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[M] Let $U$ be the $8 \times 4$ matrix in Exercise 36 in Section $6.2 .$ Find the closest point to $\mathbf{y}=(1,1,1,1,1,1,1,1)$ in $\operatorname{Col} U$ Write the keystrokes or commands you use to solve this problem.

$\hat{\mathbf{y}}=U U^{T} \mathbf{y}=\left[\begin{array}{c}{1.2} \\ {.4} \\ {1.2} \\ {1.2} \\ {.4} \\ {1.2} \\ {.4} \\ {.4}\end{array}\right]$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 3

Orthogonal Projections

Vectors

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Okay, This question wants us to find the determinant of May. So to do this we want to use co factors along the Rower column where it's easiest, and by easiest, we really mean most zeroes and good numbers to work with. So Row two is all zeros, so let's pick that. And just for reference, here's our co factor signed chart. So this says the determinative eh is equal to expanding on Grow, too. Wrote to Colin one Plus wrote You Colin to plus Row two column three their co factors. So doing this. Remember that our formula for the co factor ISS the sign that we get from our sign pattern times the matrix century, times the determinant from deleting that road code. But since it depends on that entry of the Matrix, we see that something really nice happens because everything in this rose zero. So every co factor term turns into zero, and since every term is 20 than our total determine, it must be zero. And this is another application off a fact from linear algebra. If you have a row or column of zeros, that means that you're determined in a zero so if you're if you have a full row of zeroes or a full column zeros that automatically makes your determinant zero. So we could have just seen Oh, hey, We have a row of zeroes here, so that means our determine it must be zero. But expanding these and co factors was also really simple and gave us the same answer.

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