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: Expand the determinant directly along the first column.
For a 3 by 3 matrix, expanding along column 1 gives $\Delta = 1 \cdot \begin{vmatrix} q & r+p \\ r & p+q \end{vmatrix} - 1 \cdot \begin{vmatrix} p & q+r \\ r & p+q \end{vmatrix} + 1 \cdot \begin{vmatrix} p & q+r \\ q & r+p \end{vmatrix}$.

Step 2: Work out each 2 by 2 minor.
First minor: $q(p+q) - r(r+p) = pq + q^2 - r^2 - rp$. Second minor: $p(p+q) - r(q+r) = p^2 + pq - rq - r^2$. Third minor: $p(r+p) - q(q+r) = pr + p^2 - q^2 - qr$.

Step 3: Combine the three minors with their signs.
$\Delta = (pq + q^2 - r^2 - rp) - (p^2 + pq - rq - r^2) + (pr + p^2 - q^2 - qr)$. Group the terms: the $q^2$ terms give $q^2 - q^2 = 0$, the $p^2$ terms give $-p^2 + p^2 = 0$, the $r^2$ terms give $-r^2 + r^2 = 0$. The remaining terms are $pq - rp + rq - pq + pr - qr$, and pairing $pq$ with $-pq$, $-rp$ with $pr$, and $rq$ with $-qr$ cancels everything.

Final Answer:
Every term cancels out, so the determinant reduces to zero regardless of the values chosen for $p$, $q$, and $r$.
\[ \boxed{\Delta = 0} \]