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Q18E

Expert-verifiedFound in: Page 165

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.**

** **

**atr**

**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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