નિશ્ચાયક left| {begin{array}{*{20}{c}}{{b_1} | vedclass

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(B) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\

Vedclass · Accepted Answer

$2 \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\{{a_2}}&{{b_2}}&{{c_2}}\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} ight|$

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(B) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.પ્રક્રિયા $C_1 ightarrow C_1 + C_2 + C_3$ લાગુ પાડતા:$\Delta = \left| {\begin{array}{*{20}{c}}{2(a_1+b_1+c_1)}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{2(a_2+b_2+c_2)}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{2(a_3+b_3+c_3)}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.$C_1$ માંથી $2$ સામાન્ય લેતા:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{a_1} + {b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{a_2} + {b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{a_3} + {b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.$C_1 ightarrow C_1 - C_2$ લાગુ પાડતા:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{b_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.$C_2 ightarrow C_2 - C_3$ લાગુ પાડતા:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1}}&{{a_1} + {b_1}}\{{b_2}}&{{c_2}}&{{a_2} + {b_2}}\{{b_3}}&{{c_3}}&{{a_3} + {b_3}}\end{array}} ight|$.$C_3 ightarrow C_3 - C_1$ લાગુ પાડતા:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1}}&{{a_1}}\{{b_2}}&{{c_2}}&{{a_2}}\{{b_3}}&{{c_3}}&{{a_3}}\end{array}} ight|$.$C_1$ અને $C_2$ ની અદલાબદલી કરતા,ત્યારબાદ $C_2$ અને $C_3$ ની અદલાબદલી કરતા (બે વાર અદલાબદલી,તેથી ચિહ્ન ધન રહેશે):$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\{{a_2}}&{{b_2}}&{{c_2}}\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} ight|$.

નિશ્ચાયક $\left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} \right|$ બરાબર શું થાય?

Similar Questions

નિશ્ચાયકનું મૂલ્ય શોધો: $\left| \begin{array}{ccc} a_1 & m a_1 & b_1 \\ a_2 & m a_2 & b_2 \\ a_3 & m a_3 & b_3 \end{array} \right|$

જો $\left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array} \right| = \alpha (a+b)(b+c)(c+a) \neq 0$ હોય,તો $\alpha$ ની કિંમત શોધો.

નિશ્ચાયકના ગુણધર્મોનો ઉપયોગ કરીને સાબિત કરો કે:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

નિશ્ચાયકનું મૂલ્ય શોધો: $\left| {\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}} \right|$

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