We see immediately that A = 2, thus B = 0. Since B = 0, D = -1. Since A = 2 and B = 0, then C = 1. So the decomposition is
Use Heavyside's method to find A:
Therefore
Clearing the fractions yields
So B = 1 and C = 3. So the decomposition is
We can find A and C, but not B, by Heavyside's Method:
Thus
Clearing the fractions by multiplying both sides by the least common denominator yields:
Thus, B = 2. So the decomposition is
We can find B, but not A, using Heavyside's Method:
Thus,
Next, we multiply through by the least common denominator to clear the fractions:
So A = 1. The decomposition is
Using Heavyside's Method:
So the partial fraction decomposition is
2 x - 5 y + 2 z = 0 x + 3 y + z = -1 y - 3 z = 0
Using Cramer's Rule
5 x + 3 y = - 2 -3 x + y = 4
Using Cramer's Rule
S1 | x + y + z = 0 |- x + 2 y = -1 | x + z = 1> The corresponding matrix is:> 1 1 1 0 -1 2 0 -1 E2 --> E1 + E2 1 0 1 1 E3 --> -E1 + E3 1 1 1 0 0 3 1 -1 0 -1 0 1 E2 <--> E3> 1 1 1 0 0 -1 0 1 E2 --> -E2 0 3 1 -1 1 1 1 0 0 1 0 -1 0 3 1 -1 E3 --> -3E2 + E3 1 1 1 0 Since this is triangularized, we begin the back substitution: 0 1 0 -1 0 0 1 2 E1 --> -E3 + E1 1 1 0 -2 0 1 0 -1 0 0 1 2 E1 --> -E2 + E1 1 0 0 -1 0 1 0 -1 0 0 1 2 Which corresponds to a solution x = -1 y = -1 z = 2Return
S1 | x - y + z = 1 | x + z = 1 E2 --> -E1 + E2 | x + y + z = 2 E3 --> -E1 + E3 S2 | x - y + z = 1 | y = 0 | 2y = 1 E3 --> -2E2 + E3 S3 | x - y + z = 1 | y = 0 | 0 = 1 Thus the system is inconsistent.Return
S1 | x + 2 y + z = 0 | y - z = 2Insert a third equation to represent all possible values of z.
S2 | x + 2 y + z = 0 E1 --> -E3 + E1 | y - z = 2 E2 --> E3 + E2 | z = t S3 | x + 2 y = -t E1 --> -2E2 + E1 | y = 2 - t | z = t S4 | x = -4 + t | y = 2 - t | z = twhich is the 'solution'. There are actually an infinite number of solutions, one for each value of t. The system is dependent.
S1 | x + 2y - z = 1 | x + z = 3 E2 --> -E1 + E2 | y + z = 1 S2 | x + 2y - z = 1 | - 2y + 2z = 2 E2 <--> E3 | y + z = 1 S3 | x + 2y - z = 1 | y + z = 1 | - 2y + 2z = 2 E3 --> 2E2 + E3 S4 | x + 2y - z = 1 | y - z = -1 | 4z = 4 E3 --> (1/4)E3 S5 | x + 2y - z = 1 | y - z = -1 | z = 1Solving for y by back-substitution, y = 0. Using the values of z and y, we solve the first equation for x, getting x = 2.
S1 | x + 2 y - z = 0 | - 3 y + 3 z = 0 E2 --> -(1/3)E2 | 7 y - z = -6 S2 | x + 2 y - z = 0 | y - z = 0 | 7 y - z = -6 E3 --> -7E2 + E3 S3 | x + 2 y - z = 0 | y - z = 0 | 6 z = -6Solving the last equation, z = -1. Substitution into the second equation, y = -1. Substituting these two values into the first equation yields x = 1.
S1 | x + 2 y - z = 0 | 2 x + y + z = 0 E2--> - 2 E1 + E2 | - 3 x + y + 2 z = - 6 E3--> 3 E1 + E3 S2 | x + 2 y - z = 0 | - 3 y + 3 z = 0 | 7 y - z = - 6Return
S1 | x + 2 y - z = 3 | y - z = 1 | z = 2Substituting z = 2 into the second equation gives y - 2 = 1, so y = 3. Substituting these two values into equation 1 yields x + 6 - 2 = 3, so x = -1.
S1 | x + y = 1 |2x + y = -1Solving the first equation for y in terms of x yields y = 1 - x.
Substituting y = 1 - x into the second equation and simplifying
yields x +1 = -1. So x = -2. Since
y = 1 - x, then y = 3.
S1 | x + y = 0 | x + y2 = 0Solving the first equation for y in terms of x yields y = - x. Substituting into the second equation and simplifying yields x + x2 = 0. Factoring the expression on the left yields
x ( 1 + x ) = 0
So, either x = 0 and y = 0, or x = -1 and y = 1.