Question 1165251
<pre>
{{{matrix(1,3,

A,""="",
(matrix(3,3,
3,1,6,
-6,0,-16,
0,8,-17)))}}}

{{{matrix(1,3,

L,""="",
(matrix(3,3,
1,0,0,
a,1,0,
b,c,1))
)}}}

{{{matrix(1,3,
U,""="",
(matrix(3,3,
d,g,h,
0,e,j,
0,0,f)))}}}

{{{matrix(1,5,
LU,""="",
(matrix(3,3,
1,0,0,
a,1,0,
b,c,1))(matrix(3,3,
d,g,h,
0,e,j,
0,0,f)),
""="",
(matrix(3,3,
d,g,h,
ad,ag+e,ah+j,
bd,bg+ce,bh+cj+f))
)}}}

{{{matrix(1,3,
LU,""="",A)}}}

{{{matrix(1,3,
(matrix(3,3,
d,g,h,
ad,ag+e,ah+j,
bd,bg+ce,bh+cj+f)),""="",

(matrix(3,3,
3,1,6,
-6,0,-16,
0,8,-17))

)}}}

d=3, g=1, h=6, ad=-6,      ag+e=0,      ah+j=-16 
             a(3)=-6  (-2)(1)+e=0  (-2)(6)+j=-16
                a=-2       -2+e=0      -12+j=-16
                              e=2          j=-4

  bd=0,        bg+ce=8,          bh+cj+f=-17
b(3)=0   (0)(1)+c(2)=8  (0)(6)+(4)(-4)+f=-17
   b=0            2c=8             -16+f=-17 
                   c=4                 f=-1

[Yes, I know I could have gone to online solvers and
not have to do that by hand.  But the whole reason for
doing this problem is NOT to get the answer. Nobody in 
their right mind would solve a 3x3 system using this
method!  But a computer would solve many 20+ x 20+ 
systems this way!  The reason for solving this is to 
learn what the computers do.]

{{{matrix(1,3,

L,""="",
(matrix(3,3,
1,0,0,
-2,1,0,
0,4,1))
)}}}

{{{matrix(1,3,
U,""="",
(matrix(3,3,
3,1,6,
0,2,-4,
0,0,-1)))}}}

{{{matrix(1,5,
LU,""="",
(matrix(3,3,
1,0,0,
-2,1,0,
0,4,1))(matrix(3,3,
3,1,6,
0,2,-4,
0,0,-1)),
""="",
(matrix(3,3,
3,1,6,
-6,0,16,
0,8,-17))
)}}}
 
Now we solve the lower triangular system, Ly<sub>k</sub> = b<sub>k</sub>

{{{matrix(1,3,

(matrix(3,3,
1,0,0,
-2,1,0,
0,4,1))(matrix(3,1,y[1],y[2],y[3])),""="",(matrix(3,1,0,4,17))
)}}}

y<sub>1</sub> = 0     -2y<sub>1</sub>+y<sub>2</sub> = 4     0y<sub>1</sub>+4y2+y<sub>3</sub> = 17
          -2(0)+y<sub>2</sub> = 4        4(4)+y<sub>3</sub> = 17
                y<sub>2</sub> = 4          16+y<sub>3</sub> = 17 
                                   y<sub>3</sub> = 1

Finally we solve U<sub>(x,y,z)</sub> = y<sub>k</sub>

{{{matrix(1,3,

(matrix(3,3,
3,1,6,
0,2,-4,
0,0,-1))(matrix(3,1,x,y,z)),""="",(matrix(3,1,0,4,1))
)}}}

     3x+y+6z = 0     2y-4z = 4     -z = 1
                                    z = -1
                  2y-4(-1) = 4
                      2y+4 = 4
                        2y = 0
                         y = 0
  3x+0+6(-1) = 0
        3x-6 = 0
          3x = 6
           x = 2

So the solution to the system is (x,y,z) = (2,0,-1)
    
Edwin</pre>