Find the inverse of the given matrix by elementary transformation (3x3 matrix)
[0 1 2]
A= |1 2 3|
[3 1 3]
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A=[ 32107 ]
We know that A=IA (using row operations)
[ 32 107 ]=[ 1001 ]A
Using operation R 1
⇒R 1 −R 2
[ 1237 ]=[ 10−11 ]A
Using operation R 2
⇒R 2 −2R 1
[ 1031 ]=[ 1−2−13 ]A
Using operation R 1
⇒R 1 −3R 2
[ 1001 ]=[ 7−2−103 ]A
Therefore,
A −1 =[ 7−2−103 ]
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