# Thread: Are these right? (Determinants)

1. ## Are these right? (Determinants)

$\displaystyle \left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|$
Hello all! I'd like to check these as I'm not very confident that I got them right, and I'd really appreciate your input, especially as I've scrambled my brains with trilingual madness! So much fun

Anywho here's the questions with my answers, are they right?

Evaluate the following determinants.
$\displaystyle \left|\begin{array}{ccc}-5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|$
and
$\displaystyle 3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|$

$\displaystyle \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|$
$\displaystyle =(5)(3)+(4)+(-2)-(3)-(-5)(4)(-2)=-56$
and
$\displaystyle 3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|$
$\displaystyle =3(-2\left|\begin{array}{cc}5 & -2 \\ 8 & -1 \end{array}\right|$$\displaystyle +1\left|\begin{array}{cc}-3 & -2 \\ 0 & -1 \end{array}\right|$$\displaystyle +0\left|\begin{array}{cc}-3 & 5 \\ 0 & 8 \end{array}\right|)$
$\displaystyle =3(-22 + 3 + 0) = -57$

And one more question;
Expand by minors
$\displaystyle \left|\begin{array}{ccc}3 & 4 & -2 \\ 6 & 4 & 3 \\ 0 & 4 & 2 \end{array}\right|$
This question really bothers me considering that supposedly it doesn't matter which row or column you expand- you're supposed to get the same answer everytime, I'm getting a different answer each time.

Example, expanding the first row;
$\displaystyle =3\left|\begin{array}{cc}4 & 3 \\ 4 & 2 \end{array}\right|$$\displaystyle -4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right|$$\displaystyle -2\left|\begin{array}{cc}6 & 4 \\ 0 & 4 \end{array}\right|$
$\displaystyle =3(8-12)-4(12-0)-2(24-0)=-108$

And if I expand the middle column;
Example, expanding the first row;
$\displaystyle =-4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right|$$\displaystyle +4\left|\begin{array}{cc}3 & -2 \\ 0 & 2 \end{array}\right|$$\displaystyle -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right|$
$\displaystyle =-4(12-0)+4(6-0)-4(9-12)=-12$

So... what am I doing wrong? D:

Thanks in advance for any help

2. Hi

Originally Posted by StonerPenguin
$\displaystyle \left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|$
Hello all! I'd like to check these as I'm not very confident that I got them right, and I'd really appreciate your input, especially as I've scrambled my brains with trilingual madness! So much fun

Anywho here's the questions with my answers, are they right?

Evaluate the following determinants.
$\displaystyle \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|$
and
$\displaystyle 3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|$

$\displaystyle \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|$
$\displaystyle =(5)(3)+(4)+(-2)-(3)-(-5)(4)(-2)=-56$
No, how did you calculate it anyway?

Have you ever heard of "rule of Sarrus". I get : det() = +40

Originally Posted by StonerPenguin
and

$\displaystyle 3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|$
$\displaystyle =3(-2\left|\begin{array}{cc}5 & -2 \\ 8 & -1 \end{array}\right|$$\displaystyle +1\left|\begin{array}{cc}-3 & -2 \\ 0 & -1 \end{array}\right|$$\displaystyle +0\left|\begin{array}{cc}-3 & 5 \\ 0 & 8 \end{array}\right|)$
$\displaystyle =3(-22 + 3 + 0) = -57$

This is correct, well done

Originally Posted by StonerPenguin
And one more question;
Expand by minors
$\displaystyle \left|\begin{array}{ccc}3 & 4 & -2 \\ 6 & 4 & 3 \\ 0 & 4 & 2 \end{array}\right|$
This question really bothers me since supposedly it doesn't matter with row or column you expan you're supposed to get the same answer everytime and I get a different answer everytime.

Example, expanding the first row;
$\displaystyle =3\left|\begin{array}{cc}4 & 3 \\ 4 & 2 \end{array}\right|$$\displaystyle -4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right|$$\displaystyle -2\left|\begin{array}{cc}6 & 4 \\ 0 & 4 \end{array}\right|$
$\displaystyle =3(8-12)-4(12-0)-2(24-0)=-108$
This is correct, too

Originally Posted by StonerPenguin
And if I expand the middle column;
Example, expanding the first row;
$\displaystyle =-4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right|$$\displaystyle +4\left|\begin{array}{cc}3 & -2 \\ 0 & 2 \end{array}\right|$$\displaystyle -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right|$
$\displaystyle =-4(12-0)+4(6-0)-4(9-12)$
No, it is $\displaystyle -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right| = -4(9-(-12)) = -4(9+12)$

So it should be -4*21+24-48 = -108

Originally Posted by StonerPenguin
So... what am I doing wrong? D:

Thanks in advance for any help
By the way, nice solution!

Rapha

3. Ah thank you so much! Yeah, I had a serious case of brainfart here, how embarassing O: For the first question I used $\displaystyle a_1 b_2 c_3 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_3 b_2 c_1 - a_2 b_1 c_3 - a_1 b_3 c_2$
but like an idiot I cast out some of the zeros D: e.g. $\displaystyle (5)(3)(0)=0$ not 15. However re-evaluating it, I get NEG 40, because
$\displaystyle (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (-5)(4)(-2)$

So then det = $\displaystyle - (-5)(4)(-2)= -40$ Right? Or am I being stupid again?

4. Originally Posted by StonerPenguin
Ah thank you so much! Yeah, I had a serious case of brainfart here, how embarassing O: For the first question I used $\displaystyle a_1 b_2 c_3 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_3 b_2 c_1 - a_2 b_1 c_3 - a_1 b_3 c_2$
but like an idiot I cast out some of the zeros D: e.g. $\displaystyle (5)(3)(0)=0$ not 15. However re-evaluating it, I get NEG 40, because
$\displaystyle (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (-5)(4)(-2)$

So then det = $\displaystyle - (-5)(4)(-2)= -40$ Right? Or am I being stupid again?
It is actually
$\displaystyle (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (5)(4)(-2)$
because $\displaystyle a_1=5$

5. Originally Posted by running-gag
It is actually
$\displaystyle (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (5)(4)(-2)$
because $\displaystyle a_1=5$
Ah crap! I wrote my original question wrong! The question was $\displaystyle \left|\begin{array}{ccc}-5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|$ so $\displaystyle a_1 = -5$ I'm so sorry you guys, I'm usually not this bad about making mistakes D: Like I said, I think I scrambled my brains a bit, I stayed up 22 hours straight after only 4 hours of sleep (Nice excuse amirite? I'm retarded BAAWWW)

So it is -40 then right? Thank you so much for helping my stupid ass D':

6. Yes -40 is the right result

7. Looks like you need to lay off that reefer, son ^_^