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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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    Not exactly: it is $-qz-y$. – 2017-01-04
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    But we could multiply both sides by $-1$, right? – 2017-01-04
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    For an "easy" way to compute $3\times 3$ determinants, look at [The Rule of Sarrus](https://en.wikipedia.org/wiki/Rule_of_Sarrus). – 2017-01-04
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    Those equations are equivalent though – 2017-01-04
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    @Mark: not this one! – 2017-01-04
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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. – 2017-01-04
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    @Supermario: Yes. I had not seen you were solving an equation. – 2017-01-04
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    Do you know Laplace expansion? – 2017-01-05

3 Answers 3

3

You may use Wolfram Alpha to check this kind of computations.

2

Expanding the determinant about the first row, we obtain: $$\begin{vmatrix} 1 & 0 & 0 \\ p & q & -1 \\ a-x & -y & -z \end{vmatrix} = \begin{vmatrix} q & -1 \\ -y & -z \end{vmatrix} = q(-z) -(-1)(-y) = -(qz+y)$$

2

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:

$$\begin{vmatrix}1 & 0 & 0 \\ p & q & -1 \\a-x & -y & -z\end{vmatrix} = -qz - y = -(qz + y)$$

Determinant expansion