I need help with these! I got B for 1. C for 2. and D for 3. Thanks for the help!
1.) Find the values of a, b, and c that will make the statement true.
[5 a -2] [5 102]
[3 4 –b] = [-3 -4 3]
[2c 5 9] [-2 5 9]
a.) 8R1-R2--> =R2
b.) -8R1-R2 -> =R2
c.) –8R1 +R2 -> =R2
d.) 8R1-R2 -> =R2
2.) Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
-x+y+z=0
3x+2y+5z=18
15x+10y+25z=89
A.) (3,2,1)
B.) (3,2,0)
C.) Coincident dependence; {(x,y,z) |-x+y-z=0
D.) No solution, inconsistent
3.) Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
X+y-z=3
x-y-2z=6
2x-y+z=3
A.) (0,0,-4)
B.) (0,0,-3)
C.) Coincident dependence; {(x,y,z) | -x+y-z=3
D.) No solution, inconsistent
1. I do not understand the question , nor your notation
Are the [5 a -2] [5 102] etc supposed to be rows of the first matrix ?
Why are there only 2 elements in [5 102] , when all others have 3 elements?
What does 8R1 - R2 --> = R2 mean
does it say: 8 times row 1 - row 2 becomes a new row 2 ?
2.
-1 1 1 0
3 2 5 18
15 10 25 89
1 -1 -1 0
3 2 5 18
15 10 25 89
notice if we divide the third by 5 we get
3 2 5 89/5
So the last two are parallel planes and the first plane will cut these two, resulting in 2 parallel lines
(I am not familiar with your terminology in C and D
we can rule out A and B, and since there is a solution of two parallel lines, we can rule out D
So by the process of elimination....
3.
1 1 -1 3
1 -1 -2 6
2 -1 1 3
1 1 -1 3
0 2 1 -3 -- I did #1 - #2
0 3 -3 3 -- I did 2 x #1 - #3
1 1 -1 3
0 1 -4 6 -- I did #3 - #2
0 1 -1 1 -- I did #3 ÷ 3
1 1 -1 3
0 1 -4 6
0 0 3 -5 -- I did #3 - #2
from that I read: 3z = -5
z = -5/3
from the 2nd: y - 4z = 6
y + 20/3 = 6
y = -2/3
in the 1st:
x +y-z=3
x - 2/3 + 5/3 = 3
x = 2
I get (2, -2/3 , -5/3)
which I subbed into all three equations and it worked!