InterviewSolution
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InterviewSolution
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What will be the value of \(\begin{vmatrix}2bc – a^2 & c^2 & b^2 \\c^2 & 2ac – b^2 & a^2 \\b^2 & a^2 & 2ab – c^2\end {vmatrix}\) if given another determinant \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)?(a) (a^3 + b^3 + c^3 + 3abc)^2(b) –(a^3 + b^3 + c^3 + 3abc)^2(c) (a^3 + b^3 + c^3 – 3abc)^2(d) –(a^3 + b^3 + c^3 – 3abc)^2This question was addressed to me in semester exam.This interesting question is from Determinant topic in portion Determinants of Mathematics – Class 12 |
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Answer» CORRECT choice is (c) (a^3 + b^3 + c^3 – 3abc)^2 For explanation I would say: Now, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\) Interchanging 2^nd and 3^rd columns, = –\(\begin{vmatrix}a & c & b \\b & a & c \\c & b & a \end {vmatrix}\) = \(\begin{vmatrix}-a & c & b \\ -b & a & c \\-c & b & a \end {vmatrix}\) So, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}^2\) = \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)\(\begin{vmatrix} -a & c & b \\ -b & a & c \\-c & b & a \end {vmatrix}\) = {–(a^3 + b^3 + c^3 – 3abc)}^2 = \(\begin{vmatrix}-a^2 + bc + bc & -ab + ab + c^2 & -AC + b^2 + ca \\-ab + ab + c^2 & -ac – b^2 + ca & -a^2 + bc + bc \\-ac + b^2 + ca & -a^2 + bc + bc & -ab + ab – c^2\end {vmatrix}\) => \(\begin{vmatrix}2BC – a^2 & c^2 & b^2 \\c^2 & 2ac – b^2 & a^2 \\b^2 & a^2 & 2AB – c^2\end {vmatrix}\) = (a^3 + b^3 + c^3 – 3abc)^2 |
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