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Q18E
Expert-verifiedThe expansion of a \({\bf{3}} \times {\bf{3}}\) determinant can be remembered by the following device. Write a second type of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals.
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Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises 15-18. Warning: This trick does not generalize in any reasonable way to \({\bf{4}} \times {\bf{4}}\) or larger matrices.
\(\left| {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{4}}\\{\bf{2}}&{\bf{3}}&{\bf{1}}\\{\bf{3}}&{\bf{3}}&{\bf{2}}\end{aligned}} \right|\)
The value of the determinant is \( - 12\).
The determinant can be written as shown below:
\(\left| {\left. {\begin{aligned}{*{20}{c}}1&3&4\\2&3&1\\3&3&2\end{aligned}} \right|\begin{aligned}{*{20}{c}}1&3\\2&3\\3&3\end{aligned}} \right|\)
The determinant can be calculated as shown below:
\(\begin{aligned}{c}\det = \left( 1 \right)\left( 3 \right)\left( 2 \right) + \left( 3 \right)\left( 1 \right)\left( 3 \right) + \left( 4 \right)\left( 2 \right)\left( 3 \right) - \left( 3 \right)\left( 3 \right)\left( 4 \right) - \left( 3 \right)\left( 1 \right)\left( 1 \right) - \left( 2 \right)\left( 2 \right)\left( 3 \right)\\ = 6 + 9 + 24 - 36 - 3 - 12\\ = - 12\end{aligned}\)
So, the value of the determinant is \( - 12\).
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