Need help with determinants of matrix. Question is in the attached file

- anonymous

Need help with determinants of matrix. Question is in the attached file

- jamiebookeater

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- anonymous

##### 1 Attachment

- anonymous

So my question is can I just take the cofactor expansion of column 3. which will give me |dw:1444515136849:dw| right?

- anonymous

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## More answers

- amistre64

your pic is cut ...

- anonymous

the drawing?

- anonymous

at the end is just a question mark, its not part of the solution

- amistre64

i cant remember how to express the alternating (-1)^n in terms of the element picked from the ith row and jth column ...

- amistre64

+ - + - + -
- + - + - +
+ - + - + -
visually it is compiled like this

- anonymous

I dont know what that means

- amistre64

oh well, there was some 'summation' formula that is compact for it ...
which element are we getting the cofactor expression for?

- anonymous

in the question it just says to get the determinant using cofactor expansion

- anonymous

so in the question I took the cofactor expansion of the 3rd column and I got what is in the drawing

- amistre64

http://www.wolframalpha.com/input/?i=cofactor+expansion
it appears that you followed the formula correctly to me
-1^(3+2) (-7) |..|
yeah you did fine for that element

- amistre64

3+3 that is ...

- anonymous

So now I just have to get a cofactor expansion of any aij entry in the 3x3 matrix correct? so I took the a31 entry |dw:1444516001995:dw| is that right?

- anonymous

|dw:1444516084912:dw|

- anonymous

but when I check my answer on determinant calculator its totally off

- anonymous

what am i doing wrong?

- mathmate

One good trick to remember, diagonal is always positive!
Yes, 3 column is best!

- amistre64

\[...+(-1)^{(3+3)}~(-7)[...
+(-1)^{(3+1)}~(1)
\begin{pmatrix}
-7&-10\\
-3&-8
\end{pmatrix}
+...
]+...\]

- amistre64

you have one more cofactor to express

- amistre64

unless you simply stopped at this and decided to shortcut the 2x2 det

- anonymous

yes that what I tried to do

- Empty

Yeah you were able to throw away those other 3 determinants with your first step since you had 3 zeroes there, but now you don't have anymore zeroes so you have to fully evaluate the ones that are rest

- amistre64

oh, i thought they were just randomly checking an element in the expansion .... didnt even consider that they were thinking this was the whole of it

- anonymous

@empty so from when I have the 3x3 det in the second step. I have to get the cofactor expansion of each row or column? is that what you are trying to say?

- anonymous

*sorry not "each" row, just one row or column

- amistre64

we only need to go across one row right?
\[\begin{pmatrix}
a&b\\c&d
\end{pmatrix}\]
row 1
\[(-1)^{(1+1)}(a)(d)+(-1)^{(1+2)}(b)(c)=ad-bc\]
or row2
\[(-1)^{(2+1)}(c)(b)+(-1)^{(2+2)}(d)(a)=ad-bc\]

- anonymous

yes, one row or one column

- mathmate

|dw:1444517082554:dw|
It's a personal preference, but I prefer to expand along the first column (because of the one's).

- anonymous

ok lets try that, let me do it on paper first

- amistre64

a 3x3 has a diagonalized shortcut similar to the 2x2 .. you just have to copy the first 2 columns

- amistre64

-(gfc+hga+ieb)
a b c a b
e f g e f
g h i g h
+(afi+bgg+ceh)

- amistre64

lol, forgot how to do the alphabet on that one ...

- amistre64

hij for the bottom row of course .. and then corrected for lack of any common sense

- anonymous

|dw:1444517589850:dw|

- anonymous

70(30+24)-(70+30)+(56-30)=3706

- anonymous

that is still now right, according to a determinant calculator. can anyone see what I did wrong?

- anonymous

well igtg for now, ill leave this question open for anyone else to figure it out.

- mathmate

@m0j0jojo
I suggest you multiply by -7 at the very end, then there will be less distribution to do.
so the whole thing, instead of:
70(30+24)-(70+30)+(56-30)
should read:
-7(-10(30+24)-1(70+30)+1(56-30))
and you should get the correct answer (just a touch below 4300).

- anonymous

oh right, lol totally forgot about bedmas here. Thanks for the help!

- mathmate

You're welcome! :)

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