies to a product involving $\sin\theta$ and $\cos\theta$, whose maximum is found using the identity $\sin\theta\cos\theta \le \frac12$.

Step 1: Compute the determinant. Subtract row 1 from rows 2 and 3:

Step 2: Expanding along the first column gives: …

The problem asks for the maximum value of a $3 \times 3$ determinant that depends on $\theta$. At first glance, it looks messy — but the structure is actually quite neat. The matrix has many 1’s, with only two entries that vary: the $(2,2)$ entry is $1+\sin\theta$ and the $(3,3)$ entry is $1+\cos\theta$. This pattern suggests that subtracting rows or columns will create zeros and simplify the determinant dramatically.

The key insight: when a matrix has many repeated entries, row/column operations can reduce it to a much simpler form without changing the determinant’s value. Here, subtracting the first row from the second and third rows will turn most of the matrix into zeros, leaving only a tiny $2 \times 2$ determinant to evaluate.

Let’s work through it step by step.

1. Write the determinant

2. Simplify using row operations Subtract the first row from the second row ($R_2 \to R_2 - R_1$) and from the third row ($R_3 \to R_3 - R_1$). This does not change the determinant’s value.

Why does this work? Because subtracting a multiple of one row from another leaves the determinant unchanged. The first row stays intact; the second row becomes $(1-1,\; 1+\sin\theta-1,\; 1-1) = (0,\sin\theta,0)$, and the third becomes $(0,0,\cos\theta)$.

3. Evaluate the triangular determinant The matrix is now upper triangular (all entries below the main diagonal are zero). The determinant of a triangular matrix is simply the product of its diagonal entries. $D(\theta) = 1 \cdot \sin\theta \cdot \cos\theta = \sin\theta \cos\theta$ …

Method: Reducing a Parameter-Dependent Determinant to Triangular Form, Then Optimising

Use this whenever asked for the maximum/minimum of a determinant whose entries involve a variable like $\theta$.

Steps

Step 1: Spot the repeated-entry structure

Look for a pattern such as "mostly $1$'s, with only one or two entries that differ" — this signals that subtracting the first row from the others will create a wall of zeros.

Step 2: Apply row operations to reach triangular form

Perform $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ (these never change the determinant's value). If the structure is right, this reduces the matrix to upper-triangular form.

Step 3: Read the determinant as the product of the diagonal entries

For an upper-triangular matrix, $\det = (\text{diagonal entry }1)\times(\text{diagonal entry }2)\times\cdots$. This gives the determinant as an explicit trigonometric function of $\theta$. …

Common Mistakes

Mistake 1: Attempting to expand the $3\times3$ determinant directly by cofactors instead of using row operations

Why it's wrong: direct cofactor expansion here produces a messy trigonometric expression that's easy to mis-simplify; the matrix's repeated-$1$'s structure is a strong hint that $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ collapses it instantly to a triangular form. Correct approach: always check for a repeated-entry pattern first and use row/column operations to reach a triangular determinant before expanding anything.

Mistake 2: Claiming the maximum value of $\sin\theta\cos\theta$ is $1$ …