should-be-solved-without-performing-row-operations-hint-write-a-mathbfxmathbf0-as-a-vector-equation-

Question

Should be solved without performing row operations. [Hint: Write $A \mathbf{x}=\mathbf{0}$ as a vector equation.
Given $A=\left[\begin{array}{rrr}{4} & {1} & {6} \ {-7} & {5} & {3} \ {9} & {-3} & {3}\end{array}ight],$ observe that the first column.
plus twice the second column equals the third column. Find a nontrivial solution of $A \mathbf{x}=\mathbf{0} .$

Viswesh Uppalapati · Accepted Answer

uh, this video Where you calling Paul? Number 32 of section 1.7 of the book, which is on linear Independence. So in this problem, we&#x27;re getting given a Matrix with Rose 416 negative. 7539 Negative three and three And we&#x27;re also given the ah, the first call them. Let&#x27;s call it a one plus a second goal in this Call it a two plus two times a second column equals the third column. 83 So here I wrote the equation, though, Um, one sec. Let me change my tip over Penn. Um, so a one plus 282 equals a three. And it&#x27;s asking us to find a nontrivial solution for the matrix equation. X equals zero. So X equals zero is just another way of writing a constant. So ah, a solution to this inspector would be a nontrivial solution Will be C one sea to N. C. Three, where it&#x27;s three scaler numbers because we have three columns in our original matrix. Therefore, we&#x27;re gonna be solving for three variables, and this solution to this equation should give us three variables. So if you multiply that by a one we should get C one times One are if you multiply X by the matrix. Avery She got C one times a one was C two times 82 waas. See, three times eight that we equals zero. So when we look at this, this is indicative of a bad equation. So if we get the first equation that is given to us into this form, we know that a one plus two a two was our, um minus 83 equals zero. And since these two questions are in the same form, one trivial solution to this equation without solving this setting this matrix to zero and I reducing it as it states in the prompted the problem, we can just assume that there are constants constants in front of, um, each of the each of a one a two and a three. So we know that one trivial solution would be X equals 12 and negative one. That is one possible trivial solution out of

Michael Jacobsen · Answer

In this example, we have a four by three matrix a Merkel. Here&#x27;s to consider the Matrix equation eight times X equals zero vector. We know that X equals the zero vector itself is the trivial solution and so automatically the system is consistent. Now the challenge is given that we have this homogeneous equation. Can we find a nontrivial solution? Well, in order to do that, what&#x27;s analyzed the columns of A If we take the 1st 2 columns, notice that we add them together we produce the third column automatically. This tells us that the columns of a are linearly dependent and there will be nontrivial solutions so we can find one in earnest now. So let&#x27;s consider again The Matrix equation. Eight times X equals the zero vector. Well, if we take a which is too negative. Five negative 31 31 negative 10 and five negative four. Negative for one and multiplied by a particular vector X, we will get the zero vector, which will have four entries. So let&#x27;s see how to find the particular entries here. Recall that when we take a matrix and multiply it by a vector were just forming a linear combination. So if I place one here, we&#x27;re taking one times the first column. One here means take one times this column and at them together. And if I take zero here, it now means take one times this column one times this column zero times this column and add together the results. Since we&#x27;re taking zero times the last column, we&#x27;re just adding together the 1st 2 But if we add together the 1st 2 we know we just produced this calm once again. So let&#x27;s do something a little bit different. Let&#x27;s multiply by a negative one. Then we would have won. Times of first column to negative five negative 31 minus one times a second column 31 Negative 10 And this here tells us we&#x27;ll take also plus zero times the last column. Five. Negative Ford Negative 41 Well, there&#x27;s no contribution here. And when we add the&#x27;s together, we do in fact, get the zero vector, So X equals one negative 10 is an example of a nontrivial solution. In fact, there are infinitely many examples. So if you didn&#x27;t find this vector X as your solution, that&#x27;s okay. Since C Times X is also a solution for all, see, in our

