If A=[2 3 1 2], prove that A3-4A​2+A=0. Hence find A-1.

We have,A = 2312A2 = A . A =  2312  2312 = 4+36+62+23+4 = 71247A3 = A2.A = 712472312 = 14+1221+248+712+14 = 26451526Now, LHS = A3 - 4A2 + A=26431526 - 471247 + 2312=26451526 + -28-48-16-28 + 2312=26-28+245-48+315-16+126-28+2 = 0000 = 0 = RHSHence,A3 - 4A2 + A = 0A2 - 4A + 1 = 0A2 - 4A + I = 0A-1AA - 4A-1A + A-1I = 0IA - 4I + A-1 = 0A-1 = 4I - AA-1 = 4004 - 2312 = 2-3-12

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A is of order 1*3 so how will you multiply A.
no. of rows will not be equal to no. of columns.......
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first find A2 that would be [ 7 12 4 7] then multiply A with A2.. A3 will be
[ 26 45 15 26 ] then applt the values in the equation A3- 4 A2+ A.. u get zero... then u get the inverse as [2 -3 -1 2] by multiplying A inverse to the equation..
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Can you once more tell me clearly how to get A-1?
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yup sure
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first by finding |A| (determinant).. n then the minors n cofactors...this will give u tha adjoint.. then by using the formula..
A inverse = 1/|A|×adjoint A.
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u can solve it another way.. fir that refer to Q 13 solution of page no 132
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I got it. Thank you so much for your help.
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Anytime :).. ur most welcome..^_^
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