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solution6

# solution6 - Assignment#6 prob 1 A=[3 2 5;7 1 6;7 3 7 B=[5 2...

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% Assignment #6 % prob 1 A=[3 2 5;7 1 6;7 3 7]; B=[5 2 9;6 8 4;6 3 1]; disp('det(A)') det(A) disp('det(B)') det(B) disp('inv(A)') inv(A) disp('inv(B)') inv(B) Anew=[3 2 5;7 1 6;10 3 11]; disp('det(Anew)') det(Anew) disp('inv(Anew') inv(Anew) det(A) ans = 23 det(B) ans = -254 inv(A) ans = -0.4783 0.0435 0.3043 -0.3043 -0.6087 0.7391 0.6087 0.2174 -0.4783 inv(B) ans = 0.0157 -0.0984 0.2520 -0.0709 0.1929 -0.1339 0.1181 0.0118 -0.1102 det(Anew) ans = 0 inv(Anew Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.316376e-017. ans = 1.0e+015 * 0.6906 0.6906 -0.6906 1.6772 1.6772 -1.6772 -1.0852 -1.0852 1.0852 The determinate of the altered matrix A (called Anew) is zero because the third row is not independent of the first two rows. In fact, the third row is the sum of the first two rows. Whenever all of the rows of a matrix are not independent, the determinant of that matrix is zero. Such a matrix is called a singular matrix. The fact that the matrix Anew is singular means that it does not have an inverse. Thus the warning after the attempt to find its inverse.

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% Prob. 2 disp('solution using matrices') A=[3 4 -2;1 -1 3;2 3 -1]; B=[3;2;1]; disp('Let X be the matrix whose entries are x, y and z repectively') X=inv(A)*B disp(' ') solution using matrices Let X be the matrix whose entries are x, y and z repectively

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