find-an-lu-factorization-of-the-matrices-in-exercises-7-16-with-l-unit-lower-triangular-note-that--5

Question

Find an LU factorization of the matrices in Exercises $7-16$ (with $L$ unit lower triangular). Note that MATLAB will usually produce a permuted LU factorization because it uses partial pivoting for numerical accuracy.
$$
\left[\begin{array}{rrr}{3} & {-6} & {3} \ {6} & {-7} & {2} \ {-1} & {7} & {0}\end{array}ight]
$$

Chris Trentman · Accepted Answer

were given a matrix and were asked to find an L U factory ization for this matrix with L Unit Lower Triangular Matrix is a three by three matrix with entries three negative 63 six Negative seven to negative 170 First, let&#x27;s roll birdies a using Onley row replacement operations. So first I&#x27;m going to subtract to over one from Road to. So I obtained three negative 63 zero five negative four and then I&#x27;m going to add one third of Rohan to Row three. So I get zero five one and next I&#x27;m going Thio. Subtract two. Wrote to from Road three to get three negative six three 05 Negative four zero zero five and this is going to be the row echelon matrix you and to obtain l first Pivot column was 36 negative one. Dividing by the pivot, which is three. We obtained that the first column novel is one to negative one half or, I mean negative. One third. Second pivot column is 55 dividing by the pivot, which is five. We get 011 and third Pivot column is five, divided by the pivot, which is five we obtained 00 one. So this is our matrix l And we have that a could be factored as l times you.

Chris Trentman · Answer

were given a matrix and rest to find an l u factory ization of this matrix with L being a unit. Lower triangular matrix matrix forgiven. Call it a is three negative 12 negative three Negative to 10. Nine Negative 56 We see that using on Leigh Road replacement operations, we can add Rwanda wrote to. So we get three negative. 12 zero negative three 12 and then subtract three of row one from row three. So we get zero negative, too. And then six minus 60 We see that we can, and then we can and to reduce. And 2 to 0. We can add or subtract two thirds of road to to Row three. And so we obtain the matrix. Zero negative. 12 zero Negative. Three 12 zero zero Negative eight. And this is going to be our role echelon matrix you and we&#x27;re going to take our first column in a This is three negative three nine. We&#x27;re going to divide it by the pivot, which is three. And we&#x27;ll put this as her first column l So we get one negative 13 for the second call. Manele. Consider these second column in the second Matrix in the sequence. So we have negative one negative three negative two. We&#x27;re really only looking at the pivot called Negative three. Negative, too. We&#x27;re going to divide by the pivot, which is negative. Three. We&#x27;ll put this as the second column and l So you have zero one two thirds. And finally, we&#x27;re looking at the pivot column in you, which is the column to 12 negative early looking at the entry entry. Eight. Negative eight. We&#x27;re going to divide by the pivot, which, of course, is just negative eight itself. And we&#x27;ll put this in as the third column. She&#x27;s going to be 001 So L is The Matrix 100 negative 11032 3rd one

Chris Trentman · Answer

given Matrix and were asked to find an l U factory ization of this matrix with l unit Lower triangular were given the Matrix A, which has entries 14 negative. 15 37 negative to nine Negative to negative. 31 negative four negative 16 Negative 17 So let&#x27;s put this matrix in Roche one form first. I&#x27;m going to subtract three of real one from road to So I get one for negative. One five zero seven minus 12 is negative. Five negative too. Plus three is one and nine minus 15 is negative. Six. I&#x27;m going to add two of row one to row three. So I get zero eight This negative three is five negative. Two plus one is negative. One and 10 plus negative four is positive six. And I&#x27;m going to add one of Rio one to real four. So I get zero 10 Negative too 12 and next Going dead. Add wrote to to row three and I will had two of route to to rule four. And so I get 14 one. When for negative 15 zero Negative 51 negative six zero zero 00 in 0000 There&#x27;s no more pivot points, so this is going to be equal to Are Groeschel on Matrix You and you know we have that. Our first pivot column is 13 negative to negative one. Writing by the pivot, which is one you obtained the first column Matrix L, which is 13 negative to negative. One second pivot column is negative. 55 10 We divide this by the pivot. Negative five will make this our second column in l. So we have 01 negative one negative, too. And because there are no pivot columns left was simply, add two columns from the identity matrix to make this a unit lower triangular matrix. So 0010 and 0001 And this is the Matrix l and we have that they could be factored as l times you

