Solve Delta=left|begin{array}{lll}1 & a & b c 1 & b & c a 1 & c & a bend{arr

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3 and 4 .Determinants and Matrices

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Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

A

$(a-b)(b+c)(c-a)$

B

$(a-b)(b-c)(c-a)$

C

$(a+b+c)(b-c)(c-a)$

D

$(a+b)(b-c)(c-a)$

Solution

Solution Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1},$ we get

$\Delta=\left|\begin{array}{ccc}
1 & a & b c \\
0 & b-a & c(a-b) \\
0 & c-a & b(a-c)
\end{array}\right|$

Taking factors $(b-a)$ and $(c-a)$ common from $R_{2}$ and $R_{y}$ respectively, we get

$\Delta=(b-a)(c-a)\left|\begin{array}{ccc}
1 & a & b c \\
0 & 1 & -c \\
0 & 1 & -b
\end{array}\right|$

$=(b-a)(c-a)[(-b+c)]$ (Expanding along first column)

$=(a-b)(b-c)(c-a)$

Standard 12

Mathematics

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