Question 174668

lets call the time spent at soup kitchen,trash pickup, and toy collecting
s, t, and c, respectively
:
c=4t...........eq 1
s=t-2..........eq 2
s+t+c=40.......eq 3
:
rewrite equations
:
 s+t+c=40......eq 3
 s-t+0c=-2.....eq 2
0s-4t+c=0......eq 1
:
now writing out the main matrix
:
{{{(matrix(3,3,1,1,1,1,-1,0,0,-4,1))}}}...lets call this matrix X
:
we need to find the determinant of this matrix to begin with 
:
we can choose any row or column to determine this ..I will work with column 1 since it has a zero in it
:
det X=(1) det{{{(matrix(2,2,-1,0,-4,1))}}}-(1)det{{{(matrix(2,2,1,1,-4,1))}}}+0=
.....1(-1)-1(5)=-6
:
s=det s/-6 where matrix s is matrix X with the 1st column replaced by constant coefficents 40,-2,0.  I will use column one for our determinant
:
det s = 40 det{{{(matrix(2,2,-1,0,-4,1))}}}-(-2)det{{{(matrix(2,2,1,1,-4,1))}}}
=40(-1)+2(5)=-30
:
s=det s/-6={{{highlight(-30/-6=5)}}}hours in the soup kitchen
:
:
t=det t/-6 where matrix t is matrix X with the 2nd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det t=0+0+(1)det{{{(matrix(2,2,1,40,1,-2))}}}=1(-42)=-42
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t=det t/-6={{{highlight(-42/-6=7)}}}hours picking up trash
:
:
c=det c/-6  where matrix c is matrix X with the 3rd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det c= 0-(-4){{{(matrix(2,2,1,40,1,-2))}}}+0=4(-42)=-168
:
c=det c/-6={{{highlight(-168/-6=28)}}}hours toy collecting


If you forgot the number of hours you volunteered you could not determine the hours you spend doing each activity....
:
you would only have 2 equations and 3 unknowns