Verify A (a d j A) = (a d j A) A = |A|I ...

Verify `A (a d j A) = (a d j A) A = |A|I " " " " " ` where `A=[(2 ,3),(-4,-6)]`

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Let `A=[(2 ,3),(-4,-6)]`

`adj A= [(-6 ,-3),(4,2)]`

`|A|=|(2 ,3),(-4,-6)|`

` " " " " =2xx(-6)-3xx(-4)=-12+12=0`

Calculating `A(adj A)`

` " " " " =[(2 ,3),(-4,-6)][(-6 ,-3),(4,2)]`

` " " " " =[[2(-6)+3(4),2(-3)+3(2)],[-4(-6)+(-6)4,-4(-3)+(-6)2]]=[[-12+12,-6+6],[+24-24,12-12]]`

` " " " " =[[0,0],[0,0]]`

Similarly, `(adj A)A`

` " " " " =[(-6 ,-3),(4,2)][(2 ,3),(-4,-6)]`

` " " " " =[[-6(2)+(-3)(-4),-6(3)+(-3)(-6)],[4(2)+2(-4),4(3)+2(-6)]]=[[-12+12,-18+18],[8-8,12-12]]`

` " " " " =[[0,0],[0,0]]`

Also, `|A|I`

` " " " " =0I=O`

`therefore` `A (a d j A) = (a d j A) A = |A|I`

Hence proved

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