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3 and 4 .Determinants and Matrices
medium
बिना प्रसरण किए और सारणिकों के गुणधर्मो का प्रयोग करके सिद्ध कीजिए।
$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$
Option A
Option B
Option C
Option D
Solution
$\Delta=\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$
By applying $C_{3} \rightarrow C_{3}+C_{2} .$ We have:
$\Delta=\left|\begin{array}{lll}1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a\end{array}\right|$
Here. Two columns $C_{1}$ and $C_{3}$ are proportional.
$\therefore \Delta=0$
Standard 12
Mathematics