unless-otherwise-specified-assume-that-all-matrices-in-these-exercises-are-n-times-n-determine-whi-7

Question

Unless otherwise specified, assume that all matrices in these exercises are $n 	imes n .$ Determine which of the matrices in Exercises $1-10$ are invertible. Use as few calculations as possible. Justify your answers.
$$
\left[\begin{array}{rrrr}{-1} & {-3} & {0} & {1} \ {3} & {5} & {8} & {-3} \ {-2} & {-6} & {3} & {2} \ {0} & {-1} & {2} & {1}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

in this example, we have a four by four matrix a provided, and our job here is to determine if this is in verbal. Well, in order to get to that, we&#x27;re going to use row reduction here, and we&#x27;re going to reduce until we are in echelon form. So to that end, we&#x27;ll start with this pivot position here and eliminate the entries in Rows two and Row three that air shaded. So let&#x27;s start off by copying in row one. It&#x27;s negative. One negative three is your on one, and the operations will be to first multiply row one by three at the result to row two will have zero, then negative four, eight and zero for the next operation. Let&#x27;s multiply row one by negative too. At that result to row three will have zero. It looks like zero here as well. A three and another zero. Now we already have a zero here, so we can just copy in this row. Now it&#x27;s analyzed the pivots. So far, we have a negative four here, and if we like, we could eliminate this entry by doing the following rule operation divide Row two by four At the results to Row four. First, let me copy The Matrix rose 12 and three. So there&#x27;s our row one. Row two is zero negative. 48 and zero. We&#x27;re copping and row three since no changes necessary there either. And now we&#x27;re ready to begin. So we&#x27;re dividing by negative for here. Adding the result to the fourth row will have 00 divided by a negative four will have a negative two plus two. So another zero and then copying in a one. Well, that was pretty quick, since this was a nice matrix toe work with were already it already initial on form and along the maid diagonal. We see that every column is now a pivot column. This implies immediately that a must be raw equivalent to the four by four identity matrix, but for using the vertebral matrix there, um, notice that we don&#x27;t even need to use that step. If a is of size four by four. And we found four pivots, then the vertical matrix theory. Um, on its own already says that a is convertible, so this example is complete

Michael Jacobsen · Answer

for this exercise were considering a four by four matrix say that&#x27;s displayed here, and our goal is to determine whether or not this matrix will be in vertebral by inspection. There doesn&#x27;t seem to be anything interesting going on. I can&#x27;t tell if the columns are linearly dependent, for example. And if we were to roll, reduce this matrix by hand, it would be a long and difficult journey, since these are not very friendly numbers. So for this particular exercise, it&#x27;s encourage that we either use a computer algebra system or a calculator. I took the approach of using occurred a computer algebra system, and I found that this matrix ISRO equivalent to the following matrix here this is the four by four identity matrix, and from the statement that the Matrix A is roll equivalent to an identity matrix, it follows immediately that a is in vertebral. If, for example, though we did not have an entry one here, suppose it came out as a zero. We would have say that a is not in vertebral in that case, but for our situation. We do have a pivot in every row, so we have a number convertible matrix

Runpeng Li · Answer

So in this problem, given Matrixes. Hey, so three by three matrix. So it&#x27;s 1 95 collective four. Second row is 03 or the third roll Connective three steaks zero. So again, we need to find a way to expand the three by three matrix. So the way to expand it is to find find a find A Row column. Her column that contains entry up zero because zero makes you makes you easier. Makes it easier to carefully. So how? Just choose the second row here. So right now we have connected one. So this is the second row and the first column. So that gives the power to be and collector is zero times determined and talked The First Matrix. I&#x27;ll just write down a one so it doesn&#x27;t matter because we have zero here and the 2nd 1 will be Nicky one to the power for times three. With we are crossing out the second row second roll in a second column. The thing we left is one naked for and next three zero. Their term will have to be next one to the power five and times or times determined are crossing. Now the second row, and they call him. So that will be one next five. And negative three and six. Okay. So to calculate this, we have first term you zero second term, three times the determined off this. So this will be one times zero is zero and minus collective. Three times. Next four, which is positive. 12. We have minders, so that is negative. 12. Okay, the third term, we have a negative one, because the negative one to the power of five, which is odd. And we have four. I don&#x27;t need to time. It&#x27;s by four. And the determinant of this matrix one times six. That&#x27;s six minus three times five, which is positive. Uh, 15. So it&#x27;s six minus 15. That&#x27;s six minus 15. All right, so three times Negative. 12. That&#x27;s Ah, negative. 36 minus four times negative. Nice. Okay. And negative. 36 minus. So I should be positive. 36. And this is Cyril. All right, so that means the determinant off, eh is zero. So a is singular, So that means a snotty convertible. So we&#x27;re done

