Dec 27, 2012 Thread starter #1 Y Yankel Active member Jan 27, 2012 398 Hello I need some help proving the next thing, I can't seem to be able to work it out.. Let A be an nxn matrix. Prove that: [tex](adj A)^{-1} = adj(A^{-1})[/tex] Thanks...
Hello I need some help proving the next thing, I can't seem to be able to work it out.. Let A be an nxn matrix. Prove that: [tex](adj A)^{-1} = adj(A^{-1})[/tex] Thanks...
Dec 27, 2012 #2 Deveno Well-known member MHB Math Scholar Feb 15, 2012 1,967 $A = IA = A^*(A^*)^{-1}A$ so: $A^* = (A^*(A^*)^{-1}A)^* = A^*((A^*)^{-1})^*A$ therefore: $A^*A^{-1} = A^*((A^*)^{-1})^*$ and multiplying on the left by $(A^*)^{-1}$ we get: $A^{-1} = ((A^*)^{-1})^*$ so $(A^{-1})^* = ((A^*)^{-1})^{**} = (A^*)^{-1}$
$A = IA = A^*(A^*)^{-1}A$ so: $A^* = (A^*(A^*)^{-1}A)^* = A^*((A^*)^{-1})^*A$ therefore: $A^*A^{-1} = A^*((A^*)^{-1})^*$ and multiplying on the left by $(A^*)^{-1}$ we get: $A^{-1} = ((A^*)^{-1})^*$ so $(A^{-1})^* = ((A^*)^{-1})^{**} = (A^*)^{-1}$
Dec 28, 2012 Thread starter #3 Y Yankel Active member Jan 27, 2012 398 thanks, took me some time to understand your proof, but now I see it, nice one !