If {a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0 s | vedclass

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(B) Given the determinant $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a}&1&1\1&{1 + b}&1\1&1&{1 + c}\end{array}} ight| = \lambda $. Applying col

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$abc$

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(B) Given the determinant $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a}&1&1\1&{1 + b}&1\1&1&{1 + c}\end{array}} ight| = \lambda $. Applying column operations ${C_2} 	o {C_2} - {C_1}$ and ${C_3} 	o {C_3} - {C_1}$,we get: $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a}&{-a}&{-a}\1&b&0\1&0&c\end{array}} ight| = \lambda $. Expanding along the third row $({R_3})$: $1(0 - (-ab)) + 0 + c((1+a)b - (-a)) = \lambda $ $ab + c(b + ab + a) = \lambda $ $ab + bc + abc + ac = \lambda $ $ab + bc + ca + abc = \lambda \quad \dots (i)$ Given that ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$,we have: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \implies \frac{bc + ac + ab}{abc} = 0$. This implies $ab + bc + ca = 0$. Substituting this into equation $(i)$,we get $\lambda = 0 + abc = abc$. Thus,the value of $\lambda$ is $abc$.

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $,then the value of $\lambda $ is

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