let-aleftbeginarrayrr2-1-6-3endarrayright-and-mathbfbleftbeginarraylb_1-b_2endarrayright-show-that-t

Question

Let $A=\left[\begin{array}{rr}{2} & {-1} \ {-6} & {3}\end{array}ight]$ and $\mathbf{b}=\left[\begin{array}{l}{b_{1}} \ {b_{2}}\end{array}ight] .$ Show that the equation $A \mathbf{x}=\mathbf{b}$ does not have a solution for all possible $\mathbf{b},$ and describe the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution.

Michael Jacobsen · Accepted Answer

in this video, we have a two by two matrix a a Vector B, and we put them together so that we can consider the Matrix equation A X equals B. Let&#x27;s see what the solutions might be for such your equation. We can put the entire system in the form of two native six negative 13 and augment with B one and B two. If we do, this will be able to reduce to echelon form to determine whether or not this system is consistent. So, first, to get this in echelon form, let&#x27;s eliminate the negative six indicated here I&#x27;ll copy Row one, which is to negative one B one, and our operation will be to multiply row one by three at the result to Row three and record the result here. So we&#x27;ll obtain altogether 00 and be too plus three B one. So that&#x27;s our system. So far, let&#x27;s analyze where the pivots might be. Well, there&#x27;s definitely a pivot here, but I say might be the A pivot here because that just depends on the value of B two and three B one. So now we can say if B two plus three B one is not equal to zero, then the right most column has a pivot when we know that when there&#x27;s a pivot in the right, most column, then a X equals B R system in matrix form is inconsistent, meaning there will not be solutions. Now, if be to plus three B one is equal to zero, then the system will be consistent. And that&#x27;s because if we have this quantity equal zero, that means there is a zero here, and this highlighted position is not a pivot position, forcing the system to be consistent. Notice that this equation could also be ascribed at described as be too equals negative three B one. So this is a line in the variables B two and B sub one, and this line would go through the origin. So as long as you choose values for B two and B one that&#x27;s on that line described here will have a consistent system for this matrix equation.

Ankit Gupta · Answer

you&#x27;ll find the solution off these equations by substitution method. We will first name this equation as one. And this question is to So we will solve the equation too. For about four b, therefore be equals three a minus six. So named this as our party question. No. What value of B which we got from part equation in equation one so weightless. The expense to you minus six equals four. This implies the way bless 18 a minus start basics equals food which implies 20 equals 40. Therefore, value of it is to now for the value of a in equation three bigger B equals three into two minus six. It is six minus six. So zero, therefore there got our solution that is aloof A and B equals do calm a zero.

Heather Zimmers · Answer

to solve the system using the elimination method. What we want to do is change one or both equations so that we have opposite coefficients on one of the variables. So what we could do is we could take the second equation and multiply it by two. And the second equation is now two a minus six p equals four. And if we leave the first equation alone, we have for a plus six p equals negative one. So we see opposite coefficients on be positive six and negative six. So when we have these equations together, we have six A equals three divide by six and we get a equals 1/2 And then to find the value be, what we can do is substitute 1/2 in for a into one of our equations. So if I go back to the first equation for a plus six p equals negative one, we have four times 1/2 plus six times b equals negative one and four times 1/2 is to And then we&#x27;re going to subtract two from both sides, and that gives us negative three and then we divide both sides by six, and we get negative 1/2. So our solution in the order a comma be would be 1/2 common negative 1/2.

Jingyun Wang · Answer

the question asked to determine the constant value off a N B. When a system has a no solution, be infinite number off solution and the sea. Unique solution. First, we need to change a system off equation into a semantic metrics form. So take the coefficient out. Remember, we need to find a reduced her or I shall inform Too funny. Final answer. So for road you we can do to time. 01 minus wrote you and for rosary we can do three times. Roll one plus Row three Copy Row one to the Adam Antero operation for Row two and the Rosary part A. Ask if the system has no solution. If we see there&#x27;s a back row, which means the Matrix is inconsistent. In other words, there&#x27;s no solution and we don&#x27;t expect to see all zeros in one row. This make the system has a flu variable, so let&#x27;s look back to row two. Suffered this case negative. Two a minus one should not equal to zero. Okay, should not equal to a negative 10 virtue. And look back to rose three for this case, negative to a plus B equal zero when a equal to negative 1/2 b roll equal negative one. So there are two possible answer for part A to satisfy the condition. What A does not equal to you in negative 1/2 or a equal to negative well over two and a be equal to negative one. The system will have no solution. Par be asked if this system has infinite number off solution. So we expect to see all zeros in Vairo. However, look back to rosary. We can never satisfy the condition Since the last column here is there is 10. So so there is no value off A and B can make. The system has infinite number off solution. Parsi asked if the system has a unique solution since this is a three by three matrix and we only have two bearable so we can have all zeros in one rope. So look back to a row two and the rosary and for this case, negative two a minus one equal zero and negative A. To a plus. B should not equal to zero so a equal to negative 1/2 Spertus here, so B should not equal to negative one. So what

Ankit Gupta · Answer

give any questions on a minus to be equal 16 on B plus three equals three. So named execution is one, and this is to reveal saw this system of equation by substitution method. Therefore, solving question number two or B bigger B equals three year minus three. Now, for this rally or be in a question number one, that is a minus tour two year minus three equal 16. Now we get a minus 60 last six equals 16 therefore minus five A equals then. So we get the value of a AZ minus two. Now, for this value of A in this equation to find value of B, so be is real. Minus two into minus minus three equals minus six minus. Steve, it is minus nine. So we have got the value of A and B, which is our solution as minus took on my minus name

Ankit Gupta · Answer

we have been given to questions on Does a B plus to a question. And 2nd 1 is is it closed toe minus T minus? Speak. So what we can do? You can go for their values. Offend me. When is it based? Toe And B zero is what for hero call. So really get a B. When is it the value of people Come smile ist here and love be zero. It becomes minus looking. Now they&#x27;ve been Just applaud these points on the piece of crap paper. So what will get the graph? Something like this. This and this are corresponding toe. This is your A And this is corresponding toe well directing the order Perez comma be so the 16 point becomes five comma minus eight from the ground. No, we need to check for the two values. No, go and see this Particular Redman is corresponding to the line. That is a big place toe a request toe and this agreement is corresponding toe a line that is a close to minus three minus B. Now for the first line that is B plus toe to toe put 20 b minus. It&#x27;s plus two into five you get with physical qualities and for the second that is a call to minus three minus B. It is a fight and minus T minus my estate with chemicals to fight. This is also kind. Therefore, the point that is five comma minus it is the correct solution.

Problem

Repeat Exercise $15 : A=\left[\begin{array}{rrr}{…

Problem 15 Medium Difficulty

Answer

$-3 b_{1}$ only for these set of values the system is consistent

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Anna Marie V.

Heather Z.

Maria G.

Kristen K.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$-3 b_{1}$ only for these set of values the system is consistent

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Heather Z.

Maria G.

Kristen K.

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

Linear Algebra and Its Applications