Determinant of the matrix $$A= \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$ is $$\det A=4.$$ So what is the determinant of \begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}

I found that the row operations that were done were $3R1, -1.R2,$ and the last one doesn't matter.

So is the determinant $3\times 4\times(-1)= -12$ or we have to do the inverse $4\times 1/3 \times 1/(-1)?$

  • 3
    Yes the answer is -12. Your approach is correct. – SchrodingersCat Nov 9 at 17:54
up vote 8 down vote accepted

Yes that's correct, indeed by the properties of the determinant we have that

$$\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\det\begin{bmatrix}a&b&c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=\\=\frac13 \det\begin{bmatrix}3a&3b&3c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=-\frac13 \det\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$

Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!

Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:

$ X = a(ei - fh) - b(di - fg) + c(dh - eg) $

$ X = aei - afh - bdi + bfg + cdh - ceg $

$ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $

$ \frac{Y}{-3} = -afh + eai -ecg -bdi + bfg + cdh $

You could re-arrange the terms of any of the equations to see that:


Since X=4,


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