# Thread: [SOLVED] Are determinants always positive?

1. ## [SOLVED] Are determinants always positive?

I got the matrix A

3 2 -1
1 6 3
2 -4 0

I'm not sure if determinants are always positive, but they must be because in my textbook, the determinant is 64.

I got this calculating the minors of row 1. I got 36+12+16. For row 2 I got -4-12-48 and row 3 I got 24-40+0.

This leads me to believe dets are always positive, but I'm not sure. What are the conditions for the sign of determinants?

2. No matter which row you expand, you should always get the same determinant. So you must've made a mistake when expanding across row 2.

As for matrices in general, of course you can have a non-positive determinant. Consider the matrix:

$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

3. I was going to give $\left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|$ as an example having a negative determinant.

And, of course, $\left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right|$ has determinant 0.

4. Originally Posted by o_O
No matter which row you expand, you should always get the same determinant. So you must've made a mistake when expanding across row 2.

As for matrices in general, of course you can have a non-positive determinant. Consider the matrix:

$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
well, $\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix}$ was a simpler example

@OP: personally I would have expanded along the 3rd column

5. I ended up with determinant = 64

$\begin{bmatrix} 3 & 2 & -1 \\ 1 & 6 & 3 \\ 2 & -4 & 0 \end{bmatrix}$

$[(3 \times 6 \times 0)+(2\times3\times2)+(-1\times-4\times1)] - [(2\times6\times-1)-(-4\times3\times3)-(0\times2\times1)]$

$16+48 =64$

By the way, determinants are not the identical to absolute values and they can be negative. If you end up with a negative that simply means that the shape is left-handed rather than right-handed.

6. Originally Posted by Roam
I ended up with determinant = 64

$\begin{bmatrix} 3 & 2 & -1 \\ 1 & 6 & 3 \\ 2 & -4 & 0 \end{bmatrix}$

$[(3 \times 6 \times 0)+(2\times3\times2)+(-1\times-4\times1)] - [(2\times6\times-1)-(-4\times3\times3)-(0\times2\times1)]$

$16+48 =64$

By the way, determinants are not the identical to absolute values and they can be negative. If you end up with a negative that simply means that the shape is left-handed rather than right-handed.
yes, i got +64 also.

and no one said they had to be positive, we all gave examples of negative determinants. the straight bars are notation: $\text{det} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix}$

7. Expanding row 2:

(-1)^2+1 * 1 det (2*0)-(-1*-4) = 4

+

(-1)^2+2 * 6 det (3*0)-(-1*2) = 12

+

(-1)^2+3 * 3 det (3*-4)-(2*2) = 48

Okay I get 64 as well. I understand my mistake. Originally I did not pay attention to the sign and sign changes depending on the location of the minors 4, 12, 48. So originally I had -4, -12 and -48 because I forgot to change the sign, or apply the correct ij in the (-1)^ij part.

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# is the value inside determinant always positive

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