જો ab + bc + ca = 0 અને left| {,begin{array}{*{20}{c}}{a - x}&c&bc&{b - x}&ab&a&{c - x}end{array},} right| = 0, તો x ની કોઈ એક

Q: જો $ab + bc + ca = 0$ અને $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, તો $x$ ની કોઈ એક કિમત મેળવો.

Applying $ {c_1}→{c_1}+{c_2}+{c_3}$$(a+b+c-x)$ $\left| {\,\begin{array}{*{20}{c}} 1&c&b \\ 1&{b - x}&a \\ 1&a&{c - x} \end{array}\,} \right|\, = 0$ $ \Rightarrow (a+b+c-x)$ $[{(b-x)(c-x)-a^2}$ $ + c(a-c+x) + {b({a-b+x)}}] =0$ $ \Rightarrow (a+b+c-x)$ $[(bc-cx+bx+x^2-a^2 + ca - c^2 + cx +ab -b^2 +bx] =0$ $ \Rightarrow (a+b+c)[x^2 -(a^2+b^2+c^2)+ab+bc+ca]=0$ $(a+b+c-x)[x^2-(a^2 + b^2 +c^2] = 0$ $[ \because \ ab + bc + ca = 0]$ $x=a+b+c $ and $(a^2+b^2+c^2)^{1/2} $

MCQ

ગુજરાતી માધ્યમ

જો $ab + bc + ca = 0$ અને $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, તો $x$ ની કોઈ એક કિમત મેળવો.

A${({a^2} + {b^2} + {c^2})^{\frac{1}{2}}}$
B${\left[ {\frac{3}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
C${\left[ {\frac{1}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
D
એકપણ નહી.

Difficult

ધોરણ 12 સાયન્સ
ગણિત
3 and 4 . determinant and metrices
JEE

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