1
,
then
¶
¸
=
1
3
1 .
and if
¼
¸
=
2
1
4 ,
then
¼
¸
¸
=
2
1
4
¸
=
2
1
4
= ¼
.
Obviously, for any column vector
¶
:
¶
¸
¸
= ¶
.
If
µ =
1
2
3
D
E
F
−1
1
0
1
F
1
(3 × 4),
then
µ
¸
=
1
D
−1
2
E
1
3
F
0
1
F
1
4 × 3 .
or
µ =
¶
·
¸
I
F
J
¶
K
¸
,
µ
¸
=
¶
·
I
F
¶
K
.

9/11/2015
9
Transpose of
³ × ¾
matrix
µ =
¶
··
⋯
¶
·@
⋮
⋱
⋮
¶
º·
⋯
¶
º@
is
¾ × ³
matrix
µ
¸
=
¶
··
⋯
¶
º·
⋮
⋱
⋮
¶
·@
⋯
¶
º@
.
Note that
µ»
¸
= »
¸
µ
¸
.
You can go through the proof of this at the end of
Notes X.

9/11/2015
10
YOU CALCULATE ONE:
Write down the transpose of
µ =
1
2
5
7
6
9
3
2
1
1
−1
1
0
1
0
2
.
THE ZERO MATRIX
A zero matrix is any matrix of which all entries are 0.
Examples:
0
0
0
0
0
0
,
0
0
0
0
,
0
0
0
0
0
0
0
0
0
,
0
0
0
0
0
0
.
In equations and expressions: Symbol
0.
Properties:
I.
For any
³ × ¾
matrix
µ
,
µ − µ = 0.
Example:
1
2
5
6
7
1
2
3
4
−
1
2
5
6
7
1
2
3
4
=
1 − 1
2 − 2
5 − 5
6 − 6
7 − 7
1 − 1
2 − 2
3 − 3
4 − 4
=
0
0
0
0
0
0
0
0
0
= 0.

9/11/2015
11
II. For any matrix
5
and the zero matrix with the same
number of rows and columns
µ + 0 = µ.
Example:
1
2
5
6
7
1
2
3
4
+
0
0
0
0
0
0
0
0
0
=
1 + 0
2 + 0
5 + 0
6 + 0
7 + 0
1 + 0
2 + 0
3 + 0
4 + 0
=
1
2
5
6
7
1
2
3
4
III. Multiplying any
° × ±
matrix
5
with the
± × ²
zero
matrix 0, results in a
° × ²
zero matrix.
Zero matrix acts like the number 0 in the multiplication of
numbers.
EXAMPLE
Example 1.
If
µ =
1
2
3
4
1
2
and
0 =
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
calculate
µ × 0
.
Solution:
µ × 0 =
1
2
3
4
1
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
=
0
0
0
0
0
0
0
0
0
0
= 0.

9/11/2015
12
THE IDENTITY MATRIX
Always a square matrix
:
± × ±
.
The diagonal from top left to bottom right always has a
value 1 and the rest of the matrix is zero’s.
M =
1
0
…
0
1
…
…
…
…
0
0
…
0
0
…
1
.
Examples:
2 × 2:
M =
1
0
0
1
3 × 3:
M =
1
0
0
0
1
0
0
0
1
EXAMPLES
Example 2.
For
µ =
1
2
1
3
4
2
,
and
M =
1
0
0
0
1
0
0
0
1
,
show
that
µM = µ
.
Solution:
µM =
1
2
1
3
4
2
1
0
0
0
1
0
0
0
1
=
µ
1
0
0
µ
0
1
0
µ
0
0
1
=
1 + 0 + 0
0 + 2 + 0
0 + 0 + 1
3 + 0 + 0
0 + 4 + 0
0 + 0 + 2
=
1
2
1
3
4
2
= µ.

9/11/2015
13
Example 3
.
For
µ =
1
2
1
3
4
2
,
and
M =
1
0
0
1
,
show that
Mµ = µ.
Solution:
Mµ =
1
0
0
1
1
2
1
3
4
2
=
M
1
3
M
2
4
M
1
2
=
1 + 0
2 + 0
1 + 0
0 + 3
0 + 4
0 + 2
=
1
2
1
3

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