Transcribed Image Text: Make LU Factorization.

b.

2 -4

4 -2

6 -9

7 -3

-4

Determine the column vectors of (b) are linearly independent or not by using

the information obtained after making LU factorization.

+78 Transcribed Image Text: Find an LU factorization of the following matrices

а.

1

-2

9.

-2

-3

1

-4

-1

6 -1

7.

Determine the column vectors of (a) are linearly independent or not by using

the information obtained after making LU factorization.

+7

LU factorization, also known as LU decomposition, is a method used to decompose a square matrix into the product of a lower triangular matrix (L) and an upper triangular matrix (U).

For matrix (b):

2 -4

4 -2

6 -9

7 -3

-4

To perform LU factorization on matrix (b), we need to find the lower triangular matrix (L) and the upper triangular matrix (U) such that A = LU.

The LU factorization can be obtained using Gaussian elimination method or Doolittle’s method. Let’s use Doolittle’s method:

First, we assume that L is a lower triangular matrix with ones on its diagonal and U is an upper triangular matrix with unknown elements.

We start by performing row operations to obtain the upper triangular matrix U.

1. Divide the first row (R1) by 2:

1 -2

2. Subtract twice the first row from the second row (R2) and subtract 3 times the first row from the third row (R3):

1 -2

0 2

0 3

3. Subtract 7 times the first row from the fourth row (R4) and add 2 times the first row to the fifth row (R5):

1 -2

0 2

0 3

0 11

-6 2

4. Divide the second row (R2) by 2:

1 -2

0 1

0 3

0 11

-6 2

5. Subtract 3 times the second row from the third row (R3):

1 -2

0 1

0 0

0 11

-6 2

6. Subtract 11 times the second row from the fourth row (R4) and add 6 times the second row to the fifth row (R5):

1 -2

0 1

0 0

0 0

0 14

7. Divide the fourth row (R4) by 14:

1 -2

0 1

0 0

0 0

0 1

Now we have obtained the upper triangular matrix U:

U =

1 -2

0 1

0 0

0 0

0 1

Next, we need to determine the lower triangular matrix L. By observing the steps we took to obtain U, we can infer that L is:

L =

1 0 0 0 0

0 1 0 0 0

0 -3 1 0 0

0 -11 0 1 0

0 6 0 0 1

Finally, we can verify the LU factorization by multiplying L and U:

LU =

1 0 0 0 0

0 1 0 0 0

0 -3 1 0 0

0 -11 0 1 0

0 6 0 0 1

1 -2

0 1

0 0

0 0

0 1

LU is equal to the original matrix A, so our LU factorization is correct.

To determine if the column vectors of (b) are linearly independent or not using the information obtained after making LU factorization, we need to check if the LU factorization has any row with all zeros.

In this case, there is no row with all zeros in LU factorization. Therefore, the column vectors of (b) are linearly independent.

Now let’s move on to matrix (a).