# Lab3 - Exercise 9 A=[2 7 3 2 8 1 3 7 9 A=[A eye(3 A[1...

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Exercise 1 Det(B)= -622 Det(D)= Cannot find due to dimensions (i.e. Not a sqr matrix). Exercise 2 By entering D(3,2)=-7 into matrix D, the entry in the third row-second column has been replaced by a -7. Exercise 3 B 1 , B 2 , B 3 , are simply matrix B with the column matrix V replacing columns 1, 2, and 3 respectively. Exercise 4 Det(A)=6; ( 29 ( 29 5 6 30 det det 1 = = = A A x , ( 29 ( 29 2 6 12 det det 2 - = - = = A A y , ( 29 ( 29 1 6 6 det det 3 = = = A A z , ( 29 ( 29 4 6 24 det det 4 = = = A A w Exercise 5 - - - = 14 4 10 1 22 21 26 18 3 AB - - - - - = 16 6 22 26 27 1 26 28 14 BA Exercise 6 - - - = 11 13 33 15 AB - - - - - = 10 5 18 12 6 26 14 7 10 BA Exercise 7 - = 8 19 32 56 2 116 AD - - = 28 136 50 112 244 20 BD - - = + 20 155 82 168 246 156 BD AD - = + 11 6 22 20 B A ( 29 - - = + 20 155 82 168 246 156 D B A One can not find either A+D or DA because of their dimensions. Exercise 8

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B and D are inverse matrixes because B*D and D*B give the identity matrix.
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Unformatted text preview: Exercise 9 A=[2 7 3; 2 8 1; 3 7 9] A=[A eye(3)] A([1 3],:)=A([3 1],:) A(1,:)=A(1,:)/3 A(2,:)=A(2,:)-2*A(1,:) A(3,:)=A(3,:)-2*A(1,:) A(2,:)=(3/10)*A(2,:) A(3,:)=A(3,:)-(7/3)*A(2,:) A(3,:)=(2)*A(3,:) A(2,:)=A(2,:)+(3/2)*A(3,:) A(1,:)=A(1,:)+(3)*A(3,:) A(1,:)=A(1,:)-(3)*A(3,:) A(1,:)=A(1,:)-(3)*A(3,:) A(1,:)=A(1,:)-(7/3)*A(2,:) -----=-5 / 2 5 / 7 2 5 / 4 5 / 9 3 5 / 17 5 / 42 13 1 A B=[1 -3 0; -1 2 -1; 0 -2 -2] Det(B)=0 A=[B eye(3)] A(2,:)=A(2,:)+A(1,:) A=[B eye(3)] A(3,:)=A(3,:)+2*A(2,:) A(1,:)=A(1,:)+3*A(1,:) B*inverseB Does not equal B No inverse of B; determinate equals 0....
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Lab3 - Exercise 9 A=[2 7 3 2 8 1 3 7 9 A=[A eye(3 A[1...

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