If \(A = \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right]\) find |A|.
Answer
Given: \(A = \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right]\)
\( \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right|\)
Expanding along first row, \(1\left| {\begin{array}{*{20}{c}} 1&{ - 3} \\ 4&{ - 9} \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ 5&{ - 9} \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 2&1 \\ 5&4 \end{array}} \right|\)
\(= \left\{ { - 9 - \left( { - 12} \right)} \right\} - \left\{ { - 18 - \left( { - 15} \right)} \right\} - 2\left( {8 - 5} \right)\)
\( = - 9 + 12 - \left( { - 18 + 15} \right) - 2\left( 3 \right)\)
\(= 3-\left( { - 3} \right) - 6\)
\( = 3 + 3 - 6 = 0\)
\( \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right|\)
Expanding along first row, \(1\left| {\begin{array}{*{20}{c}} 1&{ - 3} \\ 4&{ - 9} \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ 5&{ - 9} \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 2&1 \\ 5&4 \end{array}} \right|\)
\(= \left\{ { - 9 - \left( { - 12} \right)} \right\} - \left\{ { - 18 - \left( { - 15} \right)} \right\} - 2\left( {8 - 5} \right)\)
\( = - 9 + 12 - \left( { - 18 + 15} \right) - 2\left( 3 \right)\)
\(= 3-\left( { - 3} \right) - 6\)
\( = 3 + 3 - 6 = 0\)
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