Exercise 12 — Linear Algebra

Many people have learned this mnemonic for the determinant of a $3\!\times\!3$ matrix: copy the first two columns to the right side of the matrix, then take the products down the forward diagonals and add them together, and then take the products on the backward diagonals and subtract them. That is, first write

\left(\begin{array}{ccc|cc}h_{1,1}&h_{1,2}&h_{1,3}&h_{1,1}&h_{1,2}\\ h_{2,1}&h_{2,2}&h_{2,3}&h_{2,1}&h_{2,2}\\ h_{3,1}&h_{3,2}&h_{3,3}&h_{3,1}&h_{3,2}\end{array}\right)

and then calculate this.

\begin{array}{l}h_{1,1}h_{2,2}h_{3,3}+h_{1,2}h_{2,3}h_{3,1}+h_{1,3}h_{2,1}h_{3,2}\\ \>-h_{3,1}h_{2,2}h_{1,3}-h_{3,2}h_{2,3}h_{1,1}-h_{3,3}h_{2,1}h_{1,2}\end{array}

Check that this agrees with the formula given in the preamble to this section.
Does it extend to other-sized determinants?

Hefferon, Linear Algebra, Determinants, Exploration, Exercise 12

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