Using property of determinants and without expanding, prove that: left|begin{array}{lll}1 & b c

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3 and 4 .Determinants and Matrices

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Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

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

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