# 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 x_{1} − 2 x_{2} + 4 x_{3} = 1,

x_{1} + x_{2} − 2 x_{3} = 3,

2 x_{1} − 3 x_{2} + 6 x_{3} = 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 x^{2} + 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) .