Question
If $\left|\begin{array}{lll}-2 a & a+b & c+a \\ b+a & -2 b & b+c \\ c+a & c+b & -2 c\end{array}\right|=k(b+c)(c+a)(a+b)$, then $k$ is equal to(a) 1(b) 2(c) 6(d) 4
Step 1
Step 1: We are given the determinant \[ \left|\begin{array}{lll} -2 a & a+b & c+a \\ b+a & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right| \] and we are told that it equals $k(b+c)(c+a)(a+b)$. Show more…
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