Jingyun Wang · Answer

the question asks us to find a solution off a X equal. Be first. We need to change human history into automatic of metrics for furrow to we can do three times. Row one minus wrote you for rosary We can do to times Roll one minus rose three. And for no for you can do two times. Roll one first role for copy Roll one to the elementary row operation for row to Row three in a row. For For a rosary, we can do negative 03 After dead, we can sort a rosary. Where is the road to for rosary? We can do to times Roll Chu minus a rosary for role, for We Can Do five times wrote to minus a role for every time a racy all zeros in wire. Oh, which means there&#x27;s a free viable In this case, X three and x four are the free marbles, so x four can be any number. Accessory can be any number acts two. You call negative one minus look X ray x one equal Tu minus acts for why this acts two. Put the X two. You&#x27;ll get three minus X four plus X ray. This is your final answer

Nez Nikoo · Answer

okay. We want to find the value of X. We know what the matrix of a is is a three by three and bees a three by one. And we want to use Cramer&#x27;s rule because that&#x27;s what the question is asking us to do. So in order to find X one on, what we need to do is we need to find the determinant of a. So I put here Determine of a determine of a is a three by three matrix and you can look at it and go through it yourself. It&#x27;s You should know how to do that by now. And that basically is to 71. I&#x27;ve done that on a piece of paper, so I&#x27;m just gonna write it here t to 71. Now the Cramer&#x27;s rule says the matrix A. When you want to find the first value you&#x27;ve plugging to be, which is right here into the first call. Okay, so be is three minus one four. Then write the A as iwas three, four, five in six, minus seven and nine. Now you have to find a determinant on the top, which is the same but again by three. By three. And if I&#x27;m just gonna show you how you start the determinant you don&#x27;t need I&#x27;m not going to write all the details here because you need to know. You practice yourself. I give you the answer is three minus one for here. You repeat the first column, you repeat the second column and you do your magic off diagonal multiplication and adding and subtraction the other way around. So when you do all of that, the top becomes 30. And we know where the bottom is to 71 truth. So now we have X one. That&#x27;s X one. We&#x27;re gonna find X two. We do the same thing. Cramer&#x27;s rule on X two We have to not substitute be for second column. So it becomes three minus 14 for second column. We write to 71 on the bottom because that&#x27;s determinant of a. And then you just repeat the outer columns from Matrix a 5 to 2 six minus seven and nine. When you find a determinant on the top is 59 over to 71. Okay, last item is X three. I&#x27;m gonna write the X three right here. Start X three, right here equal to now for X three, you need to substitute The third column would be okay. So I like to kind of right to be first when I whatever I substitute. So it&#x27;s minus three minus 14 So I kind of verify that first knowing that that&#x27;s the substitution be. Then I go put the rest of the a whatever is therefore to to 345 Now I like to substitute by column since we substituting my column, I write it that way also. So it&#x27;s less chance of error in the bottom 2 71 Okay, so the X three again You right here. 5 to 2, 345 Whatever. You feel the most comfortable mitt to find a determinant, you do it that way. I like it this way that I go on, we&#x27;ll fight this but this at these three. And then I multiplied these three. Subtract, subtract, multiply, subtract. When you do all of that, you&#x27;re going to get a 81 over to 71. So those are the three answers. This is our X one x two on x three. Okay. Thank you.

Jingyun Wang · Answer

to a question asked to find a solution off a X equal. Be first. We need to change a given system into all romantic and metrics form for row to we can do to times roll. One minus wrote you and for over three we can do three times. Roll one minus roastery Copyright one. But for rosary, we can do row to minus three times a role. Three Every time when we see zeros here, but a constant. The number here, which means is this is a better room. The matrix is inconsistent, which means there&#x27;s no solution.

Jack Chen · Answer

So we were given a Matrix A, which is a three by three metrics and, uh, the ah see stuff in question can be writing this Make the full eight times a vector X equals toward undirected peak. So in this case, a is a three by three matrix. That means our choice off X should be x one x two x three Mixture the lens off this vector He crossed the second decks uh, in the Matrix A and the in this case because 231 faction. So our results phone the metrics modification will be two excellent prospects to past three x three because 23 four x one minus x two plus two x three Close to one seven x one past six x two plus three x three close to five in the lead tape.

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Each statement in Exercises 33–38 is either true …

Problem 32 Hard Difficulty

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

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Linear Algebra and Its Applications

Catherine R.

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Absolute Value - Example 1

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

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications