Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 9 - Systems and Matrices - 9.3 Determinant Solution of Linear Systems - 9.3 Exercises - Page 885: 35

Answer

$$ - 144 - 8\sqrt {10} $$

Work Step by Step

$$\eqalign{ & \left| {\matrix{ {\sqrt 2 } & 4 & 0 \cr 1 & { - \sqrt 5 } & 7 \cr { - 5} & {\sqrt 5 } & 1 \cr } } \right| \cr & {\rm{Calculating\, the\, determinant\, by \,expanding\, the\, third\, column}} \cr & \left| {\matrix{ {\sqrt 2 } & 4 & 0 \cr 1 & { - \sqrt 5 } & 7 \cr { - 5} & {\sqrt 5 } & 1 \cr } } \right| = 0\left| {\matrix{ 1 & { - \sqrt 5 } \cr { - 5} & {\sqrt 5 } \cr } } \right| - 7\left| {\matrix{ {\sqrt 2 } & 4 \cr { - 5} & {\sqrt 5 } \cr } } \right| + 1\left| {\matrix{ {\sqrt 2 } & 4 \cr 1 & { - \sqrt 5 } \cr } } \right| \cr & {\rm{Solving}} \cr & \left| {\matrix{ {\sqrt 2 } & 4 & 0 \cr 1 & { - \sqrt 5 } & 7 \cr { - 5} & {\sqrt 5 } & 1 \cr } } \right| = - 7\left( {\sqrt {10} + 20} \right) + \left( { - \sqrt {10} - 4} \right) \cr & {\rm{Simplifying}} \cr & \left| {\matrix{ {\sqrt 2 } & 4 & 0 \cr 1 & { - \sqrt 5 } & 7 \cr { - 5} & {\sqrt 5 } & 1 \cr } } \right| = - 7\sqrt {10} - 140 - \sqrt {10} - 4 \cr & \left| {\matrix{ {\sqrt 2 } & 4 & 0 \cr 1 & { - \sqrt 5 } & 7 \cr { - 5} & {\sqrt 5 } & 1 \cr } } \right| = - 144 - 8\sqrt {10} \cr} $$
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