If \( |\begin{matrix}1&1&1\\bc&ca&ab\\a(b+c)&b(c+a)&c(a+b)\end{matrix}|=k, \) then the value of k is:
(A) \(ab+bc+ca\)
(B) \( (ab+bc+ca)^{2} \)
(C) \(2(ab+bc+ca) \)
(D) 0
View Dynamic Solution & Explanation ▼
Correct Solution: Option D
Step-by-step Solution:
Solution
We are given:
\[
\Delta =
\begin{vmatrix}
1 & 1 & 1 \\
bc & ca & ab \\
a(b+c) & b(c+a) & c(a+b)
\end{vmatrix}
\]
Expand the third row terms:
\[
a(b+c) = ab + ac, \quad
b(c+a) = bc + ab, \quad
c(a+b) = ca + cb
\]
So the determinant becomes:
\[
\Delta =
\begin{vmatrix}
1 & 1 & 1 \\
bc & ca & ab \\
ab+ac & bc+ab & ca+bc
\end{vmatrix}
\]
Notice that the third row is the sum of the first two rows:
\[
(ab+ac, \, bc+ab, \, ca+bc) = (ab,bc,ca) + (ac,ab,bc)
\]
which is simply a linear combination of Row 1 and Row 2.
Therefore, the rows of the determinant are linearly dependent.
Hence:
\[
\Delta = 0
\]