### 2.69 Obtain the LU decomposition of a matrix

Problem: Given a matrix $$A$$, ﬁnd matrix $$L$$ and $$U$$ such that $$LU=A$$. The matrix $$L$$ will have $$1$$ in all the diagonal elements and zeros in the upper triangle. The matrix $$U$$ will have $$0$$ in the lower triangle as shown in this diagram.

 Mathematica Remove["Global*"] mat = {{1, -3, 5}, {2, -4, 7}, {-1, -2, 1}}; {lu,pivotVector,conditionNumber}= LUDecomposition[mat]  {{{1,-3,5}, {2,2,-3}, {-1,-(5/2),-(3/2)}}, {1,2,3},1}  lower=lu*SparseArray[{i_,j_}/;j1,{3,3}]+ IdentityMatrix[3]  {{1,0,0}, {2,1,0}, {-1,-(5/2),1}}  upper=lu SparseArray[{i_,j_}/;j>=i->1,{3,3}]  {{1,-3,5}, {0,2,-3}, {0,0,-(3/2)}}  lower.upper  {{1,-3,5}, {2,-4,7}, {-1,-2,1}} 

 Matlab clear all; A=[1 -3 5; 2 -4 7; -1 -2 1]; [L,U,p]=lu(A)  L = 1.0000 0 0 -0.5000 1.0000 0 0.5000 0.2500 1.0000 U = 2.0000 -4.0000 7.0000 0 -4.0000 4.5000 0 0 0.3750 p = 0 1 0 0 0 1 1 0 0  L*U  ans = 2 -4 7 -1 -2 1 1 -3 5 

 Maple A:=Matrix([[1,-3,5],[2,-4,7],[-1,-2,1]]); p,l,u:=LinearAlgebra:-LUDecomposition(A); ` $\left [ \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ] , \left [\begin {array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & -\frac {5}{2} & 1 \end {array}\right ] , \left [\begin {array}{ccc} 1 & -3 & 5 \\ 0 & 2 & -3 \\ 0 & 0 & -\frac {3}{2} \end {array}\right ] \right ]$