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Evaluate 1, 2, 3; 2, 3, 1; 3, 1, 2. Each row is the row above it moved one place to the left…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Evaluate $

\displaystyle \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}

$. Each row is the row above it moved one place to the left. Show that the determinant has a common-factor structure by adding all the columns to the first.

Need a reminder? The key ideaShow

Swapping two rows changes the sign. Multiplying a row by $k$ multiplies the determinant by $k$ . Adding a multiple of one row to another changes nothing. Equal or proportional rows give $0$ . Also $|kA| = k^n|A|$ and $|AB| = |A||B|$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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