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Let A = [2 -4; 2 5] Find A^{-1} and use it to solve AX=B where B=[3; 2] X=[x{1}; x{2}] where x_{1}=?? and x_{2}=?? .
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say A=[a b ; c d] then the inverse of A is B=[d -b; -c a]/(ad-bc); all the elements are divided by determinant of A
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