From todays Quiz (#11):

A = {{2,0,0},{1,1,0},{-1,3,-3}};
B = {{2,0,0},{0,-1,0},{0,0,3}};
Cm = {{-2,4,2},{0,4,2},{0,0,3}};

A//MatrixForm

B//MatrixForm

Cm//MatrixForm

Transpose[A]//MatrixForm

Det[A.B]

Det[A]*Det[B]

Det[Transpose[A]]*Det[B]

Det[Cm]

Det[Inverse[Cm]]

The following matrix A was taken from the lecture on 2016.02.08 (see the notebook).  We then broke down A into the product of a lower triangular matrix and an upper triangular matrix by performing elementary matrix type 3 matrix multiplications:

A = {{4,2,-1,7},{1,-2,3,-2},{3,-5,6,6},{4,-2,3,9}};
Lm = {{1,0,0,0},{1/4,1,0,0},{3/4,13/5,1,0},{1,8/5,12/17,1}};
Um = {{4,2,-1,7},{0,-(5/2),(13/4),-(15/4)},{0,0,-(17/10),(21/2)},{0,0,0,(10/17)}};

A//MatrixForm
Lm//MatrixForm
Um //MatrixForm

Lm.Um//MatrixForm

If we want to solve AX=B, and we have A = LU, then we are trying to solve:

(LU)X = B
L(UX) = B

To do this, we can first define Y = UX, and solve:

LY = B

for Y.  Now that we know Y, we can solve for the original unknown column matrix X in :

UX = Y.

B = {{5},{7},{-3},{1/3}};

B//MatrixForm

Y = {{y1},{y2},{y3},{y4}};

(Lm.Y )//MatrixForm

B//MatrixForm

y1 = 5;
y2 = 7-5/4;
y3=-3-(((3 y1)/4)+((13 y2)/5));
y4 = 1/3 -(y1+((8 y2)/5)+((12 y3)/17));

B//MatrixForm

Lm.Y//MatrixForm

X = {{x1},{x2},{x3},{x4}};

Um.X//MatrixForm

Y//MatrixForm

Solve[((10 x4)/17) == (74/51),x4]

x4 = 37/15;

Solve[-((17 x3)/10)+((21 x4)/2)==-(217/10),x3]

x3 = 28;

Solve[-((5 x2)/2)+((13 x3)/4)-((15 x4)/4)==(23/4),x2]

x2 = 152/5;

Solve[4 x1+2 x2-x3+7 x4 == 5,x1]

x1 = -169/15;

X//MatrixForm

A.X//MatrixForm