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3 and 4 .Determinants and Matrices
hard
જો ${a^2} + {b^2} + {c^2} = - 2$ અને $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$ તો $f(x)$ એ . . . . બહુપદી ઘાતાંક છે .
A
$3$
B
$2$
C
$1$
D
$0$
(AIEEE-2005)
Solution
(b) Applying ${C_1} \to {C_1} + {C_2} + {C_3}$
$f(x) = \left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\1&{1 + {b^2}x}&{(1 + {c^2})x}\\1&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$,
$(\because {a^2} + {b^2} + {c^2} + 2 = 0$)
[Applying ${R_2} \to {R_2} – {R_1}$,${R_3} \to {R_3} – {R_1}$]
$ = \left| {\,\begin{array}{*{20}{c}}1&{(1 + {b^2})x}&{(1 + {c^2})x}\\0&{1 – x}&0\\0&0&{1 – x}\end{array}\,} \right| = {(1 – x)^2}.$
Hence degree of $f(x) = 2.$
Standard 12
Mathematics
