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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
Question
If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to
✓

Answer

c
$\Delta  = \left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right|$

Multiply ${C_1}$ by $a,{C_2}$by $b$ and ${C_3}$ by $c$ and hence divide by $abc$.

$ = \frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}
{a\left( {{b^2} + {c^2}} \right)}&{a{b^2}}&{a{c^2}}\\
{{a^2}b}&{b\left( {{c^2} + {a^2}} \right)}&{b{c^2}}\\
{{a^2}c}&{{b^2}c}&{c\left( {{a^2} + {b^2}} \right)}
\end{array}} \right|$

Take out $a,b,c $ common fopre ${R_1},{R_2}$ and ${R_3}$ respectively.

$\Delta  = \frac{{abc}}{{abc}}\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{{b^2}}&{{c^2}}\\
{{a^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{{a^2}}&{{b^2}}&{{a^2} + {b^2}}
\end{array}} \right|$

Apply ${C_1} \to {C_1} - {C_2} - {C_3}$

$\Delta  = \left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{ - 2{c^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{ - 2{b^2}}&{{b_2}}&{{a^2} + {b^2}}
\end{array}} \right|$

$ =  - 2\left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{{c^2}}&{{c^2} + {a^2}}&{{c^2}}\\
{{b^2}}&{{b^2}}&{{a^2} + {b^2}}
\end{array}} \right|$

Apply ${C_2} - {C_1}$ and ${C_3} - {C_1}$

$ =  - 2\left| {\begin{array}{*{20}{c}}
0&{{b^2}}&{{c^2}}\\
{{c^2}}&{{a^2}}&0\\
{{b^2}}&0&{{a^2}}
\end{array}} \right|$

$ =  - 2\left[ { - {b^2}\left( {{c^2}{a^2}} \right) + {c^2}\left( { - {a^2}{b^2}} \right)} \right]$

$ = 2{a^2}{b^2}{c^2} + 2{a^2}{b^2}{c^2} = 4{a^2}{b^2}{c^2}$

But $\Delta  = k{a^2}{b^2}{c^2}\therefore k = 4$

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