use-cramers-rule-to-compute-the-solutions-of-the-systems-in-exercises-16-beginaligned-x_13-x_2x_3-4-

Question

Use Cramer’s rule to compute the solutions of the systems in Exercises 1–6.
$$
\begin{aligned} x_{1}+3 x_{2}+x_{3} &=4 \-x_{1}+& 2 x_{3}=2 \ 3 x_{1}+x_{2} &=2 \end{aligned}
$$

Viswesh Uppalapati · Accepted Answer

in this video, I&#x27;m gonna be going over how to do problem number six. Uh, session They&#x27;re going three, which is based on determines in the Cramer&#x27;s rule. And I did everything beforehand because the determinants of three by three take are very time consuming and very long, So I&#x27;ll just go through how I did this problem. Um, I didn&#x27;t go over it once, but the video, then upward properly. So, uh, I&#x27;ll just go over roughly. Um, it&#x27;s pretty simple is just like solving a problem, too. And for except this time instead of two by two, it&#x27;s a three by three. So that&#x27;s more mathematically, a little bit difficult. So to start off were given this set of system of equations and that would make this problem this have a problem a lot easier if you were toe. Visualize it in a X equals B matrix form. Like this, uh, that I underline just now, um, so here the problem is asking us to solve, uh, system using the Cramer&#x27;s rule, which is basically says that if we were a substitute 4 to 2, which is, uh, here is be label be into each column of a and take it. The determinant. The rest of the matrix is a, um, and divided by the initial initial determinant when, without substituting be into it. Um, then we would get every single solution to the system equations. So here to start off, I first took the initial determinant of A without swapping anything. I&#x27;m to take the determinant of the three right to be, um, you&#x27;re basically for the for the first moment you do one times the determinant of this square two by two, right here and then minus three times the determinant of the outer two by two closer, one times the determinant of the 1st 2 columns. Right here. Well, sorry if it&#x27;s a little bit messy, cause I keep going over what I&#x27;ve drawn over previously. So here we got, um, negative to minus negative 18 minus one, which is which ends up being a negative too close. 18 minus one, which is 15 for the initial determinant. And then we, um and then for the first time of the creamers are always swap the first row with 4 to 2 and take it to determine which comes out to be six, and we repeat the same for the other two rows where when we swapped the second row with 4 to 2, Um, which is which is Ah Bee column. We got the determinant to be 12. And when we do so with the third column, we get the determined to be 18 um, and take out the solution to the system of equations. We just divide these determinant values by the initial determinant to get each value of the X column vector. That would be this. That was service a solution to the system equations he wrecks. One is six, divided by 15 extra. Strolled the water by 15 Extras 18 Divided by 15 um, and which gives us these Nissen, while is, respectively.

Chris Trentman · Answer

were given a system and were asked to use Cramer&#x27;s rule to compute. The solution of this system system is negative or sorry x one plus x two equals x three Negative three x one plus two x three equals zero and x two minus two x three equals two. This system is equivalent to the equation. A X equals B where a is the three by three matrix 110 negative 302 01 Negative, too, and B is the column Vector 302 So to use Cramer&#x27;s rule first, let&#x27;s find a one of B. This is the Matrix A with the column one replaced by B So we have 30 to here and then 10021 negative, too. A two of B. This is the Matrix A with column to replaced by B So one negative. 30 302 0 to 2. And finally a three of B. This is Matrix A with column three replaced by Vector B. So this is 11 negative 3001 and then 302 And now computing determinants, we have the determinant of a Can you sound in this case, I&#x27;m going to find it by expanding across the top row. So we have the determinant of 0 to 1 negative too. Minus the determinant of negative 3 to 0. Negative too. This is equal to well, we have negative. Too minus. And then we have six minus zero. So minus six, which is negative. Eight. The determinant of a one of B. We&#x27;ll find this by expanding across the second row. So really, we have Let&#x27;s see positive. Negative, positive, negative. So negative. Two times the determinant of 3121 This is negative. Two times three minus two, which is one so simply negative too. And tournament of a two of B. Find this. I&#x27;ll expand across the top row again. So we have determined of 0222 minus three times the determinant of negative 3 to 02 So we get negative four minus three times. This is negative. Six negative four plus 18 which is simply 14 and the determined of a three of B We confined bikes being across the second row. So we get the opposite of negative three, which is simply aerated health times. The termine int of the Matrix 1312 So we get three times two minus three, which is three times negative one or negative three. So now we have all that. We need to calculate the solution using Cramer&#x27;s rule. So it follows that Cramer&#x27;s Rule X one is the determinant uh, a one of B over the determinants of a call that we calculated the determinant of a one of b was negative. Two. So this is negative. Two over negative eight or 1/4 x two is the determinant of a to B over determinant of a call we calculated the determinant of a two of B was 14. This is 14 over negative eight, just the same as negative 7/4. Finally, x three is the determinant of a three of the over the determinant today, and we calculated that the determinant of a three b is negative. Three. So this is negative three over negative eight, which is simply 3/8. And so it follows that the answer by Cramer&#x27;s rule to this system is the vector. 1/4 negative 7/4 3/8

