1001 Algebra Problems.PDF

d.Note that (2F)(–2E) = -4FE. Since the inner
dimensions ofFand Edo not match, this
product is not defined. As such, the entire
computation is not well-defined.

Set 55 (Page 130)

d.

c.

a.

d.

b.

c.

a.

c.

b.

a.

c.

d.

Set 56 (Page 131)

d.First, extract the coefficients from the variable
terms on the left sides of the equations to form
a 2 2 coefficient matrix. Multiplying this by

x yand identifying the right side as a 2^1 constant matrix, we can rewrite the system

as the following matrix

equation:

a.Extract the coefficients from the variable
terms on the left sides of the equations to form
a 2 2 coefficient matrix. Multiplying this
by xyand identifying the right side as a 2 1
constant matrix, we can rewrite the system

as the following matrix equation:

b.Begin by extracting the coefficients from
the variable terms on the left sides of the equa-
tions to form a 2 2 coefficient matrix. Multi-
plying this by xyand identifying the right side
as a 2 1 constant matrix, we can rewrite the

system as the following matrix

equation:

x y

1

2

3

4

>>>HH H= 2

xy 24 23 2xy

+=

*

x y

a b

1

0

>>> 1 HH H=

xa yb

=

*

x y

3

1

7

5

2

8

–

>>>HH H=

37 2xy xy 58

–+ =

* +=

det>^0420 H==()() ()()00 42–– 8

det^1 ()()()() 1

1

0

– 10 1 1 1

–

– ==–––––

> H

det^3 ()( ) ()( ) 9

2

6

– 36 92 0

–

> H==–– –

det^3 ()() ()() 3

2

> H==32 32 0–

det^1 ()()()() 2

0

2

– 11 20 1

–

> H==–––

det^0 ()()()() 2

1

01 21 2

––

> H==–––

det^2 ( )() ( )() 12

0

3

– 23 120 6

–

> H==––– –

det^3 ()( ) ()( ) 1

1

2

– 32 11 5

–

> H==–– – –

det^1 ()( ) ()( ) 0

4

25

> – H==125 0 4 25––

det^3 ( )() ()() 4

4

2

>– H==––32 44 –^14

det^6 ()() ()() 2

3

1

> H==61 23 0–

det^1 ()()()() 2

2

4

– 14 22 0

–

> H==–––

det^2 ()() ()() 1

21 13 1

1

(^3) ==––

H
det^1 ()() ()()
2

2

3

> H==13 22––^1

det a ()() ()() b

ab ab 0

(^0) ==– 00

H
det^3 ( )() ()()
1

7

5

>– H==––35 17 –^22

ANSWERS & EXPLANATIONS–