left| {begin{array}{*{20}{c}} {{(b + c)}^2} & | vedclass

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(C) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \ {{b^2}} & {{(a + c)}^2} & {{b^2}} \ {{c^2}} & {{c^2}} & {{

Vedclass · Accepted Answer

$2abc(a + b + c)^3$

Vedclass · Answer

(C) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \ {{b^2}} & {{(a + c)}^2} & {{b^2}} \ {{c^2}} & {{c^2}} & {{(a + b)}^2} \end{array}} ight|$.$C_1 ightarrow C_1 - C_3$ અને $C_2 ightarrow C_2 - C_3$ લાગુ કરતા:$\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2 - a^2} & 0 & {{a^2}} \ 0 & {{(a + c)}^2 - b^2} & {{b^2}} \ {{c^2 - (a + b)^2}} & {{c^2 - (a + b)^2}} & {{(a + b)}^2} \end{array}} ight|$$\Delta = \left| {\begin{array}{*{20}{c}} {(b+c-a)(b+c+a)} & 0 & {{a^2}} \ 0 & {(a+c-b)(a+c+b)} & {{b^2}} \ {(c-a-b)(c+a+b)} & {(c-a-b)(c+a+b)} & {{(a + b)}^2} \end{array}} ight|$$C_1$ અને $C_2$ માંથી $(a+b+c)$ સામાન્ય લેતા:$\Delta = (a+b+c)^2 \left| {\begin{array}{*{20}{c}} {b+c-a} & 0 & {{a^2}} \ 0 & {a+c-b} & {{b^2}} \ {-(c-a-b)} & {-(c-a-b)} & {{(a + b)}^2} \end{array}} ight|$નિશ્ચાયકનું વિસ્તરણ કરતા,આપણને $\Delta = 2abc(a+b+c)^3$ મળે છે.

$\left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \\ {{b^2}} & {{(a + c)}^2} & {{b^2}} \\ {{c^2}} & {{c^2}} & {{(a + b)}^2} \end{array}} \right|$ નું મૂલ્ય શોધો.

Similar Questions

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?

જો $\begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} = (x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ હોય, તો $k$ ની કિંમત શોધો.

જો $A$ એ $n$ કક્ષાનો ચોરસ શ્રેણિક હોય અને $A = kB$ હોય,જ્યાં $k$ એ અદિશ છે,તો $|A|=$

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