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ELEMENTARY LINEAR ALGEBRA PROGRESS CHECK-I

Example 4. Let us consider a system of linear equations
x − y = 1,
−2x + 2y = 5.

Again, geometrically, each of the two equations represents a line in the xy-plane.
However, unlike Example 3, the two lines do not intersect with each other.
The two lines are parallel . This indicates that the given system has no solution. In
fact, we can logically prove that the given system has no solution. Namely, multiply
2 to the first equation in the given system:
2x − 2y = 2.

We use this, and the second part of the given system of equations:
−2x + 2y = 5.

We add them together sidewise, and obtain
0 = 7,

which is an absurd identity. Hence we arrive at a conclusion that, the original system
of linear equations has no solutions.

• A system of linear equations is called consistent , if it has at least one solution.
Otherwise it is called inconsistent . For example,
x + y = 10,
2x − y = 5

(in Example 3) is a consistent system of equations, whereas
x − y = 1,
−2x + 2y = 5

(in Example 4) is an inconsistent system of equations.

Example 5. Let us consider a system of linear equations
x + y = 10,
2x + 2y = 20.

We notice that, the second equation is obtained by multiplying 2 to the both sides.

Hence the given system of equation is indeed equivalent to the single equation
x + y = 10,

meaning that, the original system consisting of two equations has precisely the same
amount of information as the above last equation, which in turn is a single equation.

We already know that this latter equation has many solutions:
(x, y) = (t, 10 − t ).

In particular, the given system of equations is consistent .

• In this last example, geometrically, the two equations in the original system
correspond to an identical line in the xy-plane.

[III] Solve each of the following system of linear equations.

Decide whether each of the above systems is consistent or not.

• We give an example of a system of linear equations involving three variables .

• Three-Variable Case. Gaussian Elimination.

Example 6. Consider the following system of linear equations
x + y + z = 2,
−x + 3y + 2z = 8,
4x + y = 4.

We attempt to solve this system. It goes step-by-step .

tep 1. Multiply 2 to the first equation in the system sidewise. The result is
2x + 2y + 2z = 4.

Step 2. Subtract it from the second equation in the given system sidewise. The
result is
−3x + y = 4.

Step 3. Subtract it from the third equation in the given system sidewise. The
result is
7x = 0.

Step 4. Multiply 1/7 to the both sides. The result is
x = 0.

Step 5. Go back to Step 2. Apply the result of Step 4. The result is
y = 4.

Step 6. Go back to the original equation. Apply the result of Step 4–5. The
result is
z = −2.

In sum, we have obtained the solution (x, y, z) = (0, 4, −2) .

• The process we have demonstrated in Example 6 is called the Gaussian elimination .

[IV] Decide whether each of the above systems is consistent or not.
(1)
3 x₁ − 2 x₂ + 4 x₃ = 1,
x₁ + x₂ − 2 x₃ = 3,
2 x₁ − 3 x₂ + 6 x₃ = 8.

(2)
x − 3 y + 2 z = 18,
5 x − 15 y + 10 z = 18.

[V] Find the value(s) k such that each of the following systems of linear equations
has an infinite number of solutions.
(1)
4 x + k y = 6,
k x + y = −3.

(2)
k x + y = 4,
2 x − 3 y = −12.

[VI] Find the value(s) k such that the following system of linear equations has exactly
one solution.
x + k y = 0,
k x + y = 0.

[VII] Let
f(x) = a x² + b x + c.

(1) Write out the conditions
f(0) = 0,
f(1) = 0,
f(2) = 0.

(2) Regard your the result of (1) as a system of linear equations with variables
a, b, c. Solve the system of equations.

[VIII]* Compare the following two systems of equations:

and

Agree that (*) is a non-linear system, whereas (**) is a linear system.
(1) Is the system (**) homogeneous?
(2) Find the general solution for (**).
(3) Find the exact condition for the general solution which you found in (2) to
also become a solution for (*).

Solution for problems [I–VII].
[I] The equations (1) and (2) are non-linear. The equation (3) is linear.

[II] (1) (x, y) = (t, 6 t − 18) .
Alternatively,

(2) (x, y, z) = (t, s, 1 − t − s) .
Alternatively, (x, y, z)= (t, 1 − t − s, s ).
Alternatively,(x, y, z) = (1 − t − s, t, s) .

Alternatively, .
In particular, both systems (1) and (2) are consistent .

[IV] Both systems (1) and (2) are inconsistent .

[V] (1) k = −2. (2) k = −2/3.

[VI] k ≠ 1 and k ≠ −1.

[VII]

We may regard the above set of three equations as a system of linear equations in
a, b, c. We may solve it as (a, b, c) = (0, 0, 0) .