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In Exercises 15 and $16,$ use the factorization $A=Q R$ to find the least-squares solution of $A \mathbf{x}=\mathbf{b} .$$$A=\left[\begin{array}{rr}{1} & {-1} \\ {1} & {4} \\ {1} & {-1} \\ {1} & {4}\end{array}\right]=\left[\begin{array}{rr}{1 / 2} & {-1 / 2} \\ {1 / 2} & {1 / 2} \\ {1 / 2} & {-1 / 2} \\ {1 / 2} & {1 / 2}\end{array}\right]\left[\begin{array}{rr}{2} & {3} \\ {0} & {5}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{-1} \\ {6} \\ {5} \\ {7}\end{array}\right]$$

$\hat{x}\left[\begin{array}{c}{\frac{29}{10}} \\ {\frac{9}{10}}\end{array}\right]$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 5

Least-Squares Problems

Vectors

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we are given the cue are the composition of a I want to find the least choir solution to a X equal Be so we first started computing que transposed being you do the multiplication and turns out to be 17 house and nine hours. And now since we have the cure the composition, we can find the least square solution By solving the system, Our ex hat is equal to you. Transpose be now our ease up square two by two metrics So we can just all way by inverting are so x hot is equal to well, the inverse of our turns out to be 1/10 off five months 302 That multiplies u transpose being so 17 house and nine house So 1/10 off we do the multiplication So we're left with 29 9 So the solution for X hat he's the vector 2.90 point nine And this is the least square solutions

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