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Mathematics 10 Online
OpenStudy (anonymous):

Help needed

OpenStudy (anonymous):

@freckles

OpenStudy (anonymous):

@Directrix

OpenStudy (amistre64):

what does it mean for a matrix to be singular?

OpenStudy (amistre64):

and for that matter, what does A and B have to do with B^* being singular?

OpenStudy (anonymous):

a matrix is singular iff its determinant =0

OpenStudy (anonymous):

and im not sure

OpenStudy (anonymous):

it was 2 parts of the question part A was to find AB part A was this part

OpenStudy (amistre64):

right, so we need the determinant of the 3x3 to be 0 ... how far have you gotten?

OpenStudy (amistre64):

.... by working the determinant process id imagine. of course, for a 3x3 i do the shortcut process similar to a 2x2

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i only know to do the 2x2 so far

OpenStudy (amistre64):

the shortcut on a 3x3 is to copy the first 2 columns onto the end, and multiply the diagonals

OpenStudy (amistre64):

write down the values of ab and c ... where do abc come from?

OpenStudy (anonymous):

idk that what the question ask

OpenStudy (amistre64):

you say this question is part of another one?

OpenStudy (anonymous):

thats the whole question

OpenStudy (amistre64):

and you sure that when you drew it up ... you made the correct entries? with pq and r?

OpenStudy (anonymous):

yep thats exactly whats there

OpenStudy (amistre64):

well im out of ideas on how to make this sensible then lol.

OpenStudy (amistre64):

i can make an equation of a plane with pqr as variables ... but ive got no idea how that helps in finding some rather ambiguous abc values.

OpenStudy (anonymous):

@amistre64

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

have you worked the determinant?

OpenStudy (anonymous):

6(8c-8b)+2a(32)

OpenStudy (amistre64):

and when we set that equal to 0 we get an infinite number of solutions .... all the points in a plane. are they wanting the solution set, or something more concrete?

OpenStudy (amistre64):

(a,b,c) = (3,3,-1) is a solution, since that makes the last column the same as the first (a,b,c) = (0,2,2) is a solution, since that makes the last column the same as the second and any linear combination of the first 2 columns would be a solution as well

OpenStudy (amistre64):

6x +0y = 2a 6x+4y = 2b -2x+4y = 2c

OpenStudy (anonymous):

o ok

OpenStudy (anonymous):

can it = to 0,0,0 also?

OpenStudy (amistre64):

yes it can :) let x=0 and y=0

OpenStudy (anonymous):

ok thanks

OpenStudy (amistre64):

youre welcome

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