# Taking signs out of columns of matrix

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?

• The determinant is also linear in every column, so you can just multiply the first and the third column with $-1$.
– user592521
Oct 2, 2018 at 8:28
• 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 $- \det A.$ Oct 2, 2018 at 8:30
• @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 $1$ Oct 2, 2018 at 8:37
• @ab123 correct, thank you. Oct 2, 2018 at 9:39

Rather than premultiplying the matrix, try post multiplying the matrix by $$diag(-1, 1, -1)$$ to achieve that result.
$$\begin{bmatrix}-1 & 1&-2\\-1&-1&-4\\-1&1&-7\end{bmatrix}\begin{bmatrix}-1 & 0&0 \\ 0&1&0\\0&0& - 1\end{bmatrix}= \begin{bmatrix}1&1&2\\1&-1&4\\1&1&7\end{bmatrix}$$
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 $$-1$$