Solution Manual Jaan Kiusalaas Numerical Methods In Engineering With Matlab 2nd 58 -

b_perm = P*b; y = forwardSub(L, b_perm); x = backSub(U, y); disp(x); ( x \approx [0.7234, -0.6809, -1.1064]^T ) Step 3: Compute inverse using LU decomposition For ( A^-1 ), solve ( A X = I ), column by column, reusing ( L, U, P ):

(Values are approximate, matching typical pivot choices.) First permute ( b ): ( b' = P b ). Then forward substitution: ( L y = b' ). Then back substitution: ( U x = y ). b_perm = P*b; y = forwardSub(L, b_perm); x

Manual’s MATLAB code:

The decomposition yields (as shown in manual): Manual’s MATLAB code: The decomposition yields (as shown

I’ve put together an explanatory piece based on the context of (and its solution) from the Solution Manual for Jaan Kiusalaas’ Numerical Methods in Engineering with MATLAB , 2nd Edition. -2 4 1

A = [3 -1 2; -2 4 1; 5 2 -3]; b = [1; 2; 3]; [L, U, P] = luDecomp(A); % P is permutation matrix

I = eye(3); invA = zeros(3); for j = 1:3 b_col = I(:, j); b_perm = P * b_col; y = forwardSub(L, b_perm); invA(:, j) = backSub(U, y); end disp(invA);