S 4.1 (P 2.7)
_____________

1st product:

	A = 	2	-1	(3*2)
		3	-2
		1	 5

	B = 	-2	-3	 3	(2*3)
		 4	 0	 7

	A*B =	   -8    -6    -1	(3*3)
   		  -14    -9    -5
    		   18    -3    38

	    
2nd product:

	A = 	 2	 0	 0
		-2	 4	 0
		 3	-5	 1

	B =	 2	-2	 3
		 0	 4	-5
		 0	 0	 1

	A*B =    4    -4     6
    		-4    20   -26
     		 6   -26    35

Note that B = A', so the answer is symmetric.




S 4.2 (P 2.7)
_____________

The column vector v = [0 ; 1 ; 0] selects the second column
of A:

	A*v = [b ; e ; h]

The row vector u.' = [1 0 0] selects the first row of A*v : 

	[1 0 0]*[b ; e ; h] = b

Thus u = [1 ; 0 ; 0] .


S 4.3 (P 2.10)
______________

Both P*x and Q*x are row-permuted forms of the vector x.  Thus

	P =	0 0 1 0
		1 0 0 0
		0 1 0 0 
		0 0 0 1

	Q = 	1 0 0 0 
		0 0 1 0
		0 0 0 1
		0 1 0 0

and 
	P*Q =	0 0 0 1
		1 0 0 0
		0 0 1 0
		0 1 0 0 

(row selection on Q or column selection on P; either works)


S 4.4 (P 2.11)
______________

	A = 	1 2 3
		4 5 6
		7 8 9

	B = 	9 8 7
		6 5 4
		3 2 1

If	C =	0 0 1
		0 1 0
		1 0 0 
then
	A*P = 	3 2 1
		6 5 4
		9 8 7
and
	C*A*P = 9 8 7 = B
		6 5 4
		3 2 1

Therefore P = Q = C


S 4.5 (P 2.13)
______________

We can take, for example:

	A = [1 ; 2 ; -1 ; -2]*[a b c d]

	B = [1 ; 1/t ; 1/t^2; 1/t^3]*[1 t t^2 t^3]

(other choices are also possible).

Thus in each case, the 4x4 matrix is the product of a column vector (4x1) 
and a row vector (1x4), in that order.

Using the forms for A and B above, we see that the product A*B has the form

	col*row*col*row = col*(row*col)*row

The middle term (in parentheses) equals

	row*col = scalar = a+b/t+c/t^2+d/t^3 

Thus

	A*B = (a+b/t+c/t^2+d/t^3) * [1 ; 2 ; -1 ; -2] * [1 t t^2 t^3]

(The scalar coefficient can be absorbed in either of the two vectors.)


S 4.6
______

(i) By straightforward matrix multiplication,

	A = L*U =

	 1	 1	 0	 3
	 2	 1	-1	 1
	 3	-1	-1	 2
	-1	 2	 6	-1

(ii)  A*x = b is equivalent to L*(U*x) = b. Let U*x = y; then L*y = b.

To solve L*y = b, we use forward substitution:

	y1 = 4

	2*y1 + y2 = 1			: y2 = 1-8 = -7

	3*y1 + 4*y2 + y3 = -3		: y3 = -3-12+28 = 13

	-y1 - 3*y2 + y3 + y4 = 4	: y4 = 4+4-21-13 = -26

To solve U*x = y, we use backward substitution:

	-26*x4 = -26			: x4 = (-26)/(-26) = 1

	3*x3 + 13*x4 = 13		: x3 = (13-13)/3 = 0

	-x2 - x3 - 5*x4 = -7		: x2 = 7-0-5 = 2

	x1 + x2 + 3*y4 = 4		: x1 = 4-2-3 = -1

Thus x = [-1 2 0 1].'


S 4.7
______	

m       x1      x2      x3      x4
________________________________________
	 2*	 7	-7	-3
-1/2	 1	 5	-5	-3
-1	 2	 3	 1	 3
-1/2	 1	 4	-2	 2
________________________________________
	 2 	 7	-7	-3
	 0	 3/2*	-3/2	-3/2
 8/3	 0	-4	 8	 6
-1/3	 0	 1/2	 3/2	 7/2
________________________________________
	 2 	 7	-7	-3
	 0	 3/2	-3/2	-3/2
	 0	 0	 4*	 2
-1/2	 0	 0	 2       4
________________________________________
	 2 	 7	-7	-3
	 0	 3/2	-3/2	-3/2
	 0	 0	 4	 2
	 0	 0	 0	 3

The last matrix is U.

L = [	 1	 0	 0	 0
	 1/2	 1	 0	 0
	 1	-8/3	 1	 0
	 1/2     1/3	 1/2 	 1  ]

(ii)  To solve L*y = b:

m       x1      x2      x3      x4	b
______________________________________________
	 1*	  	  	  	 10
-1/2	 1/2	 1	  	  	 5
-1	 1	-8/3	 1	 	-6
-1/2	 1/2	 1/3	 1/2	 1	 3
______________________________________________
	 1	  	  	  	 10
	 0	 1*	  	  	 0 * makes next step redundant
 8/3	 0	-8/3	 1	 	-16
-1/3	 0	 1/3	 1/2	 1	-2
______________________________________________
	 1	  	  	  	 10
	 0	 1	  	  	 0
	 0	 0	 1*	  	-16
-1/2	 0	 0	 1/2	 1	-2
______________________________________________
	 1	  	  	  	 10
	  	 1	  	  	 0
	  	  	 1	  	-16
	  	  	  	 1	 6

and thus y = [10 0 -16 6].'   (Note that the
solution of Ly*b involved:

* forward substitution (solve for y1, then y2, etc.)

* multipliers which were read off directly from L

* nontrivial changes in the b column only

* the same changes in the b column as if we had
  carried out a forward elimination for A*x = b)

U*x = y is the standard backward substitution:

m       x1      x2      x3      x4	y
________________________________________________
	 2	 7	-7	-3	 10
	 	 3/2	-3/2	-3/2	 0
	 	 	 4	 2	-16     
	 	 	 	 3	 6   
________________________________________________
 3/2	 1	 7/2	-7/2	-3/2	 5
 1	 	 1	-1	-1	 0
-1/2		 	 1	 1/2	-4   
	 	 	 	 1*	 2      
________________________________________________
 7/2	 1	 7/2	-7/2	 0	 8
 1	 	 1	-1	 0	 2
	 	 	 1*	 0	-5      
	 	 	 	 1	 2      
________________________________________________
-7/2	 1	 1	 0	 0	-19/2
	 	 1*	 0	 0	-3
	 	 	 1	 0	-5
	 	 	 	 1	 2
________________________________________________
	 1	 	 	 	 2
	 	 1	 	 	-3
	 	 	 1*	 	-5     
	 	 	 	 1	 2      

and thus the solution is x = [2 -3 -5 1].'