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In Exercises $1-6,$ solve the equation $A \mathbf{x}=\mathbf{b}$ by using the LU factorization given for $A .$ In Exercises 1 and $2,$ also solve $A \mathbf{x}=\mathbf{b}$ by ordinary row reduction.$$A=\left[\begin{array}{rrr}{3} & {-7} & {-2} \\ {-3} & {5} & {1} \\ {6} & {-4} & {0}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{-7} \\ {5} \\ {2}\end{array}\right]$$$$A=\left[\begin{array}{rrr}{1} & {0} & {0} \\ {-1} & {1} & {0} \\ {2} & {-5} & {1}\end{array}\right]\left[\begin{array}{rrr}{3} & {-7} & {-2} \\ {0} & {-2} & {-1} \\ {0} & {0} & {-1}\end{array}\right]$$

$y=\left[\begin{array}{c}{-7} \\ {-2} \\ {6}\end{array}\right]$$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$

Algebra

Chapter 2

Matrix Algebra

Section 5

Matrix Factorizations

Introduction to Matrices

Missouri State University

Campbell University

Lectures

01:32

In mathematics, the absolu…

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In Exercises $1-6,$ solve …

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Solve the matrix equation …

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Use the LU factorization o…

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In Exercises $17-26,$let

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we were asked to solve the equation A X equals be using the l U factories ation give it from the Matrix A and to solve the equation X plus B for X equals B by ordinary rail reduction as well. So the Matrix forgiven a is the Matrix three negative seven. Negative to negative. 3516 Negative for zero and the column. Vector B is negative. 752 and we see that the l U factories ation given for a is a three by three matrix 100 negative 110 to negative 51 times the three by three matrix three negative seven negative 20 negative to negative. 100 Negative one. So, looking at the factory ization, we see that the unit lower triangular matrix l He's going to be 100 negative 110 and to negative 51 And we have that The row echelon matrix you is three negative seven negative too. Zero negative to negative one and 00 negative one. And once again we have the column. Vector B is the three by one matrix. Negative. 752 So first we're going to solve the equation. El y equals B for why so To do this, we use rail reduction. So we have that l. B. This is three augmented day tricks, which is 100 negative. Seven negative 1105 and to negative 512 And we see that after performing arithmetic in column four. This can be row reduced to the matrix. 100 Negative seven 010 Negative too. And zero negative. 51 16 and we see that it's a matrix can be reduced with arithmetic. Onley in column four to the Matrix 100 Negative seven 010 Negative too. And zero zero 16 And we see that this is the Matrix I and then a column vector. And so it follows that. Why is this column vector which is negative seven negative to six and now will solve the equation You exit was why using back substitution. So we have the augmented matrix you why this is going to be the matrix three negative seven negative two negative seven zero negative to negative one negative too. And 00 negative 16 This could be row reduced to the Matrix. Three. Negative seven Negative two Negative. Seven zero Negative to negative. One negative, too. 001 negative six. This could be row reduced 23 Negative 70 Negative. 19 zero Negative 20 Negative eight and zero zero one Negative six. This matrix can be reproduced to three negative 70 Negative 19 zero 104 and 001 Negative six and this could be reproduced to the Matrix. Three zero zero nine 0104 001 negative six and finally, this could be reduced to the Matrix. 1003 0104 001 negative. Six. You notice that this is the augmented matrix I followed by a column vector, which is X and therefore it follows that the column Vector X is three for negative six. This is one way to find solution to the equation. X equals B in other ways to simply ro reduce a. So consider the augmented matrix A B. This is the Matrix three negative seven Negative two negative seven Negative 3515 six Negative 402 And this could be road reduced to the Matrix three Negative seven Negative. Two negative. Seven zero Negative to negative. One. Negative too. Zero 10. 4. 16. This could be reproduced. 23 negative. Seven Negative. Two negative. Seven zero Negative. Too negative. One negative too. 00 negative. One six. And so notice that this is the same as the matrix. You Why And so following the row reduction, you will obtain I X again as above.

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