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Let $A=\left[\begin{array}{rr}{3} & {4} \\ {-2} & {1} \\ {3} & {4}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{11} \\ {-9} \\ {5}\end{array}\right], \mathbf{u}=\left[\begin{array}{r}{5} \\ {-1}\end{array}\right],$ and $\mathbf{v}=$ $\left[\begin{array}{r}{5} \\ {-2}\end{array}\right] .$ Compute $A \mathbf{u}$ and $A \mathbf{v},$ and compare them with $\mathbf{b}$ . Could $\mathbf{u}$ possibly be a least-squares solution of $A \mathbf{x}=\mathbf{b} ?$ (Answer this without computing a least-squares solution.)

u cannot be a least squares solution of $A \mathbf{x}=\mathbf{b}$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 5

Least-Squares Problems

Vectors

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Lectures

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In mathematics, a vector (…

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In Exercises $9-12,$ find …

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In Exercises 5 and $6,$ de…

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were given A and B and also the vectors U and we and were asked to compute first a U and U V and then decide whether either of them could be one of the least square solutions to a X equal. Be so we begin by computing a u do the multiplication you get Ah 382 and their will to do a V and he becomes 7 to 8. Now we compute the distance from me so a U minus b is minus till four minus two and the baby is mine. It was meant to be is to minus two and four now. Hey, you would be Ali Square Solutions to x equal be if a human is being so the vector in the bottom left where our focus now to the columns of a But we can immediately check that it is not so that factor minus 24 months, too. It's not orthogonal to the columns of a so we know that a you cannot be Ali square solution. Similarly, the vector two months to four So a B minus B. He's no door for go now to the columns of a either and That means that a bee cannot be one of the least square solutions to the system a X equal be

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