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$[\mathbf{M}]$ In Exercises $14-17,$ follow the instructions given for Exercises $3-6 .$$$3 x_{1} x_{2}+5 x_{1} x_{3}+7 x_{1} x_{4}+7 x_{2} x_{3}+5 x_{2} x_{4}+3 x_{3} x_{4}$$

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Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 3

Constrained Optimization

Introduction to Matrices

Oregon State University

Baylor University

Idaho State University

Lectures

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In Exercises $1-16,$ solve…

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02:34

In Exercises $5-14,$ solve…

Okay, so for problem 14. By the Given call Jessica form, we can write down the matrix, which is one have, uh, zeros on the diagonal. And you have three and five and seven on first row. So should be the same on the first column. And 307 and five. I'm second Go save it for the second, Colin and the rest will be three and three. Okay, Now, since this is a four by four matrix and these kind of hard to find out the arguments by hand, it's I would recommend to do this by the way over Mel va. So are biggest. Biggest. Ah, I value will be some 0.5 and the corresponding again. Victor, which is the the unit vector unit Vector we take this case should be in negative 0.5 connective 15 negative 150.5 and negative 0.5. So is this our, um, maximum two attained with the I was with the constraint x x transpose times X equals one and the unit vector will be this vector. Now when we are given x transpose times ags, you close one and extra suppose times u is zero. That is to find. Excuse me. That is too. Find a second maximum. I value. So I'll say s second, my maximum. I mean value, which in this case should be connective one a half. And this the, um by our definition, a power theory, Um, seven in our textbook. So we're done.

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