2. Write a function usolve, analogous to function 1solve in section 7.2.2,to solve an upper triangular system Ux = y. [Hint: Loops can be runbackwards, say, from n down to 1, by typing in MATLAB: for i=n: -1:1. Remember also that the diagonal entries in U are not necessarily 1.]I at 140matrix below the main diagonal) are zeroed out during Gaussian elimination,Therefore this space could be used to store the strict lower triangle of L, andsince the diagonal entries in L are always 1, there is no need to store these.The entries in L are simply the multipliers (the variable called mult in theprevious code) used to eliminate entries in the lower triangle of A. Thereforethe Gaussian elimination code can be modified as follows:% LU factorization without pivoting.for j=1:n-1for i=j+1:nmult = A(i, j)/A (j,j);end;A(i, j) = mult;end;% Loop over rows below j.% Subtract this multiple of% row j from row i% to make A (i, j)=0.A(i, j+1:n) = A (i, j+1:n) - mult*A (j, j+1:n); % This works on% columns j+1% through n.% Since A(i, j)becomes 0, use% the space to store L(i, j).a linear system Ax = LUx =As noted previously, once A has been factored in the form LU, we can solveb by first solving Ly = b and then solving11 0l21 22UxTo solve Ly= b, we solve the first equation for y1, then substitute= y.this value into the second equation in order to find y2, etc., as shown below:90-0-::l n 1 l2n% Loop over columns....y₁ = b₁/l11,y2 = (b2l21 y1)/22,:⠀⠀CHAPTER 7Y₁ =i-1bi - Σlijv; /lii.j=1The procedure simplifies slightly when the diagonal entries in L are all 1s.The following MATLAB function solves Ly - b, when L has a unit diagonaland the strict lower triangle of L is stored:function y = 1solve (L, b)%% Given a lower triangular matrix L with unit diagonal% and a vector b,% this routine solves Ly = b and returns the solution y.DIRECT

In this question we have to develop a MATLAB function named usolve, which is used to solve a system…

2. Write a function usolve, analogous to function 1solve in section 7.2.2, to solve an upper triangular system Ux y. [Hint: Loops can be run backwards, say, from n down to 1, by typing in MATLAB: for i=n: - 1:1. Remember also that the diagonal entries in U are not necessarily 1.] =

2. Write a function usolve, analogous to function 1solve in section 7.2.2, to solve an upper triangular system Ux y. [Hint: Loops can be run backwards, say, from n down to 1, by typing in MATLAB: for i=n: - 1:1. Remember also that the diagonal entries in U are not necessarily 1.] =

Database System Concepts

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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

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Question

2. Write a function usolve, analogous to function 1solve in section 7.2.2,
to solve an upper triangular system Ux = y. [Hint: Loops can be run
backwards, say, from n down to 1, by typing in MATLAB: for i=n: -
1:1. Remember also that the diagonal entries in U are not necessarily 1.]
I at

140
matrix below the main diagonal) are zeroed out during Gaussian elimination,
Therefore this space could be used to store the strict lower triangle of L, and
since the diagonal entries in L are always 1, there is no need to store these.
The entries in L are simply the multipliers (the variable called mult in the
previous code) used to eliminate entries in the lower triangle of A. Therefore
the Gaussian elimination code can be modified as follows:
% LU factorization without pivoting.
for j=1:n-1
for i=j+1:n
mult = A(i, j)/A (j,j);
end;
A(i, j) = mult;
end;
% Loop over rows below j.
% Subtract this multiple of
% row j from row i
% to make A (i, j)=0.
A(i, j+1:n) = A (i, j+1:n) - mult*A (j, j+1:n); % This works on
% columns j+1
% through n.
% Since A(i, j)
becomes 0, use
% the space to store L(i, j).
a linear system Ax = LUx =
As noted previously, once A has been factored in the form LU, we can solve
b by first solving Ly = b and then solving
11 0
l21 22
Ux
To solve Ly= b, we solve the first equation for y1, then substitute
= y.
this value into the second equation in order to find y2, etc., as shown below:
90-0-
::
l n 1 l2n
% Loop over columns.
...
y₁ = b₁/l11,
y2 = (b2l21 y1)/22,
:
⠀⠀
CHAPTER 7
Y₁ =
i-1
bi - Σlijv; /lii.
j=1
The procedure simplifies slightly when the diagonal entries in L are all 1s.
The following MATLAB function solves Ly - b, when L has a unit diagonal
and the strict lower triangle of L is stored:
function y = 1solve (L, b)
%
% Given a lower triangular matrix L with unit diagonal
% and a vector b,
% this routine solves Ly = b and returns the solution y.
DIRECT

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