Question

The value of \[\begin{vmatrix} a+pd & a+qd & a+rd \\ p & q & r \\ d & d & d \end{vmatrix}\] is:

• A. 0
• B. \(-1\)
• C. 1
• D. \(p+q+r\)

Hint:

Split the first row as \(a(1,1,1) + d(p,q,r)\) and use linearity of the determinant; each resulting determinant has proportional or identical rows.

Answer 1

The Correct Option is A

Step 1: Write the first row as a combination of two simpler rows using linearity of the determinant along a row: \((a+pd,\ a+qd,\ a+rd) = a\cdot(1,1,1) + d\cdot(p,q,r)\).

Step 2: By linearity of the determinant in the first row, split the determinant into two determinants: \[\begin{vmatrix} a+pd & a+qd & a+rd \\ p & q & r \\ d & d & d \end{vmatrix} = a\begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ d & d & d \end{vmatrix} + d\begin{vmatrix} p & q & r \\ p & q & r \\ d & d & d \end{vmatrix}\]

Step 3: In the second determinant, row 1 and row 2 are identical (both equal \((p,q,r)\)), so that determinant is \(0\).

Step 4: In the first determinant, row 3 is \((d,d,d) = d\cdot(1,1,1)\), which is a scalar multiple of row 1. Rows that are proportional make the determinant \(0\), so this determinant is also \(0\).

Step 5: Therefore the whole expression is \(a\cdot 0 + d\cdot 0 = 0\).

\[\boxed{0}\]