Chris Trentman · Answer

were given a system and were asked to use Cramer&#x27;s rule to compute the solution of the system system is three x one minus two x two equals three and negative. Four x one plus six x two equals negative. Five So this system is equivalent to the equation. A X equals B where a is the matrix three. Negative to negative 46 and be is Thekla Vector three Negative five. So we have that a one of B is replacing the first column of A with B, so we get three negative to negative 56 and we have a two of B Replace the second column of A With B. We get 33 negative four, negative five and now continue determinants. We have the determinant of A This is 18 minus gate, which is 10 determinant of a one of B is 18 minus 10 which is eight, and the determinant of a two of the, he says. Negative 15 plus 12 which is negative. Three. So, using Cramer&#x27;s rule, we have that X one. It&#x27;s the determinant of a one of B over the determinant of a, which is 8/10 or 4/5 and X two again, by Cramer&#x27;s rule, is the determinant of a two of B over the determinant of a, which is negative 3/10. So it follows that the answer X is the vector 4/5 negative, 3/10.

Viswesh Uppalapati · Answer

this video really solving problem number four of section 3.3, which is based on Kramer&#x27;s ruling. Determinants. Um, so we&#x27;re given this system equations and are asked to use the Cramer&#x27;s rule to solve for solutions. So making this are thinking about this problem. And, um, and an X equals B form will make it easier. So we have matrix a year. Negative 5 to 3 negative one and be here to be a nine and negative for cream. Israel basically says that, um, if you substitute be for every column of a and take those individual determinants and divide them by, uh, divide them by the determined of the original A, then we should get the solution to the system of equations. So, um, determinant of a a simple is just negative. Five times negative. One terms negative. Born minus two times three. Um, which is five minus six equals negative warm. And, um so if you take the determinant of of a one B, it will be re substitute. The first column with B and a So we&#x27;re good. Oh, sorry. Really good. Nine. Negative for to negative one for the matrix. Ciro di Cool. Um, nine times negative one minus two times negative for equals. Negative nine minus negative. Eight. Which is negative. One and negative one divided by negative one, which is the initial determinant. You give those one so x one equals negative. One divided by negative one. Here&#x27;s one. So x two would equal the tournament of A to B over determinant of a which is now we substitute the second. Call him with the nine and negative for on the first column Stays the same. Get, um, negative five times negative for minus, um nine times three over negative one was equal to 20 minus 27 divided by negative one because seven. So the solution is, um 17 because X one is one of x two is said.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be selling problem number three of section 3.3, which is based on Cramer&#x27;s rule and determinants. So the question gives us, uh, these two system equations for X one plus X two equals six and three x one close to x two equals seven and asked us a final solution using the Cramer&#x27;s rule, which is listed right here. Um, so the Cramers real basic states that if you, um if you substitute every column of A in this case, it&#x27;s is 4132 Um, what the B, which is 67 thing that&#x27;s thinking of this as an X equals B equation gives us this matrix. So if you substitute this, call them into each other columns and find the determine and divided by the determinant of the initial matrix, we should get the answer to the system&#x27;s system. Equations. So determined of a is, um, pretty simple for times two minus one times three, which is, um, eight minus eight, minus three because five, um, and then for a one big, we would get So we substitute the first column with six and seven. So we got 6712 So the determiner of that is so x one equals six times two minus one times seven divided by five, which is the initial determinant. So x one is 12 minus 27 is five over five equals one x two is is the same thing. Except now the second column is six and seven, while the first column remains four and three. So you get four times seven minus six times three over five. So really good. Clear. So x two is, um 28 minus 18 28 minus, which is 10 over five equals two. So the solution is one, too. That&#x27;s it.

Chris Trentman · Answer

were given a system of equations were asked to use Cramer&#x27;s rule to compute the solution of the system system is five x one plus seven x two equals three and two x one plus four x two equals warrant. While the system is equivalent to the equation, a X equals B where A is the matrix 57 to 4 and B is three column Vector 31 So let&#x27;s first compute a one of B So a one of B. This is the matrix 3714 and a two of B. This is the matrix 5321 We have the determinant of our Matrix A is 20 minus 14 which is six. We have the determinant of a one of B is 12 minus seven, which is five, and the determinant of a two of the is five minus six, which is negative one. And so it follows that X one, which, by Cramer&#x27;s rule, is the determinant of a one be over the determinants of a This is 5/6 and we have again. By Cramer&#x27;s rule, X two is the determinant of a two of the over a determinant of A, which is negative 1/6

Problem

In Exercises 7–10, determine the values of the pa…

Problem 6 Medium Difficulty

Use Cramer’s rule to compute the solutions of the systems in Exercises 1–6.
$$
\begin{aligned} x_{1}+3 x_{2}+x_{3} &=4 \\-x_{1}+& 2 x_{3}=2 \\ 3 x_{1}+x_{2} &=2 \end{aligned}
$$

Answer

$x_{1}=0.4$
$x_{2}=0.8$
$x_{3}=1.2$

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

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$x_{1}=0.4$$x_{2}=0.8$$x_{3}=1.2$

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Kayleah T.

Caleb E.

Kristen K.

Michael J.

$x_{1}=0.4$
$x_{2}=0.8$
$x_{3}=1.2$

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

Linear Algebra and Its Applications