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what is the value of the determinant in the last question

I am having trouble finding the determinant in the last question.

I tried det = 15 because the first row is multiplied by 5, which multiples the original det value of 3 by 5. Then, I changed 15 into -15 because the rows are swapped.

I know that adding a scalar multiple of one row or column to another does not change the determinant. I assume if I factor out the negative sign from the last row so that it fits this rule. I end up with det = 15 again.

Where in my logic am I going awry?

Thanks,

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Interchange the first and third row, this gives a $-1$. Now remove the $5$ from the third row, so you have a $5$ coming out, and the last two rows are $[d,e,f]$ and $[g,h,i]$, as with the usual matrix. Now, you must split the first row using linearity of the determinant (in its first row in our case): $$ \det\begin{bmatrix} -3a-2d & -3b-2e & -3c - 2f \\ d&e&f \\ g&h&i \end{bmatrix} \\= \det\begin{bmatrix} -3a & -3b & -3c \\ d&e&f \\ g&h&i \end{bmatrix} + \det\begin{bmatrix} -2d & -2e & - 2f \\ d&e&f \\ g&h&i \end{bmatrix} $$

The second matrix has determinant zero, and the first has determinant $-3 \times -3 = 9$. Multiplying this with the $-1$ and $5$ we got earlier gives the result $-45$.

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  • $\begingroup$ I think this answer could use a little clarification; it's not valid in general to compute the determinant of a sum of two matrices by summing the determinants. I'm assuming you're appealing to the multilinearity of the determinant? $\endgroup$ – Theo Bendit Mar 7 at 4:28
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    $\begingroup$ Ah, then I would need a small edit. $\endgroup$ – астон вілла олоф мэллбэрг Mar 7 at 5:04
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I would start from the third question. Multiplying the top row by $5$ produces a determinant of $15$. Then, multiplying the bottom row by $-3$ (the step you missed) makes a determinant of $-45$. Finally, subtract two lots of the second row from the third, which leaves the determinant at $-45$. As this produces the matrix in question, $-45$ is your answer.

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