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$[\mathbf{M}]$ In Exercises $14-17,$ follow the instructions given for Exercises $3-6 .$$$-6 x_{1}^{2}-10 x_{2}^{2}-13 x_{3}^{2}-13 x_{4}^{2}-4 x_{1} x_{2}-4 x_{1} x_{3}-4 x_{1} x_{4}+6 x_{3} x_{4}$$

a) By theorem $6,$ the maximum value of $x^{T} A x$ subject to the constraint $x^{T} x=$ 1 is the greatest eigenvalue of $A,$ which is $\lambda_{1}=-4$b) See solutionc) By Theorem 7 , the maximum value of $x^{T} A x$ subject to the constraints $x^{T} x=1$ and $x^{T} u=0$ is the second greatest eigenvalue of $A, \lambda_{2}=-10$ .

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 3

Constrained Optimization

Introduction to Matrices

Missouri State University

Campbell University

Baylor University

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Okay, so for problem 16 by the given that he quit already formed First write down the Matrix four by four matrix Negative six Naked actor Too negative. Too naked, Too naked, too negative. 10 and zeros zero active. Too literal. Active 13 and three next to zero three and active 13. Okay, now bye. Um, calculator. Or you can use it for my alfa Or use uh, programming to find out the maximum I get value for this matrix is negative for and the corresponding you didn't factor should be point 87 and detective point 29 and connective 0.29. A negative 0.29 now. So this is the make some value we can attain. We did come strained extra suppose time zags to be one now for the mix boom with the constraint of X transpose times x one and X transpose times You because zero We just need to find a second largest dog and value, which is negative 10 in this case. So this

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