Lecture 25
Lecture 25
Lecture 25
Analysis
Lecture 25
Lecture outline
1
A-1= det(A)adj(A)
Example 2
Use theorem 1 to find the inverse of the matrix A in the
previous example
1 0 1 6 0 -3
A= -1 3 0 Adj(A)= 2 1 -1
1 0 2 -3 0 3
1 0 1 1 0
-1 3 0 -1 3
det(A)= =1*3*2-1*3*1=3
1 0 2 1 0
6 0 -3
det(A)=3 Adj(A)= 2 1 -1
-3 0 3
And since
1
A-1= det(A)adj(A)
It follows:
2 0 -1
A-1= 2/3 1/3 -1/3
-1 0 1
Class assignment:
Let:
1 6 -3
A= -2 7 1
3 -1 4
Find:
(a) The matrix of cofactors.
(b) Adj(A)
(c) A-1 using the formula from theorem 1
Theorem 2 (Cramer’s rule)
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(A1) , det(A2) , ….. , det(An)
x1= det(A) x2= det(A) xn= det(A)
where Aj is the matrix obtained by replacing the entries
in the jth column of A by the entries of the matrix B
b1 a11 a12 : a1j : a1n
B= b2 a21 a22 : a2j : a2n
: : : : :
bn an1 an2 : anj : ann
Example 3
Use Cramer’s rule to solve:
x1-3x2+ x3= 4
2x1- x2 = -2
4x1 -3x3= 0
Solution
1 -3 1 4 -3 1
det(A)= 2 -1 0 ; det(A1)= -2 -1 0
4 0 -3 0 0 -3
1 4 1 1 -3 4
det(A2)= 2 -2 0 ; det(A3)= 2 -1 -2
4 0 -3 4 0 0
det(A) = -11; det(A1) = 30
det(A2) = 38; det(A3) = 40
We get:
x1=det(A1)/det(A) = - 30/11
x2=det(A2)/det(A) = - 38/11
x3=det(A3)/det(A) = - 40/11
Class assignment:
Use Cramer’s rule to solve:
a) 4x+ 5y =2
11x+ y +2z = 3
x+5y +2z = 1
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