Question

The determinant of the matrix \( \begin{pmatrix} 1 & p & q+r \\ 1 & q & r+p \\ 1 & r & p+q \end{pmatrix} \) is ______.

• A. \( pqr \)
• B. \( p + q + r \)
• C. \( pq + qr + rp \)
• D. \( 0 \)

Hint:

Check whether adding two of the columns produces a column that is a multiple of another column.

Answer 1

The Correct Option is D

Step 1: Look for a pattern in the columns before expanding anything.
Column 1 is \( (1, 1, 1)^T \), column 2 is \( (p, q, r)^T \), and column 3 is \( (q+r, r+p, p+q)^T \). Adding columns 2 and 3 entrywise gives \( p + (q+r) = p+q+r \) in row 1, \( q + (r+p) = p+q+r \) in row 2, and \( r + (p+q) = p+q+r \) in row 3. So column 2 plus column 3 is the constant value \( p+q+r \) repeated in every row.

Step 2: Turn that pattern into a column operation.
Replace column 3 with \( C_3 \to C_3 + C_2 \). Every entry of the new column 3 becomes \( p+q+r \), the same number in all three rows. This does not change the value of the determinant, since adding a multiple of one column to another leaves the determinant unchanged.

Step 3: Spot the linear dependence.
The new column 3 is just \( (p+q+r) \) times column 1, since column 1 is \( (1,1,1)^T \). A matrix with one column equal to a scalar multiple of another column has linearly dependent columns, and its determinant is always zero.

Final Answer:
Because column 3 becomes a multiple of column 1 after a determinant-preserving operation, the determinant is exactly zero for any values of \( p, q, r \).
\[ \boxed{\Delta = 0} \]