Question

The value of the determinant \( \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ p & q & r+1 \end{vmatrix} \) is equal to:

• A. \( q - p \)
• B. \( q + p \)
• C. \( q \)
• D. \( p \)
• E. \( 0 \)

Hint:

Whenever you see two rows that are almost identical, subtract them immediately. This "Zero-Creation" strategy is the fastest way to solve competitive exam questions involving determinants.

Answer 1

The Correct Option is A

Concept: The determinant of a matrix can be simplified using elementary row or column operations. A fundamental property is that adding or subtracting a multiple of one row from another does not alter the determinant's value. Creating zeros in a row or column significantly simplifies the final expansion process.

Step 1: Applying Row Operations to simplify the third row.
To make the determinant easier to calculate, we perform the operation \( R_3 \rightarrow R_3 - R_2 \). This targets the variables \( p \) and \( q \) in the third row: \[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ p - p & q \\ - q & (r+1) - r \end{vmatrix} \] Simplifying the terms, we get: \[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ 0 & 0 & 1 \end{vmatrix} \]

Step 2: Expanding the determinant along the third row (\( R_3 \)).
Expanding along a row with the most zeros is the most efficient path. Using the signs for expansion \( (+, -, +) \): \[ \Delta = 0 \cdot \begin{vmatrix} 1 & 1 q & r \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & 1 p & r \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 p & q \end{vmatrix} \] Now, computing the resulting \( 2 \times 2 \) determinant: \[ \Delta = 1 \cdot \left( (1 \times q) - (1 \times p) \right) = q - p \]