MCQ
If $a, b, c$ are real then the value of 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
  • A
    $a + b + c = 0$
  • B
    $a + b + c = 1$
  • C
    $a + b + c = -1$
  • $a = b = c = 0$

Answer

Correct option: D.
$a = b = c = 0$
d
Multiply $R_1$ by $a, R_2$ by $b$ and $R_3$ by $c$ and divide the determinant by $abc$. Now take $a, b$ and $c$ common from $c_1, c_2$ and $c_3$. Now use $C_1 \rightarrow C_1 + C_2 + C_3$ to get
$(a^2 + b^2 + c^2 + 1)$ $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{{b^2}}&{{b^2} + 1}&{{b^2}}\\{{c^2}}&{{c^2}}&{{c^2} + 1}\end{array}\,} \right|$ $= 1$. Now use $c_1 \rightarrow c_1 - c_2$ and $c_2 \rightarrow c_2 - c_3$
we get $1 + a^2 + b^2 + c^2 = 1$ ==> $a = b = c = 0$

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