I have a few questions I need help with! Please explain if my answer was not right how you got to it. I got 1.) B 2.) A 3.) C
1. 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.
-3x+y-z=-4
6x-2y+2z=8
-12x+4y-4z=-16
A.) (1,1,2)
B.) (1,1,1)
C.) Coincident dependence; (x,y,z)|3x+yz=4}
D.) No solution, inconsistent
2. Solve the system by triangularizing the augmented matrix and using back substitution.
6x=y-4
3y=-13-7x
a.) (-1,-3)
b.) (0,-2)
C.) (0,-3)
D.) (-1,-2)
3. What row operation would produce zero beneath the first entry in the diagonal?
1 2 | -1
8 -2 | -2
A.) 8R1-R2--> R2
B.) -8R1-R2 -->R2
C.) -R1+R2 --> R2
D.) 8R1-R2 --> R2
#1 - look at the equations. They are all the same. That is, they are multiples of each other. So, (C)
#2. (D) Did you check your answer to see whether it actually fits the equations?
#3. (A) and (D) are the same, so I suspect a typo. Anyway, you want to get rid of the 8, so you know you have to multiply R1 by 8 and subtract R2.
1. To solve the system by triangularizing the augmented matrix and using back substitution, we need to perform row operations to eliminate variables one by one. Let's set up the augmented matrix:
[ -3 1 -1 | -4 ]
[ 6 -2 2 | 8 ]
[-12 4 -4 |-16 ]
We can start by eliminating x-terms in the second and third rows using row operations.
First, add 2 times Row 1 to Row 2:
[ -3 1 -1 | -4 ]
[ 0 0 0 | 0 ]
[-12 4 -4 |-16 ]
Next, add 4 times Row 1 to Row 3:
[ -3 1 -1 | -4 ]
[ 0 0 0 | 0 ]
[ 0 8 - 8 | -4 ]
Now, we have a triangular matrix. Let's express the last row in terms of the variable y:
8y - 8z = -4
This equation has infinite solutions because we still have a free variable z. To express the solution in terms of a parameter, we can set z = t (a parameter). Substituting z = t into the equation, we get:
8y - 8t = -4
y - t = -1/2
Solving for y in terms of t:
y = -1/2 + t
Therefore, the solution is (x, y, z) = (1, -1/2 + t, t), which can be rewritten as (x, y, z) = (1, -1/2, 0) + t(0, 1, 1).
The correct answer is C.) Coincident dependence; (x,y,z)|3x+yz=4}
2. Let's set up the augmented matrix for the given system:
[ 6 -1 | -4 ]
[-7 3 | -13 ]
To triangularize the matrix, we need to eliminate the x-term in the second row.
Multiplying the first row by 7 and adding it to the second row:
[ 6 -1 | -4 ]
[ 0 20 | -45 ]
Now, we can solve for y using back substitution:
20y = -45
y = -45/20
y = -9/4
Substituting y into the first equation:
6x = -4 - (-9/4)
6x = -16/4 + 9/4
6x = -7/4
x = -7/24
Therefore, the solution is (x, y) = (-7/24, -9/4), which corresponds to the answer choice D.) (-1,-2).
3. To produce zero beneath the first entry in the diagonal, we need to find the correct row operation that can achieve this. Let's look at the given matrix:
[ 1 2 | -1 ]
[ 8 -2 | -2 ]
If we perform the row operation multiplied by a factor of 8 on the first row and subtract it from the second row, we will get zero beneath the first entry in the diagonal.
The correct row operation is D.) 8R1 - R2 --> R2:
[ 1 2 | -1 ]
[ 0 -18 | 6 ]