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If  \(A = \left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 6&0&4 \\ 1&5&{ - 7} \end{array}} \right|,\)Verify that det A = det (A')

Answer

\(\left| A \right| = 2\left( {0 - 20} \right) + 3\left( { - 42 - 4} \right) + 5\left( {30 - 0} \right)\)=-28 [expanding along \(R_1\)]
\(|A' |= \left| {\begin{array}{*{20}{c}} 2&6&1 \\ { - 3}&0&5 \\ 5&4&7 \end{array}} \right|\)=2(0-20)+3(-42-4)+5(30-0) =-28 [expanding along \(C_1\)]
\(\left| {A'} \right| = - 28\)
Hence proved.
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