Runpeng Li · Answer

So the Matrix, where the inventing this problem is supplied by five tricks. I&#x27;ll press running down here, That is Bye. Three one 79 Second row is 642 eight. Activate. Um, the third rule 75 three and collective nine. Directed by or through its nine, six four and negative nine and negative five. Oh, sorry. So the third row I mess of the us to under hundreds here, so third row should be Excuse me. So the third row should be 753 tonight. The last rule use by 8585 to 11 4 eight I, too 11 for All right, So So we first observed that this is a five by five matrix, and it&#x27;s very hard to calculate that determined directly, because we have We don&#x27;t have any row or columns that that contain zero or something like that. So it&#x27;s kind of hard to calculate directly by hand. So one thing that would recommend to do is to, um, using ghost using gushing elimination too, uh, together, reduce station form off the matrix. And once we&#x27;ve done that, uh, you can do this by either by by hand or I would recommend is just to do this spot on the wall from Alfa or or men that, um you can find out that, uh, reduce station form is an identity matrix 55 identity matrix. So that means So that tells us that there are five people positions in these metrics. That means every every single row has, ah, people that position. So I over p room given in our section, we know that. So we know that this may shake season veritable if and only if, um, it has every single every single row has the people of positions. So in this case, way have 55 matrix and Heathrow. You&#x27;re a true has a people positions. So that means his matrix A It&#x27;s in vertebral. So that&#x27;s how our final conclusion, it&#x27;s incredible, and we&#x27;re done

Runpeng Li · Answer

we have four by poor matrix, which is say, Hey, is first row is 137 or second row is 0596 The third rule still was able to eight. The last room there was there was zero and Tech. Okay, So to find out a determinant again, we need to expend it along some row or some column. So in this case, I would recommend to do the first column because it has, um, three entries of zeros and this collector is one. So he&#x27;s very easy to calculate, so I&#x27;ll just do that. Expend a We pressed into the by now determined and Ave and expanded along the first column. So with the purse entry is one times since today this. Ah, first column and first row. So we have connected to power up to. But that&#x27;s the next squared, naked squared and crossing. Now the first column Members row. So we won. We left is determined off three by three matrix 596 Bye 96 0 to 8 and 00 10. All right, so negative. One squared. It&#x27;s one. So it&#x27;s just a determinant of it&#x27;s three by three matrix and 00 10. All right, so for it is three by three missions. We again, we need to expend it. Um, so right now, I would expect I would recommend to expand along the first column again because it has three as two zeros. You can also expend alone the third. The third row on it also has two zeroes here. So that&#x27;s the same where I would just do this by expending along the first column, so that would be determined. Sorry. Um, there&#x27;s this five and two connective one. Next, we once what next? One squared. So that&#x27;s one. I&#x27;ll just Ah, and we&#x27;re not write down one here, so he&#x27;s just a determined, uh, two by two matrix, which to AIDS. 00 10 280 10 minus. Because we have now the second term off the first column. So that is, uh, two plus one. So that&#x27;s, uh, three. So that&#x27;s 9 to 1 neck to onto the power of three. So he&#x27;s We have ah, minus here, and we&#x27;re crossing now. The second ruin. First column. So that is 96 Do you attend term end? Uh, 960 by the way, since we have zero as a defector. So we don&#x27;t You don&#x27;t need to care about that determined here. So should be zero. And the last term should also be zero because we have a zero point last year burning us entry on the first call him. So the only thing we need to care about is the first term, as five times determined, determined updates. Matrix two times 10 is 20 and eight times zero is zero. So that&#x27;s 20. So that would be 100 which is a non zero. So a is non singular, which means a convertible singular, singular. All right, and we&#x27;re not.

Michael Jacobsen · Answer

in this example, we have a three by three matrix a provided, and our goal here is to determine if it has an inverse for this particular case. I&#x27;m going to take the approach of role operations until it&#x27;s reduced into echelon form. So to that end, let me make row to our first row, which is 102 row three. Let&#x27;s make that the second row negative four negative 97 and the old first row could be our new third row 03 and negative five. Now, since we&#x27;re only going for echelon form from this pivot here will eliminate this entry below, so this will be raw equivalent to first copy 102 then multiply row one by four. At the result to Row two will obtain zero negative nine and let&#x27;s see eight plus seven, which is a 15 then the next row is 03 and negative five. Notice, however, that the last two rows air now suspicious. If we take negative three times the third row, then we would get to negative nine and positive 15. This means these rows are linearly dependent, and so if we make a new row operation it would be 102 copy zero negative, 9 15 and say Divide that row, Bite negative three or excuse me, divided by three and added to Row three that produce Onley zeros. So that tells us we have a pivot here and here. But the third column is not a pivot column due to this zero. So this tells us the A is not. Let&#x27;s start with a verb. A is not row equivalent to the identity matrix I. And this implies that a inverse does not exist. If we did get a pivot here, then we would have concluded the opposite. That a inverse does exist because then it would be row equivalent to the identity matrix.

Problem

Unless otherwise specified, assume that all matri…

Problem 7 Easy Difficulty

Answer

invertible

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Heather Z.

Alayna H.

Absolute Value - Example 1

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invertible

Linear Algebra and Its Applications

Grace H.

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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

Grace H.

Anna Marie V.

Heather Z.

Alayna H.

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

Linear Algebra and Its Applications