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Question

For the matrix show that A 3 − 6 A 2 + 5 A + 11 I = O. Hence, find A −1.

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Solution

Given matrix is,

A=[ 1 1 1 1 2 3 2 1 3 ]

The value of A 2 is,

A 2 =[ 1 1 1 1 2 3 2 1 3 ][ 1 1 1 1 2 3 2 1 3 ] =[ 1+1+2 1+21 13+3 1+26 1+4+3 169 21+6 223 2+3+9 ] =[ 4 2 1 3 8 14 7 3 14 ]

The value of A 3 is,

A 3 = A 2 A =[ 4 2 1 3 8 14 7 3 14 ][ 1 1 1 1 2 3 2 1 3 ] =[ 4+2+2 4+41 46+3 3+88 3+164 32442 73+28 7614 7+9+42 ] =[ 8 7 1 23 27 69 32 13 58 ]

Substitute the values of A 3 , A 2 and A in A 3 6 A 2 +5A+11I.

A 3 6 A 2 +5A+11I=[ 8 7 1 23 27 69 32 13 58 ]6[ 4 2 1 3 8 14 7 3 14 ]+5[ 1 1 1 1 2 3 2 1 3 ]+11[ 1 0 0 0 1 1 0 0 1 ] =[ 824+5+11 712+5+0 16+5+0 23+18+5+0 2748+10+11 69+8415+0 3242+10+0 13+185+0 58+84+15+11 ] =[ 0 0 0 0 0 0 0 0 0 ]

Hence, it is verified that A 3 6 A 2 +5A+11I=0.

Multiply both sides of A 3 6 A 2 +5A+11I=0 by A 1 we get,

A 1 A 3 A 1 ( 6 A 2 )+ A 1 ( 5A )+ A 1 ( 11I )=0 A 1 A( A 2 )6 A 1 A( A )+5 A 1 A+ 11A 1 I= A 1 0 I( A 2 )6I( A )+5I+ 11A 1 =0 A 2 6A+5I+ 11A 1 =0

Further simplify the above equation,

A 1 = 6A A 2 5I 11 = 1 11 ( 6[ 1 1 1 1 2 3 2 1 3 ][ 4 2 1 3 8 14 7 3 14 ]5[ 1 0 0 0 1 0 0 0 1 ] ) = 1 11 ( [ 645 620 610 6+35 1285 18+140 1270 6+30 18145 ] ) = 1 11 [ 3 4 5 9 1 4 5 3 1 ]


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