$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ એ $ . . . $ વડે વિભાજ્ય છે .
A${a^2}$
B${b^2}$
C${c^2}$
D${x^2}$
Difficult
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D${x^2}$
$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$
Multiply ${C_1},\,{C_2},{C_3}$ by $a,\,\,b,\,c$ respectively and hence divide by $abc$
$\Delta = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{a({a^2} + {x^2})}&{a{b^2}}&{{c^2}a}\\{{a^2}b}&{b({b^2} + {x^2})}&{b{c^2}}\\{c{a^2}}&{{b^2}c}&{c({c^2} + {x^2})}\end{array}\,} \right|$
Now take out $a, b $ and $c $ common from ${R_1},\,{R_2}$ and ${R_3}$,
$\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{{b^2}}&{{c^2}}\\{{a^2}}&{{b^2} + {x^2}}&{{c^2}}\\{{a^2}}&{{b^2}}&{{c^2} + {x^2}}\end{array}\,} \right|$
Now applying ${C_1} \to {C_1} + {C_2} + {C_3}$
$ \Rightarrow $ $\Delta = ({a^2} + {b^2} + {c^2} + {x^2})\,\left| {\,\begin{array}{*{20}{c}}1&{{b^2}}&{{c^2}}\\1&{{b^2} + {x^2}}&{{c^2}}\\1&{{b^2}}&{{c^2} + {x^2}}\end{array}\,} \right|$
$ \Rightarrow \Delta = {x^4}({a^2} + {b^2} + {c^2} + {x^2})$
Hence, it is divisible by ${x^2}$.
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જો $\text{ABC} = I$ હોય તો $tr(\text{ABC} + \text{BCA} + \text{CAB})$ મેળવો. $($કે જ્યાં $\text{A, B, C}$ ની કક્ષા $3$ છે અને $tr(A)$ એ $A$ ના વિકર્ણના ઘટકોનો સરવાળો છે.$)$