Show that if A_(n×n) is invertible then the inverse is unique
Since A is a "n\\times n" matrix, the linear transformation "T:x\\rightarrow Ax" is one-to-one
"\\Rightarrow" linear transformation "T:x\\rightarrow Ax" is invertible
"\\Rightarrow" A is invertible.
Suppose that "Ax=b" and "Av=Au=0,v\u2260u."
Then "A(x+v)=A(x+u)=b" , whereby "x+v\u2260x+u."
"\\therefore" "Ax=b" for some unique "x"
"\\Rightarrow" either there is no "v" such that "Av=0" or there is a unique "v" such that "Av=0" .
Since "A0=0" , we conclude that there is a unique "v" such that "Av=0" , and thus "T" is one-to-one.
Therefore, A is invertible.
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