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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Lecture 23. Adjoint formula for A -1 Cramer’s rule. Lecture Outline. C 11 C 12….… C 1n C 21 C 22….… C 2n : : : : : : C n1 C n2….… C nn . Definition.

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 23

  2. Adjoint formula for A-1 Cramer’s rule. Lecture Outline

  3. C11 C12….… C1n C21 C22….… C2n : : : : : : Cn1 Cn2….… Cnn Definition IfAis anynxnmatrix andCijis the cofactor ofaij,then the matrix is called the matrix of cofactors fromA. The transpose of this matrix is called the adjoint ofAand is denoted byadj (A).

  4. The cofactors ofAare C11=6; C12=2; C13=-3 C21=0 ; C22=1; C23=0 C31=-3; C32=-1;C33=3 1 0 1 -1 3 0 1 0 2 A= It follows that the matrix ofcofactors ofAis 6 2 -3 0 1 0 -3 -1 3 Example 1

  5. 6 2 -3 0 1 0 -3 -1 3 6 0 -3 2 1 -1 -3 0 3 Adj(A)= matrix of cofactors ofA And the adjoint ofAis: Transpose

  6. If A is an invertible matrix, then A-1= 1 adj(A) det(A) Theorem 1

  7. 1 0 1 -1 3 0 1 0 2 6 0 -3 2 1 -1 -3 0 3 A= Adj(A)= 1 0 1 -1 3 0 1 0 2 1 0 -1 3 1 0 det(A)= =1*3*2-1*3*1=3 Example 2 Use theorem 1 to find the inverse of the matrix A in the previous example

  8. And since 6 0 -3 2 1 -1 -3 0 3 A-1= Adj(A)= It follows: 2 0 -1 2/3 1/3 -1/3 -1 0 1 1 adj(A) A-1= det(A) det(A)=3

  9. 1 6 -3 -2 7 1 3 -1 4 A= Class Assignment: Exercise set 2.3-1 please submit to TA at the end of the lecture Let: • Find: • The matrix of cofactors. • Adj(A) • A-1 using the formula from theorem 1

  10. If AX=B is a system of linear equations in n unknowns such that det(A)=0, then the system has a unique solution. This solution is: det(An) det(A1) det(A2) , , ….. , x1= x2= xn= det(A) det(A) det(A) Theorem 2 (Cramer’s rule) where Aj is the matrix obtained by replacing the entries in the jth column of A by the entries of the matrix B b1 b2 : bn a11 a21 an1 a12 a22 an2 : : : : a1j a2j : anj : : : : a13 a23 : ann B=

  11. 1 -3 1 2 -1 0 4 0 -3 • -3 1 • -2 -1 0 • 0 0 -3 1 4 1 2 -2 0 4 0 -3 1 -3 4 2 -1 -2 4 0 0 Solution ; det(A1)= ; det(A3)= det(A)= det(A2)= Example 3 Use Cramer’s rule to solve: x1-3x2+ x3=4 2x1- x2=-2 4x1-3x3=0

  12. det(A1) det(A2) det(An) , , ….. , x1= x2= xn= det(A) det(A) det(A) det(A)=-11; det(A1)=30 det(A2)=38; det(A3)=40 Applying Cramer’s rule We get: x1=det(A1)/det(A)=-30/11 x2=det(A2)/det(A)=-38/11 x3=det(A3)/det(A)=-40/11

  13. Class Assignment: Exercise set 2.3-6 and 2.3-11 please submit to TA at the end of the lecture Use Cramer’s rule to solve: 4x+ 5y =2 11x+ y+2z =3 x+5y+2z=1 a) b) Use Cramer’s rule to solve for z without solving for x,y, and w 4x+ y+ z+ w = 6 3x+7y- z+ w = 1 7x+3y-5z+8w=-3 x+ y+ z+2w = 3

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