Linear systems are particularly easy to solve when theyare in triangular form (i.e., all entries above or belowthe diagonal are zero). a. Solve the lower triangular system | x 1 = − 3 − 3 x 1 + x 2 = 14 x 1 + 2 x 2 + x 3 = 9 − x 1 + 8 x 2 − 5 x 3 + x 4 = 33 | by forward substitution, finding x 1 first, then x 2 ,then x 3 , and finally x 4 . b. Solve the upper triangular system | x 1 + 2 x 2 − x 3 + 4 x 4 = − 3 x 2 + 3 x 3 + 7 x 4 = 5 x 3 + 2 x 4 = 2 x 4 = 0 |
Linear systems are particularly easy to solve when theyare in triangular form (i.e., all entries above or belowthe diagonal are zero). a. Solve the lower triangular system | x 1 = − 3 − 3 x 1 + x 2 = 14 x 1 + 2 x 2 + x 3 = 9 − x 1 + 8 x 2 − 5 x 3 + x 4 = 33 | by forward substitution, finding x 1 first, then x 2 ,then x 3 , and finally x 4 . b. Solve the upper triangular system | x 1 + 2 x 2 − x 3 + 4 x 4 = − 3 x 2 + 3 x 3 + 7 x 4 = 5 x 3 + 2 x 4 = 2 x 4 = 0 |
Solution Summary: The author explains how to find the solution of the linear equation using elimination method.
Linear systems are particularly easy to solve when theyare in triangular form (i.e., all entries above or belowthe diagonal are zero). a. Solve the lower triangular system
|
x
1
=
−
3
−
3
x
1
+
x
2
=
14
x
1
+
2
x
2
+
x
3
=
9
−
x
1
+
8
x
2
−
5
x
3
+
x
4
=
33
|
by forward substitution, finding
x
1
first, then
x
2
,then
x
3
, and finally
x
4
. b. Solve the upper triangular system
|
x
1
+
2
x
2
−
x
3
+
4
x
4
=
−
3
x
2
+
3
x
3
+
7
x
4
=
5
x
3
+
2
x
4
=
2
x
4
=
0
|
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