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It is given that determinant of the below matrix is -3\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}

What is the determinant of the following: \begin{bmatrix}-8d+a&-8e+b&-8f+c\\d&e&f\\g&h&i\end{bmatrix}

I know that swapping rows negates the determinant, and multiplying a row by a scalar scales the determinant. But I can't get this question correct. I thought it would be 24, because adding one row to another shouldn't affect the determinant, only the multiplication by -8 would, so the determinant would be -8 * -3 = 24. But that isn't the correct answer, and neither is -24. Can someone help me understand what I'm missing?

jmsCoder
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    one possible solution for the first determinant is $a=-3,e=i=1$ and all the others zero. Evaluate the second determinant for that solution. – WW1 Jun 24 '21 at 02:35
  • @WW1 Interesting, so it's -3, the determinant isn't changed at all. Why doesn't the scalar multiplication change the determinant in this case? Thank you for your help.

    Edit: I think I get it now. Row 1 = -8 * Row 2 + Row 1, which wouldn't change the determinant, because that row operation wouldn't have any effect.

     – jmsCoder Jun 24 '21 at 02:56

2 Answers2

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Thanks to a helpful comment, I was able to work out the answer. Adding a multiple of one row to another has no effect on the determinant. In the matrix above, Row 1 = -8* Row 2 + Row 1. It isn't a scalar multiplication operation like I originally thought. So the determinant is unchanged.

jmsCoder
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    I think this is better suited as a comment than an answer. Anyway, you can check out my answer, I think you might find it helpful! :) – Soham Jun 24 '21 at 04:03
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Although the problem has been solved for the OP apparently, I would add a more concise answer for future readers:

Invariance Property: The determinant is unchanged if any row (or column) is multiplied by a scalar and then added or subtracted from another row (or column).

Basically this translates to

$$\text{det}\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\text{det}\begin{bmatrix}a+kd&b+kd&c+kd\\d&e&f\\g&h&i\end{bmatrix}$$

Why is this true?

$$\text{det}\begin{bmatrix}a+kd&b+kd&c+kd\\d&e&f\\g&h&i\end{bmatrix}=\text{det}\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}+\text{det}\begin{bmatrix}kd&kd&kd\\d&e&f\\g&h&i\end{bmatrix}$$

Can you follow it up from here?

Soham
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