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Question

If verify that A 3 − 6 A 2 + 9 A − 4 I = O and hence find A −1

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Solution

Given matrix is,

A=[ 2 1 1 1 2 1 1 1 2 ]

The value of A 2 is,

A 2 =[ 2 1 1 1 2 1 1 1 2 ][ 2 1 1 1 2 1 1 1 2 ] =[ 6 5 5 5 6 5 5 5 6 ]

And, the value of A 3 is,

A 3 = A 2 A =[ 6 5 5 5 6 5 5 5 6 ][ 2 1 1 1 2 1 1 1 2 ] =[ 22 21 21 21 22 21 21 21 22 ]

Now substitute the value of A 2 and A 3 in A 3 6 A 2 +9A4I, we get,

A 3 6 A 2 +9A4I=[ 22 21 21 21 22 21 21 21 22 ]6[ 6 5 5 5 6 5 5 5 6 ]+9[ 2 1 1 1 2 1 1 1 2 ]4[ 1 0 0 0 1 0 0 0 1 ] =[ 2236+184 21( 30 )9+0 2130+9+0 21+309+0 2236+184 22+308+0 2130+9 21+309 2236+184 ] =[ 0 0 0 0 0 0 0 0 0 ]

Hence, it is verified that A 3 6 A 2 +9A4I=0.

Now, multiply both sides of A 3 6 A 2 +9A4I=0 by A 1 we get,

A 1 A 3 A 1 ( 6 A 2 )+ A 1 ( 9A ) A 1 ( 4I )=0 A 1 A( A 2 )6 A 1 A( A )+9 A 1 A 4A 1 I= A 1 0 I( A 2 )6I( A )+9I 4A 1 =0 A 2 6A+9I 4A 1 =0

Further simplify the above equation,

A 1 = A 2 6A+9I 4 = 1 4 ( [ 6 5 5 5 6 5 5 5 6 ]6[ 2 1 1 1 2 1 1 1 2 ]+9[ 1 0 0 0 1 0 0 0 1 ] ) = 1 4 [ 612+9 5+6+0 56+0 5+6+0 612+9 5+6+0 56+0 5+6+0 612+9 ] = 1 4 [ 3 1 1 1 3 1 1 1 3 ]


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