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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
If $D =\left| {\begin{array}{*{20}{c}}{{a^2}\, + \,\,1}&{ab}&{ac}\\{ba}&{{b^2}\, +\,\,1}&{bc}\\{ca}&{cb}&{{c^2}\, + \,\,1}\end{array}} \right|$ then $D =$
  • ✓
    $1 + a^2 + b^2 + c^2$
  • B
    $a^2 + b^2 + c^2$
  • C
    $(a + b + c)^2$
  • D
    none
✓

Answer

Correct option: A.
$1 + a^2 + b^2 + c^2$
a
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

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

Now use $c_1 \rightarrow c_1 - c_2 \, and \, c_2 \rightarrow c_2 - c_3$ to get the value as $1$.

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