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Suppose x LS(A, b) and y N(A) Show that x + ty is a solution to LS

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Suppose x LS(A, b) and y N(A). Show that x + ty is a
solution to LS(A, b) for all t C.
Solution
x e LS(A,b) => Ax =b y e N(A) => Ay = 0 =>
t(Ay)=0 => A(ty )= 0 addind the two equations Ax =b A(ty
)= 0 we get Ax + Aty =b => A(x+ty) = b => x+ty e
LS(A,b)

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Suppose x LS(A, b) and y N(A). Show that x + ty is a solution to LS(A, b) for all t C. Solution x e LS(A,b) => Ax =b y e N(A) => Ay = 0 => t(Ay)=0 => A(ty )= 0 addind the two equations Ax =b A(ty )= 0 we get Ax + Aty =b => A(x+ty) = b => x+ty e LS(A,b) Name: Description: ...
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