If a, b, c are real,then the value of the det | vedclass

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\{ab}&{{b^2} + 1}&{bc}\{ac}&{bc}&{{c^2} + 1}\end{array}}ight|$.Multiply $R_

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$a = b = c = 0$

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\{ab}&{{b^2} + 1}&{bc}\{ac}&{bc}&{{c^2} + 1}\end{array}}ight|$.Multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$,and divide the determinant by $abc$:$\Delta = \frac{1}{abc} \left| {\begin{array}{*{20}{c}} {a(a^2+1)}&{a^2b}&{a^2c}\{ab^2}&{b(b^2+1)}&{b^2c}\{ac^2}&{bc^2}&{c(c^2+1)}\end{array}}ight|$.Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:$\Delta = \frac{abc}{abc} \left| {\begin{array}{*{20}{c}} {a^2+1}&{a^2}&{a^2}\{b^2}&{b^2+1}&{b^2}\{c^2}&{c^2}&{c^2+1}\end{array}}ight| = \left| {\begin{array}{*{20}{c}} {a^2+1}&{a^2}&{a^2}\{b^2}&{b^2+1}&{b^2}\{c^2}&{c^2}&{c^2+1}\end{array}}ight|$.Applying $C_1 ightarrow C_1 + C_2 + C_3$:$\Delta = \left| {\begin{array}{*{20}{c}} {a^2+b^2+c^2+1}&{a^2}&{a^2}\{a^2+b^2+c^2+1}&{b^2+1}&{b^2}\{a^2+b^2+c^2+1}&{c^2}&{c^2+1}\end{array}}ight| = (a^2+b^2+c^2+1) \left| {\begin{array}{*{20}{c}} 1&{a^2}&{a^2}\{1}&{b^2+1}&{b^2}\{1}&{c^2}&{c^2+1}\end{array}}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = (a^2+b^2+c^2+1) \left| {\begin{array}{*{20}{c}} 1&{a^2}&{a^2}\{0}&{1}&{0}\{0}&{0}&{1}\end{array}}ight| = (a^2+b^2+c^2+1)(1) = 1+a^2+b^2+c^2$.Given $\Delta = 1$,we have $1+a^2+b^2+c^2 = 1$,which implies $a^2+b^2+c^2 = 0$. Since $a, b, c$ are real,this is only possible if $a = b = c = 0$.

If $a, b, c$ are real,then the value of the determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right| = 1$ if

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