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My textbook presents the following matrix and asks to find its determinant:

$$d=\left|\begin{array} --1&1&-2\\-1&-1&-4\\-1&1&-7\end{array}\right|$$

And then says that it takes the negative sign from the first column and the negative sign from the third column so that the matrix becomes:

$$d=\left|\begin{array}11&1&2\\1&-1&4\\1&1&7\end{array}\right|$$

I understand that it is possible to multiply any row by -$1$, but if I were to multiply every row by $-1$ then the second column would have the signs inverted. Any hints on how this is possible to take the negative signs out of the matrix?

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  • $\begingroup$ The determinant is also linear in every column, so you can just multiply the first and the third column with $-1$. $\endgroup$
    – user592521
    Oct 2, 2018 at 8:28
  • $\begingroup$ Multiplying a column (row) by $c\neq 0$ gives $c \det A.$ If you multiply two columns by $(-1),$ the determinant doesn't change. If you multiplied all columns of this $3\times 3$ matrix by $(-1)$ you would get $- \det A.$ $\endgroup$
    – user376343
    Oct 2, 2018 at 8:30
  • $\begingroup$ @Cesare I think there is a typo in your edit. Didn't you mean to write the $(1,1)$ entry after taking negative signs as $-1$ instead of $1$ $\endgroup$
    – ab123
    Oct 2, 2018 at 8:37
  • $\begingroup$ @ab123 correct, thank you. $\endgroup$
    – Cesare
    Oct 2, 2018 at 9:39

2 Answers 2

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Rather than premultiplying the matrix, try post multiplying the matrix by $diag(-1, 1, -1)$ to achieve that result.

$$\begin{bmatrix}-1 & 1&-2\\-1&-1&-4\\-1&1&-7\end{bmatrix}\begin{bmatrix}-1 & 0&0 \\ 0&1&0\\0&0& - 1\end{bmatrix}= \begin{bmatrix}1&1&2\\1&-1&4\\1&1&7\end{bmatrix}$$

In terms of determinant, we can perform elementary column operations. You can view it as taking transpose (transpose doesn't change the determiant), perform row operations, then taking transpose back again.

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Since $det(A^T) = det (A)$, you can apply operations to columns in just the same way as you apply them to rows, so just multiply first and third columns by $-1$

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