Question

eg a11a22a33-a11a32a23-a12a21a33+a12a31a23+a13a21a32-a13a31a22 that...my teacher didnt quite get through to me with this one when do u use this in finding determinant?she explained it as something separate from cofactors so idk

Answer 1

i think that is what your refering to

Answer 2

\begin{pmatrix}
x&y&z\\
n&1&2\\
m&3&4
\end{pmatrix}

lets try to use this instead, might be easser to read and keep track of

Answer 3

\begin{pmatrix}
x&&\\
&1&2\\
&3&4
\end{pmatrix}

x(1*4-3*2)

lets move the n & m over to help out the pattern

\begin{pmatrix}
&y&\\
&&2&n\\
&&4&m
\end{pmatrix}

-y(2m-4n)

\begin{pmatrix}
&&z\\
n&1&\\
m&3&
\end{pmatrix}

z(3n-1m)

these are the parts of the overall determinate of the 3x3 matrix

add them together to get the whole thing

Answer 4

now, those arent to be regarded as numbers; only placeholders for clarity

Answer 5

? this is my face right now o_O

Answer 6

firstly it is not a cofactor method right? or ero? but u still multiply along main diagonal?

Answer 7

\begin{pmatrix}\Large
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{pmatrix}

\[\Large D = a_{11}(a_{22}a_{33}-a_{32}a_{23})-...+...\]

if your looking for the determinate of a 3x3 matrix; its alot to try to type out :)

Answer 8

what is a cofactor method? its been awhile

Answer 9

yh i get that part that u just typed was cofactor but the method in my question seems like cofactor

Answer 10

the method in your example, is the same from what I can parse

Answer 11

a11a22a33-a11a32a23-a12a21a33+a12a31a23+a13a21a32-a13a31a22

a11(a22a33-a32a23) -a12(a21a33-a31a23) +a13(a21a32-a31a22)

oy thats hard on the eyes

Answer 12

look at that slide 66 there about

Answer 13

write, since the det is not 0, it is possible that there is an inverse to be found

Answer 14

the next test is to see if they are linearly independant i believe

Answer 15

or maybe not lol; the next slide has an adjoin that I am not familiar with yet