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Q18E

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Found in: Page 165

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

The 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. atrAdd 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 $$- 12$$.

See the step by step solution

Step 1: Write the first two columns to the right of the matrix

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|

Step 2: Calculate the determinant

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 $$- 12$$.