Machinery's Handbook, 31st Edition
128 USING MATRICES TO SOLVE A SYSTEM OF EQUATIONS
1 0 0 x 1 0 1 0 x 2 0 0 1 x 3
–4 8 12 16 3 –1 2 5 1 7 6 10
⇔
The coefficient matrix is transformed so that element c 11 is 1 and all other elements in the first column are 0, as follows: a) divide row 1 ( R 1 ) by - 4; b) multiply new R 1 by - 3, then add to R 2 ; and c) multiply R 1 by - 1, then add to R 3 . 4 –4 –--- 8 –4 --- 12 –4 --- 16 –4 --- 1 –2 –3
–4 3 – 3 ( ) – 1+6 ( ) 2+9 ( ) 5+12 ( ) 1– 1 ( ) 7+2 ( ) 6+3 ( ) 10+4 ( )
1 –2 –3 –4 0 5 11 17 0 9 9 14
⇒
⇒
3 –1 2 5 1 7 6 10
The resulting matrix is transformed so that element c 22 is 1 and all other elements in the second column are 0, as follows: a) divide R 3 by 9; b) multiply new R 3 by - 5, then add to R 2 ; c) multiply R 3 by 2, then add to R 1 ; and d) swap R 2 and R 3 .
1 –2+2 ( ) –3+2 ( ) –4 28 9 + --- 0 5– 5 ( ) 11– 5 ( ) 17 70 9 – --- 0 1 1 14 9 ---
1 0 –1 8 9 –-- 0 0 6 83 9 --- 0 1 1 14 9 ---
1 0 –1 8 9 –-- 0 1 1 14 9 --- 0 0 6 83 9 ---
1 –2 –3 –4 0 5 11 17
⇒
⇒
⇒
0 9 -- 9 9
-- 9 9
-- 14 9 ----
The resulting matrix is finally reduced so that element c 33 is 1 and all other elements in the third column are 0, as follows: a) divide R 3 by 6; b) multiply new R 3 by –1, then add to R 2 ; and c) add R 3 to R 1 .
8 9 – -- 83 54 + ---
1 0 –1 + 1 ( )
1 0 –1 8 9 –-- 0 1 1 14 9 --- 0 0 6 6 -- 83 9 6 ( ) ------
1 0 0 35 54 --- 0 1 0 1 54 --- 0 0 1 83 54 ---
14 9 --- 83 54 – ---
⇒
⇒
0 1 1– 1 ( )
83 54 ---
0 0 1
When the identity matrix has been formed, the last column contains the values of x 1 , x 2 , and x 3 that satisfy the original equations. x 1 x 2 x 3 83 54 = --- Checking that the solutions satisfy the original system: –4 ( ) + 8 ( ) + 12 ( ) = = 16 35 54 = --- 1 54 = --- 35 54 1 54 83 54 864 54
3 ( ) – 1 ( ) + 2 ( ) = 35 54 1 54 83 54 270 54 1 ( ) + 7 ( ) + 6 ( ) = = 10 = 5 35 54 1 54 83 54 540 54
Copyright 2020, Industrial Press, Inc.
ebooks.industrialpress.com
Made with FlippingBook - Share PDF online