I want to expand the determinant on the left side of the following equation: $$ \left| \begin{array}{ccc} 1 & 0 & 0 \\ p & q & -1 \\ \text{a}-x & -y & -z \end{array} \right|=0.$$
I am getting $qz+y=0.$ Please verify if this is correct.
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2Not exactly: it is $-qz-y$. – Bernard Jan 04 '17 at 23:43
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But we could multiply both sides by $-1$, right? – Student Jan 04 '17 at 23:44
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For an "easy" way to compute $3\times 3$ determinants, look at The Rule of Sarrus. – Mark Schultz-Wu Jan 04 '17 at 23:44
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Those equations are equivalent though – Ben Grossmann Jan 04 '17 at 23:45
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@Mark: not this one! – Bernard Jan 04 '17 at 23:48
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@Bernard I understand expansion here is simple, it can just be good to know two methods to do something to check your work. – Mark Schultz-Wu Jan 04 '17 at 23:49
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1@Supermario: Yes. I had not seen you were solving an equation. – Bernard Jan 04 '17 at 23:51
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Do you know Laplace expansion? – Huang Jan 05 '17 at 00:11
3 Answers
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Expanding the determinant about the first row, we obtain: $$\begin{vmatrix} 1 & 0 & 0 \\ p & q & -1 \\ a-x & -y & -z \end{vmatrix} = \begin{vmatrix} q & -1 \\ -y & -z \end{vmatrix} = q(-z) -(-1)(-y) = -(qz+y)$$
LtSten
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The determinant of a matrix can be written by a cofactor expansion across a specific row or column (any row of column for that matter).
In your case, the determinant can be chosen to be expanded on the first row (which is probably the easiest), yielding:
$$\begin{vmatrix}1 & 0 & 0 \\ p & q & -1 \\a-x & -y & -z\end{vmatrix} = -qz - y = -(qz + y)$$
q.Then
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