# Matrix determinant effect

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 $$4\times 1/3 \times 1/(-1)?$$

• Yes the answer is -12. Your approach is correct. – SchrodingersCat Nov 9 '18 at 17:54

## 2 Answers

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:

$$X=\frac{Y}{-3}$$

Since X=4,

$$Y=-12$$