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Question: The value of \[\left| {\begin{array}{*{20}{c}} {1 + {a^2} - {b^2}}&{2ab}&{ - 2b} \\\ {2ab}&...

The value of \left| {\begin{array}{*{20}{c}} {1 + {a^2} - {b^2}}&{2ab}&{ - 2b} \\\ {2ab}&{1 - {a^2} + {b^2}}&{2a} \\\ {2b}&{ - 2a}&{1 - {a^2} - {b^2}} \end{array}} \right| is equal to
A. (1+a2+b2)3{\left( {1 + {a^2} + {b^2}} \right)^3}
B. (1+a2+b2)2{\left( {1 + {a^2} + {b^2}} \right)^2}
C. (1−a2+b2)2{\left( {1 - {a^2} + {b^2}} \right)^2}
D. (a2−b2−1)3{\left( {{a^2} - {b^2} - 1} \right)^3}

Explanation

Solution

Hint: First of all, write the given determinant and take the common terms out by using simple row and column operations. Then expand the remaining determinant to obtain the required answer.

Complete step-by-step answer:
Given \left| {\begin{array}{*{20}{c}} {1 + {a^2} - {b^2}}&{2ab}&{ - 2b} \\\ {2ab}&{1 - {a^2} + {b^2}}&{2a} \\\ {2b}&{ - 2a}&{1 - {a^2} - {b^2}} \end{array}} \right|
By applying the operations C1→C1−bC3,C2→C2+aC3{C_1} \to {C_1} - b{C_3},{C_2} \to {C_2} + a{C_3}, we have

= \left| {\begin{array}{*{20}{c}} {1 + {a^2} - {b^2} + 2{b^2}}&{2ab - 2ab}&{ - 2b} \\\ {2ab - 2ab}&{1 - {a^2} + {b^2} + 2{a^2}}&{2a} \\\ {2b - \left( {b - {a^2}b - {b^3}} \right)}&{ - 2a + a - {a^3} - a{b^2}}&{1 - {a^2} - {b^2}} \end{array}} \right| \\\ = \left| {\begin{array}{*{20}{c}} {1 + {a^2} + {b^2}}&0&{ - 2b} \\\ 0&{1 + {a^2} + {b^2}}&{2a} \\\ {b + {a^2}b + {b^3}}&{ - a - {a^3} - a{b^2}}&{1 - {a^2} - {b^2}} \end{array}} \right| \\\ = \left| {\begin{array}{*{20}{c}} {1 + {a^2} + {b^2}}&0&{ - 2b} \\\ 0&{1 + {a^2} + {b^2}}&{2a} \\\ {b\left( {1 + {a^2} + {b^2}} \right)}&{ - a\left( {1 + {a^2} + {b^2}} \right)}&{1 - {a^2} - {b^2}} \end{array}} \right| \\\

Taking 1+a2+b21 + {a^2} + {b^2} common in first and second column, we have

= \left( {1 + {a^2} + {b^2}} \right)\left( {1 + {a^2} + {b^2}} \right)\left| {\begin{array}{*{20}{c}} 1&0&{ - 2b} \\\ 0&1&{2a} \\\ b&{ - a}&{1 - {a^2} - {b^2}} \end{array}} \right| \\\ {\left( {1 + {a^2} + {b^2}} \right)^2}\left| {\begin{array}{*{20}{c}} 1&0&{ - 2b} \\\ 0&1&{2a} \\\ b&{ - a}&{1 - {a^2} - {b^2}} \end{array}} \right| \\\

Expanding the determinant, we get

= {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1\left\\{ {\left( 1 \right)\left( {1 - {a^2} - {b^2}} \right) - \left( {2a} \right)\left( { - a} \right)} \right\\} - 0\left\\{ {\left( 0 \right)\left( {1 - {a^2} - {b^2}} \right) - \left( {2a} \right)\left( b \right)} \right\\} + \left( { - 2b} \right)\left\\{ {\left( 0 \right)\left( { - a} \right) - \left( 1 \right)\left( b \right)} \right\\}} \right] \\\ = {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1\left\\{ {1 - {a^2} - {b^2} + 2{a^2}} \right\\} - 0 - 2b\left\\{ { - b} \right\\}} \right] \\\ = {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1\left\\{ {1 + {a^2} - {b^2}} \right\\} + 2{b^2}} \right] \\\ = {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1 + {a^2} - {b^2} + 2{b^2}} \right] \\\ = {\left( {1 + {a^2} + {b^2}} \right)^2}\left[ {1 + {a^2} + {b^2}} \right] \\\ = {\left( {1 + {a^2} + {b^2}} \right)^3} \\\

Hence, \left| {\begin{array}{*{20}{c}} {1 + {a^2} - {b^2}}&{2ab}&{ - 2b} \\\ {2ab}&{1 - {a^2} + {b^2}}&{2a} \\\ {2b}&{ - 2a}&{1 - {a^2} - {b^2}} \end{array}} \right| = {\left( {1 + {a^2} + {b^2}} \right)^3}
Thus, the correct option is A. (1+a2+b2)3{\left( {1 + {a^2} + {b^2}} \right)^3}

Note: Since the determinants of a matrix and its transpose are equal, we can use column operations to simplify a determinant. By performing a column operation on the matrix has the same effect as performing the corresponding row operation on its transpose.