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# Solution of the linear equation using LU decomposition

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Category There are doubts that the dedicated formulas from the textbook in C++
Where `l [n] [n] `and `u [n] [n] `previously counted

``````double y [n]; // Find y [n] by formula (2.11)
Sum = 0;
for (int i = 0; i & lt; n; i ++)
For (int k = 0; k & lt; n-1; k ++) {
Sum + = L [i] [k] * y [k];
y [i] = b [i] - Sum;
}
COUT & LT; & LT; Endl & lt; & lt; "Y [n] =" & lt; & lt; Endl; // withdraw y [n]
FUNCSCREAN (Y, N);
Double x [n]; // Find X [n] by formula (2.13)
Sum = 0;
for (int i = n - 1; i & gt; = 0; i--)
For (int k = i; k & lt; n; k ++) {
Sum + = U [i] [k] * x [k];
x [i] = (y [i] - Sum) / (u [i] [i]);
}
``````

For example, formula 2.11 is much more like

``````for (int i = 0; i & lt; n; i ++) {
Sum = 0;
For (int k = 0; k & lt; i; k ++) {
Sum + = L [i] [k] * y [k];
}
y [i] = b [i] - Sum;
}
``````

than your

``````sum = 0;
for (int i = 0; i & lt; n; i ++)
For (int k = 0; k & lt; n-1; k ++) {
Sum + = L [i] [k] * y [k];
y [i] = b [i] - Sum;
}
``````

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