unless-otherwise-specified-assume-that-all-matrices-in-these-exercises-are-n-times-n-determine-whi-3

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.
$$
\left[\begin{array}{rrr}{5} & {0} & {0} \ {-3} & {-7} & {0} \ {8} & {5} & {-1}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

in this exercise were considering a three by three matrix A that&#x27;s provided here, and our job is to determine if they could be in vertebral. One way we could look at this is consider the transpose of a This is a worthwhile approach, since if a is in vertebral, so is a transposed and vice versa. So the transpose of our particular matrix will take Row one and write it as a column 500 Then for Row two, we have column to negative three negative seven and zero finally are less. Column is 85 and negative. One. Notice that the transpose of the Matrix A is now in echelon form, whereas a itself was not. And we see the A transpose has ah pivot in every row. What this means is that a transposed inverse exists. But as we said before, if the inverse of a transpose exists, then that implies immediately that a inverse exists as well. So in this case, were able to determine that a inverse exists without using any row operations whatsoever.

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.

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

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 two by two make tricks, eh? That&#x27;s provided, and our goal is to determine if it&#x27;s in vertical. One thing we can do to check this is calculate the determined of a The determined of a first involves the product of the main diagonal, which is going to be five times negative. Six. So we have fight by negative six. First, we add to that the product of the off diagonal, deft eye off diagonal is negative three and seven, so add negative three times seven. Then we find that the determinant is now negative 30 minus 21 or negative 51 altogether. The critical thing about this determined is that it&#x27;s not equal to zero, and this shows that a inverse exists.

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

Problem

Unless otherwise specified, assume that all matri…

Problem 3 Easy Difficulty

Answer

There are 3 pivots in $3 \times 3$ matrix so the matrix is invertible

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

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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There are 3 pivots in $3 \times 3$ matrix so the matrix is invertible

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.

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

Linear Algebra and Its Applications