Zero Matrix
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.
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?
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
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 A ≠ O and B ≠ O.
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
because it's not always true.
Example 2
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 BA ≠ O.
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.
