in-exercises-5-and-6-write-a-system-of-equations-that-is-equivalent-to-the-given-vector-equation-x_1

Question

In Exercises 5 and $6,$ write a system of equations that is equivalent to the given vector equation.
$x_{1}\left[\begin{array}{r}{6} \ {-1} \ {5}\end{array}ight]+x_{2}\left[\begin{array}{r}{-3} \ {4} \ {0}\end{array}ight]=\left[\begin{array}{r}{1} \ {-7} \ {-5}\end{array}ight]$

Sean Thrasher · Accepted Answer

Okay, so we got this vector equation right here and were asked to write it in on equivalent system of equations which have done here. So this is the answer How I get to that while the left inside this is six x one minus x 15 X one, Right for this vector right here. I factored this X one inside, and then this is similarly this Okay. And then the left hand side you, When you add these two vectors together, you get this. Okay? And this is equal to this on the right hand side and for two vectors to be equal, they have to be equal component by component. So this right here has to equal this. I&#x27;ve done that there. This has to equal this. That&#x27;s that. This has to equal this, and that&#x27;s that. And we&#x27;re done.

Runpeng Li · Answer

Okay, so for problem six work even the vector vector addition, that is X one times Vector Negative two and three plus X two times eight and five, which is equal to sorry. We still have X three so plus x three times 161 negative six, which is equal to zero. So if we write down this as it&#x27;ll set of Lena you questions, then this will give us connected to x one plus eight x two. Thus, X three is zero and another question is three x one thus five x two minus 663 his zero. So that is how early.

Runpeng Li · Answer

in this problem. We have three equations, and we need to find out solutions and write it down in Para mention vector borne so that your questions are x one plus three x two minus five x three It was four and x one plus four x two minus eight x three You caused seven and the last one is negative three x one minus seven x two plus nine Next three, which is negative. Six. All right, well, first we were set down the trees that it&#x27;s 13 negative five and one for 98 and negative. Seven neck of three. Negative seven on DDE nine. Remember this? Ah, nothing consistent. Sorry, This is not homogeneous. We know equation. So we have to set down the council terms. That is for, um seven. Negative six. Now we need to perform the cultural eliminations first. Uh, we first used the second roll minus the first road to cancel one here, So that gives 01 elected three and three. And we are so we can also perform that we use our three plus three times are one we used to throw three plus three times to rule the personal So that is a neck and three plus three. That&#x27;s zero. And we have to because that it&#x27;s Ah, negative. Seven plus three times three, which is nice. So that&#x27;s as to and so on, so forth. We have next six on six and the first rule keep the same. That is one really negative five. And for just for your information, I all right? I will write down the road crew operations here. So that&#x27;s road too. Minus real one and row three plus three times ago. One. All right, all right. In the next a cz we observe here. The third row is twice on the second row, so the third row can be cancelled here. So that is 13 connected five and or and 01 negative. 33 and 000 Is you all right? At last, we can use the first roll minus three times the second rule. So that is 104 and negative five. And the second road. Still the same. 01 negative. Three on three and all zeros in the natural. All right, so let&#x27;s check out. Check out the equations here. So that&#x27;s that is X one minus Sorry, it should be a plus. That&#x27;s four x for X three. He caused negative five and x two minus three x three equals three. So that means X three is a free bearable here, so that&#x27;s X Juan Can be could be represented as a negative five minus four x three and x two can be written as three plus three x three. So our solutions our solution is connected. Five minus four x three and three class three x three and extremely and we separate. First, inform off ex threes of connective four x 33 x three and that&#x27;s three plus accounts in terms will be negative. Five. So active, negative five and three and a zero. So that&#x27;s we take out the X three. So that&#x27;s negative. Four to be one times X three clothes, clothes negative. Five, three and zero. So we&#x27;re done. Now remember, we can compare our solutions with our previous solution about about exercise Six. That is, uh our previous solution is next. Four, three and one times x three. This is this is our previous solutions. And compared to this, can you notice that we have a constant term? You are a new solution. So that&#x27;s the thing to take care of

Jack Chen · Answer

in this programme, we are giving a matrix a equals 20 minus three to minus seven and 55 Well, we want to recover assistant off equation from this fact from its vector form. So eight times X equals two b. Um so look at this metric. Say it&#x27;s a three by two matrix. That means our eggs will be x one x two. He only has to entrance you be the SMS. The second index. Yeah, so, by midges money modification we have a season of equation zero times x one minus three x two equals two to minus one. Twix one minus seven, x two because two, 65 fakes one us my fakes to be close to seven and that.

Michael Jacobsen · Answer

in this exercise, we have a linear combination that&#x27;s given on the left hand side of equation set equal to a vector. So all together, we call this a vector equation. This vector equation could be written as a system of equations. It can be placed into augmented matrix, but what we want instead is a matrix equation. So let&#x27;s see how we can build a matrix equation starting from the vector equation. Well, the first thing we&#x27;re going to do is go to each of the columns we see on the left hand side and former Matrix out of them, so it&#x27;s quick enough we just have to copy four negative 17 negative four for column one negative 53 negative 51 for column, too, and seven negative. 802 for the Third column. Next, we&#x27;re going to use the scale urz heer and form a Vector. This is a vector that contains all of our unknowns, which are x one x two x three, and finally were ready to say that this is equal to Don&#x27;t forget equality since we need an equation and it&#x27;s equal to the right hand side, which we copy in as six Negative 80 negative seven. Then this is our matrix equation. But if we were to label this as a a, this as X, the unknown vector and this is B, then it&#x27;s off the form eight times X equals B. So this is our matrix equation, and you&#x27;ll often see it presented in this way as well.

Brittany Scott · Answer

So we&#x27;re gonna write this into two equations. Mito two X minus three. Why? Plus five Z equals zero when they get that often. That first line, then from the second line. I know seven acts, then we have zero for that. Why term then? Minus Z equals five.

Problem

In Exercises 5 and $6,$ write a system of equatio…

Problem 5 Easy Difficulty

In Exercises 5 and $6,$ write a system of equations that is equivalent to the given vector equation.
$x_{1}\left[\begin{array}{r}{6} \\ {-1} \\ {5}\end{array}\right]+x_{2}\left[\begin{array}{r}{-3} \\ {4} \\ {0}\end{array}\right]=\left[\begin{array}{r}{1} \\ {-7} \\ {-5}\end{array}\right]$

Answer

$6 x _ { 1 } - 3 x _ { 2 } = 1$
$- x _ { 1 } + 4 x _ { 2 } = - 7$
$5 x _ { 1 } = - 5$

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

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$6 x _ { 1 } - 3 x _ { 2 } = 1$$- x _ { 1 } + 4 x _ { 2 } = - 7$$5 x _ { 1 } = - 5$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

$6 x _ { 1 } - 3 x _ { 2 } = 1$
$- x _ { 1 } + 4 x _ { 2 } = - 7$
$5 x _ { 1 } = - 5$

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

Linear Algebra and Its Applications