Zero Matrix

Multiplying Matrices

How to solve the zero matrix problems: definition, properties, examples, and their solutions.

Zero Matrix

The zero matrix is a matrix whose elements are all 0. So AO = OA = O.

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

It's denoted by O.

Since its elements are all 0,
AO = OA = O.

Property 1: If AB = O, then A = O or B = O?

If AB = O, then A = O or B = O. This statement is false.

For two factors a and b,
if ab = 0,
then a = 0 or b = 0.

But this is not true for matrix.

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

This statement is false
because it's not always true.

Example 1

Find a counterexample for the given statement. If AB = O, then A = O or B = O.

Recall that
only one counterexample is needed
to show that the statement is false.

Counterexample

So find the matrices A and B
that make the given statement false.

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

See if AB = O.

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

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

Multiply these two matrices.

Multiplying 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

0⋅1 + 1⋅0 = 0 + 0
0⋅0 + 1⋅0 = 0 + 0

0⋅1 + 0⋅0 = 0 + 0
0⋅0 + 0⋅0 = 0 + 0

All elements are 0.

So AB = O.

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

So AO and BO.

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

Then the given conditional statment is false.

Conditional statement - Truth value

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

Property 2: If AB = O, then BA = O?

If AB = O, then BA = O. This statement is false.

If AB = O, then BA = O.

This statement is false
because it's not always true.

Example 2

Find a counterexample for the given statement. If AB = O, then BA = O.

Find the matrices A and B
that make the given statement false.

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

See if AB = O.

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

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

Multiply these two matrices.

[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

0⋅0 + 0⋅0 = 0 + 0
0⋅0 + 0⋅1 = 0 + 0

1⋅0 + 0⋅0 = 0 + 0
1⋅0 + 0⋅1 = 0 + 0

All elements are 0.

So AB = O.

Next, see if BA = O.

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

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

Multiply these two matrices.

[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

0⋅0 + 0⋅1 = 0 + 0
0⋅0 + 0⋅0 = 0 + 0

0⋅0 + 1⋅1 = 0 + 1
0⋅0 + 1⋅0 = 0 + 0

BA = [0 0 / 1 0].

So BAO.

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

Then the given conditional statment is false.

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