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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by
  • A
    ${x^3}$
  • ✓
    ${x^2}$
  • C
    $({a^2} + {b^2} + {c^2})$
  • D
    None of these
✓

Answer

Correct option: B.
${x^2}$
$\Delta = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{{a^3} + ax}&{{a^2}b}&{{a^2}c}\\{a{b^2}}&{{b^3} + bx}&{{b^2}c}\\{{c^2}a}&{{c^2}b}&{{c^3} + cx}\end{array}\,} \right|$
$=({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+x)\times \left| \,\begin{matrix} 1 & 1 & {{b}^{2}} & {{b}^{2}}+x & {{b}^{2}}  \\{{c}^{2}} & {{c}^{2}} & {{c}^{2}}+x  \\ \end{matrix}\, \right|$
$\{$Applying ${R_1} \to {R_1} + {R_2} + {R_3}\} $
$ = ({a^2} + {b^2} + {c^2} + x)\,\left| {\,\begin{array}{*{20}{c}}1&0&0\\{{b^2}}&x&0\\{{c^2}}&0&x\end{array}\,} \right|$$\left. \begin{array}{l}{C_2} \to {C_2} - {C_1}\\{C_3} \to {C_3} - {C_1}\end{array} \right\}$
$= {x^2}({a^2} + {b^2} + {c^2} + x)$.
Hence $\Delta $ is divisible by ${x^2}$ as well as by $ x.$

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