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3 and 4 .Determinants and Matrices
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If $a, b, c$ are all different from zero and $\left| {\begin{array}{*{20}{c}} {1 + a}&1&1\\ 1&{1 + b}&1\\ 1&1&{1 + c} \end{array}} \right| = 0$ , then the value of $a^{-1} + b^{-1} + c^{-1}$ is
A
$abc$
B
$a^{-1}\, b^{-1}\, c^{-1}$
C
$-a-b-c$
D
$-1$
Solution
$C_1 \rightarrow C_1 – C_2 \, \,and\, \, C_2 \rightarrow C_2 – C_3 \, \,and\, \,$ then open by $R_1$ to get $ab + abc + ac + bc = 0;$ divided by $abc$
Standard 12
Mathematics
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