solve each set of equations by the method of finding the inverse of the coefficient matrix. Hint: See Example 3. y + z = 4 2x + y - z = -1 3x + 2y + 2z
solve each set of equations by the method of finding the inverse of the coefficient matrix. Hint: See Example 3. y + z = 4 2x + y - z = -1 3x + 2y + 2z
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question

Transcribed Image Text:solve each set of equations by the method of finding the inverse of
the coefficient matrix. Hint: See Example 3.
y + z =
4
2x + y
z = -1
3x + 2y + 2z =

Transcribed Image Text:• Example 3. For the matrix M of the coefficients in equations (6.7) or (6.9), find M-1.
1
0 -1
M = -2
3
1
-3
2
We find det M = 3. The cofactors of the elements are:
3 0
= 6,
-3 2
-2 0
= 4,
-2
1st row :
3
= 3.
1
2
1
-3
1
1
1
2nd
= 3,
= 3,
= 3.
-3
row :
-3
2
2
|0
1 0
= 3.
-2 3
-1
1
-
3rd row :
- 3,
2,
-2
Then
6 4 3
C =
1
CT
det M
6 3 3
1
4 3 2
3
3 3 3
3 3 3
so M-1 -
= -
3 2 3
Now we can use M-1 to solve equations (6.9). By (6.12), the solution is given by
the column matrix r = M-'k, so we have
6 3 3
1
4 3 2
3
3 3 3,
-10
or x = 1, y = 1, z = -4. (See Problem 12.)
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