Find the inverse of
A = é1 2 2ù
ê2 1 –2ú
ë2 –2 -6û
Solution:
To find the inverse of
a square matrix A, form the huge augmented matrix (A | I) where I
is the identity matrix. You then reduce the left-hand side. If A is
nonsingular, you will get (I | B), and this B will equal A-1.
é1
2 2 ï 1 0 0ù
(2)->(2)+2(1) é1 2 2
ï 1 0 0ù
(A | I) = ê2 1 –2 ï 0
1 0ú ê0 -3
–6 ï-2 1 0ú
ë2 –2 –6 ï 0
0 1û
(3)->(3)-2(1) ë0 –6 -10 ï-2 0 1û
(2)->1/3*(2) é1
2 2 ï 1
0 0ù
ê0 1 2 ï1/3 –1/3 0ú
ë0 –6 –10 ï -2
0 1û
(1)->(1)-2(2) é1 0
-2 ï-1/3 2/3 0ù
ê0 1 2 ï 2/3 –1/3 0ú
(3)->(3)+6(2) ë0
0 2 ï 2
-2 1û
(3)->
1/2*(3) é1 0 -2 ï-1/3
2/3 0ù
ê0 1 2 ï 2/3 –1/3 0ú
ë0 0 1
ï -1 -1 1/2û
(3)-> 1/2*(3) é1 0 0 ï 5/3 -4/3
1ù
ê0 1 0 ï-4/3 5/3 -1ú
ë0 0 1 ï
-1 -1 1/2û
A-1 =
é 5/3 -4/3 1ù
ê-4/3 5/3 -1ú
ë -1 -1 1/2û
Check:
A*A-1 = I.
é1
2 2ù é 5/3 -4/3 1ù é1 0 0ù
ê2 1
–2ú ê-4/3 5/3 -1ú = ê0
1 0ú
ë2 –2 -6û ë
-1 -1 1/2û ë0
0 1û