Chris Trentman · Answer

were given a matrix and were asked to find an l U factory ization of this matrix with l&#x27;d unit. Lower triangular metrics were given is 14 negative. 15 37 negative to nine. Negative to negative. 31 negative. Four Negative. 16 Negative. 17 Call this matrix a mhm. Right. Be a mistake here. This is actually matrix from exercise. 14 matrix from exercise 15 is a little bit different. Sorry about that. So instead, the Matrix from Exercise 15 is to negative four or negative too. Six Negative 97 Negative. Three Negative one negative. Four 80 All right, now let&#x27;s put this into row echelon form. So first I&#x27;m going to subtract three of real one from road to I have to negative four for negative, too, when we have six minus 60 Negative. Nine plus 12 is three seven minus 12. This negative five. The negative three plus six is I agree, and I&#x27;m going to add one half of real one to Row three. So I get negative. One plus one half of two is zero negative. Ford, minus two, is negative. Six and eight plus two is six 10 in zero plus negative one is negative one. And next I&#x27;m going to add two of road to to row three. And so I obtained to negative four or negative, too. 03 Negative. 53 in 00 zero five. This is in row echelon form. So this is our matrix. You now to find metrics l Our first pivot column was 26 negative one by this by the pivot, which is to this will be first column in the Matrix l We have 13 negative one half for the second pivot column. We have three negative six. Divide this by the pivot, which is three. So we get our next column. Is 01 negative, too. And our final pivot column is just going to be zero. We actually don&#x27;t have any more pivot columns after this. So we&#x27;re simply going to add columns from the identity matrix to make this a unit lower triangular matrix. So we&#x27;ll add column 001 And this is our matrix l. And we have that they can be factored as L Times

Chris Trentman · Answer

were given a matrix. We were asked to find an l U factory ization for this matrix with L Unit Lower Triangular given the Matrix A with entries to negative 66 Negative. 45 Negative. Seven 35 Negative one negative 64 Negative eight eight Negative 39 First, let&#x27;s get this matrix into Roche one form so I will add two of row one to Row two. Obtained to negative 66 zero five minus 12 is negative. Seven. Negative seven plus 12 is five. Next, I will add. We&#x27;re all subtract three halves of real one from row three, so I get three minus three halves of 20 five plus nine is 14. The negative one minus nine. It&#x27;s negative. 10. Next I&#x27;m going to had three of row one to Row four, so I get negative. Six plus 60 four minus 18 is negative. 14 men negative. Eight plus 18 is 10 and next I&#x27;m going to subtract four of row one from No. Five, so I get eight minus 80 Negative. Three plus 24 is 21 and nine minus 24 is negative 15 and I will add two of road to to row three. Subtract two of road to from real four Handle. Add three of road to to row five to obtain the matrix to native 66 zero negative seven five 000000000 We see that this matrix is in row echelon form, so this is going to be our matrix to you. So to obtain the metrics, l noticed their first pivot column was to negative 43 negative 68 We&#x27;ll divide this by the pivot, which is negative, which is to this will be our first column. So this is one negative, too. Three halves negative. 34 The next pivot column is negative. 7 14 Negative 14 21. Divide this by the pivot, which is negative. Seven. This will be your second columns. We have zero one negative to to negative three, and we see that there are no more pivot columns at this point. So will make this a unit lower triangular matrix by adding columns from the identity matrix so we&#x27;ll have the column. 001 zero zero 000 10 and 00 00 one and we have that are matrix. A can be written as l times you

Chris Trentman · Answer

were given a matrix A and were asked to find an LG factory ization of the matrices with l being a unit. Lower triangular matrix A is a three by three matrix with entries negative. 534 10 Negative eight Negative nine 15 12 So to find the l u factory ization, let&#x27;s use Onley Row replacement operations to get a into Groeschel on form first going to add two of Rwanda to wrote to so get native 534 zero Negative, too negative one. And then we&#x27;re going to add three of row one to row three zero 10 14. Next going Thio. Add five of row two to Row three. So we get negative. 534 zero negative to negative one 00 nine And this is the row echelon matrix you And using this sequence of matrices, we have that our first pivot column in a This is the column negative. 5 10 15 will divide entries of this column by the pivot, which is negative. Five. And we&#x27;ll use this as the first column in Our Matrix El so he gets one negative to negative three second pivot column is negative. 2 10 more to divide this by the pivot, which is negative two. And so we obtain. Call him one negative five. So 01 negative five for l And finally looking at the therapeutic column, which is simply nine dividing it by the pivot, which is nine. We obtain 001 So the matrix l is 100 negative. 210 Negative. Three negative 51 And we have that is equal to L times. You

Problem

Find an LU factorization of the matrices in Exerc…

Problem 11 Easy Difficulty

Answer

Hence $A=L U=\left[\begin{array}{rrr}{1} & {0} & {0} \\ {2} & {1} & {0} \\ {-1 / 3} & {1} & {1}\end{array}\right]\left[\begin{array}{ccc}{3} & {-6} & {3} \\ {0} & {5} & {-4} \\ {0} & {0} & {5}\end{array}\right]$

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

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Kayleah T.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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Hence $A=L U=\left[\begin{array}{rrr}{1} & {0} & {0} \\ {2} & {1} & {0} \\ {-1 / 3} & {1} & {1}\end{array}\right]\left[\begin{array}{ccc}{3} & {-6} & {3} \\ {0} & {5} & {-4} \\ {0} & {0} & {5}\end{array}\right]$

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Kayleah T.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Heather Z.

Kayleah T.

Maria G.

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

Linear Algebra and Its Applications