mathbfm-in-exercises-14-17-follow-the-instructions-given-for-exercises-3-6-3-x_1-x_25-x_1-x_37-x_1-x

Question

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

Runpeng Li · Accepted Answer

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&#x27;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&#x27;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&#x27;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&#x27;re done.

Jack Chen · Answer

Okay, so this is the symmetric metrics Associates to the quadratic form. Um, so I can&#x27;t you later. The 1st 2 on Agon very off this metric stays on 17. And the Lambda Tweet Coast to 13. And the on the Agon vector off Lambda One will be U equals to their 10.50 point 50.5035 So this is a unit vector. Also part a, um, So the maximum value off Q X Subject two and me you need backed her ex is the largest tag umbrella, which is 17 and the puppy. When X equals to this, I get this unit, I can vet her. Um, que x achieves 17. And, uh, if we add another constraint, we require X Toby, the unit vector, and we required X is perpendicular to this. I can vent you. Then it makes it more value off. Q X is the second largest toe again. Very, which is 30

Jack Chen · Answer

Okay, so in this program, the symmetric matrix Associates, to this chaotic phone use of four by four matrix. Thank you. So by the calculator, um, B ag value off this, um, off this symmetric metric sees Lambda wine coast of 9 11 actually cost three eso. We just admit this lambda three and Lambda for because both often are less Land three and the corresponding Eigen vector for Lambda One will be you because toe one over. Route off six minus 20 While what? So based on these informations, the maximum value off the pathetic phone Q x for any you need Specter, uh, is Lamda won the cost of my life. So this is a maximum value off this quadratic phone kind of chief. So I&#x27;m the poppy. So when X equals two, um, this I get better you Q x achieves its maximum value of nine. See if if we other kinds trend are X transpose times u equals zero. That means X has to be a perpendicular to the vector. You then the maximum value maximal value. Off. Q. We will be the second largest Agon belly, which is three

Runpeng Li · Answer

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&#x27;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

Jose Avalos · Answer

1/3 times 6 a.m. I was 21 equals and minuses. I&#x27;m gonna put parentheses around the right site and I&#x27;m gonna motor Play both sides of the equation times three so that we can eliminate our fractions. Since this is just the one, I don&#x27;t need a number. So this becomes six m plus 21 equals. Distribute the three over here three times m. It&#x27;s 3 a.m. Three times naked of seven. It&#x27;s minus 21. So really do minus 21 on both sides. These cancel bring down your six m, bring down your three m, make it a 21 and negative 21 is negative. 22. We&#x27;re gonna subtract three years on both sides. Opposites will cancel. Bring down your negative 42 six M minus three m. It&#x27;s 3 a.m. both sides by three. Those cancel make it If they try to buy a positive is negative. 42 divided by three is 14. Our answer is M equals negative 14

Kerry Thornton-Genova · Answer

let&#x27;s begin by combining like terms on our left hand side, we have to him plus three in that equals five them. In addition, we have negative one plus five. This combines two plus four. We will keep our right hand signed the same for this move. Now let&#x27;s begin to move all of our M statements on to one side to do this. Let&#x27;s subtract five them from both sides. We now have four equals one m, which we will just right is M minus. Thank let&#x27;s had eight to both sides to begin to isolate them. When we do this, we have 12 on one side and m on the other because we have em alone equaling a number. We have this of the problem. Em is equal to 12.

Problem

$[\mathbf{M}]$ In Exercises $14-17,$ follow the i…

Problem 14 Medium Difficulty

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

Answer

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Linear Algebra and Its Applications

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Heather Z.

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Absolute Value - Example 1

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Linear Algebra and Its Applications

Grace H.

Heather Z.

Caleb E.

Michael J.

Absolute Value - Example 1

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Heather Z.

Caleb E.

Michael J.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications