# Zero Matrix

How to prove the zero matrix statements: definition, 2 properties, 2 examples, and their solutions.

## Definition

### Definition

A zero matrix is a matrix
whose elements are all 0.

So AO = OA = O.

## Property 1

### Property

If AB = O,
then A = O or B = O.

This statement is false.

Unlike the numbers,
AB = O doesn't mean
either A or B is a zero matrix.

## Example 1

### Solution

Recall that
a counterexample is an example
that makes the statement false.

And the given statement is a conditional statement.

So, set a counterexample
that seems to make
the hypothesis, AB = O, true
and the conclusion, A = O or B = O, false.

Set A = [0 1 / 0 0] and B = [1 0 / 0 0].

First, show that
AB = O is true.

A = [0 1 / 0 0]
B = [1 0 / 0 0]

So AB = [0 1 / 0 0][1 0 / 0 0].

Solve [0 1 / 0 0][1 0 / 0 0].

Multiply Matrices

Row 1, column 1:
0⋅1 + 1⋅0

Row 1, column 2:
0⋅0 + 1⋅0

Row 2, column 1:
0⋅1 + 0⋅0

Row 2, column 2:
0⋅0 + 0⋅0

So
[0 1 / 0 0][1 0 / 0 0]
= [0⋅1 + 1⋅0 0⋅0 + 1⋅0 / 0⋅1 + 0⋅0 0⋅0 + 0⋅0].

0⋅1 + 1⋅0
= 0 + 0

0⋅0 + 1⋅0
= 0 + 0

0⋅1 + 0⋅0
= 0 + 0

0⋅0 + 0⋅0
= 0 + 0

0 + 0 = 0

The elements of the matrix is all 0.

So this matrix is a zero matrix.

You got AB = O.

So the given hypothesis, AB = O,
is true.

Next, show that
A = O or B = O
is false.

A = [0 1 / 0 0]
So A = O is false.

B = [1 0 / 0 0]
So B = O is false.

So
A = O or B = O
is false.

Disjunction

See the given conditional statement.

The hypothesis, AB = O, is true.
The conclusion, A = O or B = O, is false.

So the given conditional statement is false.

So A = [0 1 / 0 0], B = [1 0 / 0 0]
is the counterexample.

So
A = [0 1 / 0 0]
B = [1 0 / 0 0]

## Property 2

### Property

If AB = O,
then BA = O.

This statement is false.

Recall that
AB and BA are not always equal.

Multiply Matrices

So BA is not always a zero matrix.

## Example 2

### Solution

Again, a counterexample is an example
that makes the statement false.

And the given statement is a conditional statement.

So, set a counterexample
that seems to make
the hypothesis, AB = O, true
and the conclusion, BA = O, false.

Set A = [0 0 / 1 0] and B = [0 0 / 0 1].

First, show that
AB = O is true.

A = [0 0 / 1 0]
B = [0 0 / 0 1]

So AB = [0 0 / 1 0][0 0 / 0 1].

Solve [0 0 / 1 0][0 0 / 0 1].

Row 1, column 1:
0⋅0 + 0⋅0

Row 1, column 2:
0⋅0 + 0⋅1

Row 2, column 1:
1⋅0 + 0⋅0

Row 2, column 2:
1⋅0 + 0⋅1

So
[0 0 / 1 0][0 0 / 0 1]
= [0⋅0 + 0⋅0 0⋅0 + 0⋅1 / 1⋅0 + 0⋅0 1⋅0 + 0⋅1].

0⋅0 + 0⋅0
= 0 + 0

0⋅0 + 0⋅1
= 0 + 0

1⋅0 + 0⋅0
= 0 + 0

1⋅0 + 0⋅1
= 0 + 0

0 + 0 = 0

The elements of the matrix is all 0.

So this matrix is a zero matrix.

You got AB = O.

So the given hypothesis, AB = O,
is true.

Next, show that
BA = O is false.

A = [0 0 / 1 0]
B = [0 0 / 0 1]

So BA = [0 0 / 0 1][0 0 / 1 0].

Solve [0 0 / 0 1][0 0 / 1 0].

Row 1, column 1:
0⋅0 + 0⋅1

Row 1, column 2:
0⋅0 + 0⋅0

Row 2, column 1:
0⋅0 + 1⋅1

Row 2, column 2:
0⋅0 + 1⋅0

So
[0 0 / 0 1][0 0 / 1 0]
= [0⋅0 + 0⋅1 0⋅0 + 0⋅0 / 0⋅0 + 1⋅1 0⋅0 + 1⋅0].

0⋅0 + 0⋅1
= 0 + 0

0⋅0 + 0⋅0
= 0 + 0

0⋅0 + 1⋅1
= 0 + 1

0⋅0 + 1⋅0
= 0 + 0

0 + 0 = 0

0 + 1 = 1

[0 0 / 1 0] is not a zero matrix.

So
[0 0 / 1 0] ≠ O.

You got BA ≠ O.

So the given conclusion, BA = O,
is false.

See the given conditional statement.

The hypothesis, AB = O, is true.
The conclusion, BA = O, is false.

So the given conditional statement is false.

So A = [0 0 / 1 0], B = [0 0 / 0 1]
is the counterexample.

So
A = [0 0 / 1 0]
B = [0 0 / 0 1]