- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
hard
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
A
$0$
B
$abc$
C
$-abc$
D
None of these
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}
{1 + a}&1&1\\
1&{1 + b}&1\\
1&1&{1 + c}
\end{array}\,} \right| = \lambda $
Applying ${C_2} \to {C_2}- {C_1}$ and ${C_3} \to {C_3} – {C_1},$
$\left| {\,\begin{array}{*{20}{c}}{1 + a}&{ – a}&{ – a}\\1&b&0\\1&0&c\end{array}\,} \right|$
On expanding w.r.t. ${R_3}$,
$ab + bc + ca + abc = \lambda $…….$(i)$
Given, ${a^{ – 1}} + {b^{ – 1}} + {c^{ – 1}} = 0$
==> $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$
==> $ab + bc + ca = 0$
==> $\lambda = abc$, (From equation $(i)$).
Standard 12
Mathematics
