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magiclko

May be many of u would be knowing this.... incase u dnt check this out...

Let A = [a_ij]

then adj (A) = [C_ij] whr C_ij is the cofactor of a_ij

now let A is given by | a₁₁ a₁₂ a₁₃ .......... a_1n |

| a₂₁ a₂₂a₂₃.......... a_2n |

| ..................................|

|...................................|

| a_n1 a_n2a_n3.......... a_nn |

adj (A) = | C₁₁ C₁₂ C₁₃ .......... C_1n|

| C₂₁ C₂₂C₂₃.......... C_2n|

| ................................ ..|

|.................................. .|

| C_n1 C_n2C_n3.......... C_nn |

now we have a property in detrminants which states that "if all the elemnts of a prticular row or coloumn is multiplied with its cofactor then thr algebraic sum will be same as determinant's and if they are multiplied by cofactor of any other's row (or coloumn's ) elemnt, then its value will be 0"

using this property...

we'll be having

adj (A) = | |A| 0 0 .......... 0|

| 0 |A| 0......... 0|

| ...........................|

|............................|

| 0 0 0.........|A| |

here the diagonal elemts will be |A| and rest 0 (by the above property)

thrfore adj (A) = |A| I_n

also thr's this property if all the entries of a determinant of order n is replaced by thr corresponding cofactors then value of new-determinant D is D = |A|^n-1[power cofactor formula]

thrfore using this, here

|adj (A)| = |A|^n-1

this can be genralised to |adj (adj (adj..... (n tymes)..(adj (A))))|= |A|^{[n-1)^n]}

Hopefully It helps.....

Dnt go much in proof... just rember the formula...and the property... that will do

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