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Prove the below determinants are equal without expanding them.

\begin{vmatrix} {\alpha a_2} + {a_3}&{\beta a_3} + {a_1} & {\gamma a_1} + {a_2} \\ {\alpha b_2} + {b_3}&{\beta b_3} + {b_1} & {\gamma b_1} + {b_2} \\ {\alpha c_2} + {c_3}&{\beta c_3} + {c_1} & {\gamma c_1} + {c_2} \end{vmatrix}

=

\begin{equation}(\alpha \beta \gamma +1)\end{equation}

\begin{vmatrix} {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3} \\ {c_1} & {c_2} & {c_3} \end{vmatrix}

I'm sorry this isn't correctly formatted - this is my first time using LATEX. Also, this is my first time working with determinants, so I'm not really sure where to go.

If we have this matrix:

\begin{vmatrix} {a_3} & {a_1} & {a_2} \\ {b_3} & {b_1} & {b_2} \\ {c_3} & {c_1} & {c_2} \end{vmatrix}

I noticed if we apply the following equations, we get the original L.H.S. of the equality to prove.

\begin{equation} {C_1} = \alpha{C_3} + {C_1} \end{equation} \begin{equation} {C_2} = \beta{C_1} + {C_2} \end{equation} \begin{equation} {C_3} = \gamma{C_2} + {C_3} \end{equation}

But I don't really know how to tie that back to the original question, or if it's useful to proving the above.

Also, I'm wondering if there's a property of determinants that allows us to switch two columns / rows around?

Would appreciate any help on this, thanks.

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1 Answer 1

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let $A = \pmatrix{a_1&a_2&a_3\cr b_1&b_2&b_3\cr c_1&c_2&c_3},\ B = \pmatrix{0 &1 & \gamma\cr\alpha & 0 & 1\cr1&\beta & 0}$ then by the product rule $det(AB) = det(A)det(B)$

by expanding the determinant of $B$ by the first column gibes $det(B) = 1 + \alpha\beta\gamma$ so $$det(AB) = (1+\alpha\beta\gamma)\ det(A).$$

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  • $\begingroup$ This is ingenious! I didn't know about the determinant product rule...thanks! $\endgroup$ Commented Dec 10, 2014 at 3:11

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