linear systems rref

Question

1. 1. Consider the following system of equations in the variables x, y, z with a as some unknown parameter. In this question, you are to solve by manipulating the equations directly rather than using matrices and row reduction.

x+y= 1

x+ay= 2

y+az= 10

1. (a) Suppose (for this part only) that a= 2. Find the solution to the system.
2. (b) For which values of a will this system has no solutions? (Try eliminating variables to solve for x,y and z and look for cases where no solution exists).
3. (c) If there is a solution with z= 0, what are the possible values of a?

Answer 1

Solve the following system of equations:

x+y= 1

x+ay= 2

y+az= 10

For this question, it is better to find the General Solution for x, y, z in terms of a. Then, you should be able to answer/solve any question using the General Solution for x, y, z.

In this question, you cannot create Matrix A in TI-84 Plus and solve it using rref(A) because you do not know the value of a. In fact, the solution varies with the value of a (there are different solution for different value of a)

General Solution for x, y, z in terms of a

=============================

(x + y = 1)(-a) ==> -ax - ay = -a ==> x - ax = 2 - a ==> x(1 - a) = 2 - a ==> x = (2 - a)/(1 - a)

x + ay = 2 x + ay = 2

y = 1 - x ==> y = 1 - (2 - a)/(1 - a) = (1 - a - 2 + a)/(1 - a) ==> y = (-1)/(1 - a)

y + az = 10==> az = 10 - y = 10 - (-1)/(1 - a) = (10 - 10a + 1)(1 - a) ==> az = (11 - 10a)/(1 - a)

az = (11 - 10a)/(1 - a) ==> z = (11 - 10a)/((1 - a) (a))

(a) Suppose (for this part only) that a = 2. Find the solution to the system.

x = (2 - a)/(1 - a) = (2 - 2)/(1 - 2) = (0)/(-1) ==> x = 0

y = (-1)/(1 - a) = (-1)/(1 - 2) = (-1)/(-1) = 1 ==> y = 1

z = (11 - 10a)/((1 - a) (a)) = (11 - 10(2))/((1 - 2) (2)) = (11 - 20)/(-2) = (-9)/(-2) = 9/2 ==> z = 9/2

Solution: (x, y, z) = (0, 1, 9/2); when a = 2

(b) For which values of a will this system has no solutions?

x = (2 - a)/(1 - a)

y = (-1)/(1 - a)

z = (11 - 10a)/((1 - a) (a))

It is obvious the system has no solutions, when the denominator of x, y, z is zeros

(1 - a) (a) = 0 ==> a = 0 and 1 - a = 0 ==> (a = 0 and a = 1) <== No Solution

(c) If there is a solution with z = 0, what are the possible values of a?

z = (11 - 10a)/((1 - a) (a)) ==> 0 = (11 - 10a)/((1 - a) (a)) ==> 0 = 11 - 10a ==> 10a = 11 ==> a = 11/10

Answer 2

1) a=2

x+y = 1 --> y = 1-x

x+2y =2 --> x+2-2x = 2 --> x=0 , y = 1

y+2z=10 --> z=9/2

2) y=1-x

x+a(1-x)=2 --> (1-a)x = 2-a --> x = (2-a)/(1-a)

a≠1

y+az = 10 --> 1-x+az = 10

1+(2-a)/(1-a) + az = 10

z = (10a-11)/(a(a-1))

a≠0

There are no solutions if a=0, 1

3) if z=0

(10a-11)/(a(a-1)) = 0

a = 11/10