in-exercises-15-and-16-fill-in-the-missing-entries-of-the-matrix-assuming-that-the-equation-holds-fo

Question

In Exercises 15 and $16,$ fill in the missing entries of the matrix, assuming that the equation holds for all values of the variables. 
$$
\left[\begin{array}{lll}{?} & {?} & {?} \ {?} & {?} & {?} \ {?} & {?} & {?}\end{array}ight]\left[\begin{array}{l}{x_{1}} \ {x_{2}} \ {x_{3}}\end{array}ight]=\left[\begin{array}{c}{3 x_{1}-2 x_{3}} \ {4 x_{1}} \ {x_{1}-x_{2}+x_{3}}\end{array}ight]
$$

Barsha Rana · Accepted Answer

In the current problem, we have to find out the metrics for which, if we multiply them with x one x two and next three, we should get the final answer to be three x one minus to x three in four x one and then x one minus X two Glass Extreme. Now let us Before we answer this question, let us try to do the same thing with little easy numbers. Okay, so, for example, we&#x27;re multiplying the same next one next to an extreme. And here are some e b. C. Right. So what will be the answered x one plus B x two plus see extreme. So if you see whatever coefficients x x, one extra, an extra is having those air only written in a row format. Over here, A, B and C are there were writing them in a row format. So let&#x27;s go ahead and do the same. Afford this. What is the coefficient of X? It&#x27;s excellent. It is three. What is the coefficient of X to? Well, the X two term is not there. So then it must be zero. And what is the coefficient of X three? It is minus two correct same way in the next one. For X one, it is for for extra. It is zero for ex treaties against zero and the same way. Here it is. One minus one and plus one. So one minus one. What? So this is our answered simple.

Ashley Boni · Answer

So we need to fill in the coefficients in this matrix so that when it&#x27;s multiplied by ex well, next to we get this vector on the right hand side. So first we multiply vectors. This first column is what&#x27;s multiplying X one, and the second column is what&#x27;s multiplying X two and then we add, or we have those things together. So for the first equation, we have x one minus x two. So x one just getting multiplied by one and x two is being multiplied by negative. Secondly, uh, we have negative to x one plus one times x two and lastly, just to get X one, we a multiplex one by one and x two has to go away. So that means there has to be a zero here and that&#x27;s on me.

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

Thomas Emment · Answer

Okay, We&#x27;ll be giving these. Major sees a baby. I&#x27;m gonna find the matrix a X, given the three x Plus two A equals B so we can start off by my knitting to a from both sides. So we&#x27;ll have beat minus two lots of this. So you have first off. Minus five minus minus 60 minus five plus sectors could be a one. They&#x27;re gonna have a minus one minus minus 14. It&#x27;s two lots. Circular A minus one plus 14 to be 30. Hey, we&#x27;re gonna have just zip while basically minus two lots of these three minus four minus two lots of nice 18 and then we&#x27;ll have three miles to dose of five. So three minus 10 will be minus seven a minus four minus two. Lots of zeroes. Just goody minus four. We&#x27;ve got three lots of exit to find one. A lot of people just have to divide everything by three, so we&#x27;ll have 1/3 Would have 13/3 minus 4/3. 18. Number three to be put into six minus 4/3 on minus 7/3

Elizabeth Xu · Answer

Well, if we know the matrix for A and B, we want to solve for X if negative two X is equal to four B plus three A. So if you multiply both sides my negative 1/2 what will get its ecstasy? Cool too negative to be minus three charms, eh? So we know that negative to be a minus three have say we console for this native to B is equal to 10 two 00 negative six tonight minus three halves A which is equal to nine over too. 21/2 or negative nine over to negative 21/2. Then we have three negative 27 over too. 15 over too and zero. So if we subtract thes two is to make her sees What gang is 10 minus negative? Nine of two which is equal to 29 over too to minus number 21/2 which is 25 over too. Zero minus three Negative 30 minus negative. 27 2 27 over too. I mean, you get negative six minus 15 too, which is even too negative. 27 over too. And then we have eight here as I will find a solution for X

Elizabeth Xu · Answer

So here we have three x plus a musical to be so that means that three X is equal to be minus a. And if you look for B minus A, we&#x27;ll get Is that be minus? A is equal to negative five minus negative three Which negative, too Negative, Elin. Minus negative. Seven. We get six zero minus twos, negative to zero minus. Think of nice nine. And then we have three minus five. We just negative two in negative four minus hero, which is negative for so therefore, three X is equal to all of this and every divide both sides by three or multiply both sides But 1/3 we&#x27;ll get is 1/3 uh, being gonna say which is that matrix is equal to X. So if we solve for X X is equal to we multiply every number by wonder getting ago 2/3 night of 2/3 native 2/3 two three and negative for thirds

Problem

In Exercises 15 and $16,$ fill in the missing ent…

Problem 15 Easy Difficulty

Answer

$\left[\begin{array}{ccc}{3} & {0} & {-2} \\ {4} & {0} & {0} \\ {1} & {-1} & {1}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}}\end{array}\right]=\left[\begin{array}{c}{3 x_{1}-2 x_{3}} \\ {4 x_{1}} \\ {x_{1}-x_{2}+x_{3}}\end{array}\right]$

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

Linear Algebra and Its Applications

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$\left[\begin{array}{ccc}{3} & {0} & {-2} \\ {4} & {0} & {0} \\ {1} & {-1} & {1}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}}\end{array}\right]=\left[\begin{array}{c}{3 x_{1}-2 x_{3}} \\ {4 x_{1}} \\ {x_{1}-x_{2}+x_{3}}\end{array}\right]$

Linear Algebra and Its Applications

Heather Z.

Alayna H.

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

Heather Z.

Alayna H.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications