 Multiplying Matrices How to multiply matrices: properties, examples, and their solutions.

Example 1 Multiply the elements in the following order.

Row 1, Column 1:
(Row 1, →) × (Column 1, ↓)

Row 1, Column 2:
(Row 1, →) × (Column 2, ↓)

Row 2, Column 1:
(Row 2, →) × (Column 1, ↓)

Row 2, Column 2:
(Row 2, →) × (Column 2, ↓)

To multiply matrices, this condition is needed:
(number of former matrix's row) = (number of latter matrix's column)

Example 2 Row 1, Column 1:
(Row 1, →) × (Column 1, ↓)

Row 1, Column 2:
(Row 1, →) × (Column 2, ↓)

Row 2, Column 1:
(Row 2, →) × (Column 1, ↓)

Row 2, Column 2:
(Row 2, →) × (Column 2, ↓)

For multiplying matrices,
ABBA.
(Check the previous example's answer.)

Identity Matrix The identity matrix is a matrix
whose diagonal elements are 1
and the other elements are 0.

It's denoted by I.

AI = IA = A
The order doesn't matter.

So I2 = II = I.

Example 3 AI = A

So AIA = AA.

Row 1, Column 1:
(Row 1, →) × (Column 1, ↓)

Row 1, Column 2:
(Row 1, →) × (Column 2, ↓)

Row 2, Column 1:
(Row 2, →) × (Column 1, ↓)

Row 2, Column 2:
(Row 2, →) × (Column 2, ↓)

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

It's denoted by O.

AO = OA = O

But AB = O doesn't always mean
A = O or B = O.

Even if AO and BO,
AB can be O.
(See the next example.)

Example 4 Recall that a counterexample is an example
that makes the given statement false.

And a conditional statement is false
if the hypothesis is true and the conclusion is false.

Conditional statement: truth value

So find a counterexample
that makes AB = O and BAO.

Assume A and B
that seems to be the counterexample.

A = [1 0 / 2 0], B = [0 0 / 0 1]

Show AB = O.
(true hypothesis)

Show BAO.
(false conclusion)

'A = [1 0 / 2 0], B = [0 0 / 0 1]' makes
the hypothesis (AB = O) true
and the conclusion (BA = O) false.

So the counterexample is
'A = [1 0 / 2 0], B = [0 0 / 0 1]'.