Q 90E Question 90: Consider the matrix... [FREE SOLUTION]

Chapter 2: Q 90E (page 102)

Question 90: Consider the matrix
$A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}]$ .
Find lower triangular elementary matrices E₁, E₂,...E_m such that the product E_m...E₂E₁A is an upper triangular matrix U. Hint: Use elementary row operations to eliminate the entries below the diagonal of A.
Find lower triangular elementary matrices M₁,M₂, ...., M_m and an upper triangular matrix U such that A= M₁M₂...M_mU.
Find a lower triangular matrix L and an upper triangular matrix U such that A = LU. Such a representation of an invertible matrix is called an LU-factorization. The method outlined in this exercise to find an LU-factorization can be streamlined somewhat, but we have seen the major ideas. An LU-factorization (as introduced here) does not always exist. See Exercise 92.
Find a lower triangular matrix L with 1鈥檚 on the diagonal, an upper triangular matrix with 1鈥檚 on the diagonal, and a diagonal matrix D such that A = LDU. Such a representation of an invertible matrix is called an LDU-factorization.

Short Answer

Expert verified

$E_{1} = [\begin{matrix} 1 & 0 & 0 \\ 鈭� 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], E_{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 鈭� 2 & 0 & 1 \end{matrix}], E_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}]$
$M_{1} = {E_{1}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], M_{2} = {E_{2}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}], a n d M_{3} = {E_{3}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 鈭� 1 & 1 \end{matrix}]$
$L = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}], a n d U = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 鈭� 1 \end{matrix}]$
$L = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}], D = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 鈭� 1 \end{matrix}] a n d U = [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix}] a$

Step by step solution

Given that

Given a matrix

$A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}]$

Find lower triangular elementary matrices such that Em...E2E1A=U

To find elementary matrices, reduce the matrix by applying row operations.

$A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}]$

Firstly, Apply operation on I₃:

Subtract 2 times the first row, from the second row.

$E_{1} = [\begin{matrix} 1 & 0 & 0 \\ 鈭� 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Thus,

$\begin{matrix} E_{1} A = [\begin{matrix} 1 & 0 & 0 \\ 鈭� 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] \\ = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 2 & 2 & 4 \end{matrix}] \end{matrix}$

Now, subtract 2 times the first row from third row.

So $E_{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 鈭� 2 & 0 & 1 \end{matrix}]$

Then

$\begin{matrix} E_{2} E_{1} A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 鈭� 2 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 2 & 2 & 4 \end{matrix}] \\ = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 鈭� 2 & 鈭� 2 \end{matrix}] \end{matrix}$

Now, add second row to third row.

$E_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}]$

Then

$\begin{matrix} E_{3} E_{2} E_{1} A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 鈭� 2 & 鈭� 2 \end{matrix}] \\ = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 鈭� 1 \end{matrix}] \end{matrix}$

Which forms an upper triangular matrix.

Hence, $E_{3} E_{2} E_{1} A = U$

Therefore, the lower triangular matrices are:

$E_{1} = [\begin{matrix} 1 & 0 & 0 \\ 鈭� 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], E_{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 鈭� 2 & 0 & 1 \end{matrix}], E_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}]$

Find lower triangular matrices such that A=M1M2...MmU

Part (a) gives $E_{3} E_{2} E_{1} A = U$

Using inverse laws,

$A = {E_{1}}^{鈭� 1} {E_{2}}^{鈭� 1} {E_{3}}^{鈭� 1} U$

So $M_{1} M_{2} M_{3} = E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1}$

$S i n c e, E_{1} = [\begin{matrix} 1 & 0 & 0 \\ 鈭� 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] i m p l i e s {E_{1}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] a n d E_{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 鈭� 2 & 0 & 1 \end{matrix}] i m p l i e s {E_{2}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}] a n d E_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] i m p l i e s {E_{3}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 鈭� 1 & 1 \end{matrix}] N o w, \begin{matrix} E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 鈭� 1 & 1 \end{matrix}] \\ = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}] \end{matrix} N o w, \begin{matrix} E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1} U = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 鈭� 1 \end{matrix}] \\ = [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] \\ = A \end{matrix} H e n c e E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1} U = A$

Thus, $M_{1} = {E_{1}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], M_{2} = {E_{2}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}], a n d M_{3} = {E_{3}}^{鈭� 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 鈭� 1 & 1 \end{matrix}]$

Find a lower triangular matrix L and upper triangular matrix U such that A = LU

To find A = LU use, first and second part

$A = E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1} E_{3} E_{2} E_{1} A$

From part(a), $E_{3} E_{2} E_{1} A = U$

$S o U = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 鈭� 1 \end{matrix}] A n d f r o m p a r t (b) L = E_{1}^{鈭� 1} E_{2}^{鈭� 1} E_{3}^{鈭� 1} S o L = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}] T h u s, \begin{matrix} A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] \\ = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 鈭� 1 \end{matrix}] \end{matrix}$

Find a lower triangular matrix L and upper triangular matrix U and diagonal matrix U such that A = LDU

To find L, U and D, such that A = LDU

Let,

$\begin{array}{l} [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ l_{11} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{matrix}] [\begin{matrix} d_{11} & 0 & 0 \\ 0 & d_{22} & 0 \\ 0 & 0 & d_{33} \end{matrix}] [\begin{matrix} 1 & u_{12} & u_{13} \\ 0 & 1 & u_{23} \\ 0 & 0 & 1 \end{matrix}] \\ [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] = [\begin{matrix} d_{11} & 0 & 0 \\ d_{11} l_{21} & d_{22} & 0 \\ d_{11} l_{31} & d_{22} l_{32} & d_{33} \end{matrix}] [\begin{matrix} d_{11} & 0 & 0 \\ 0 & d_{22} & 0 \\ 0 & 0 & d_{33} \end{matrix}] \\ [\begin{matrix} 1 & 2 & 3 \\ 2 & 6 & 7 \\ 2 & 2 & 4 \end{matrix}] = [\begin{matrix} d_{11} & d_{11} u_{12} & d_{11} u_{13} \\ d_{11} l_{21} & d_{22} + d_{11} l_{21} u_{12} & d_{11} l_{21} u_{13} + d_{22} u_{23} \\ d_{11} l_{31} & d_{22} l_{32} + d_{11} l_{31} u_{12} & d_{33} + d_{11} l_{31} u_{13} + d_{22} + l_{32} u_{23} \end{matrix}] \end{array}$

Equating the corresponding elements of matrics:

$\begin{array}{l} d_{11} = 1 \\ d_{11} u_{12} = 2 \\ d_{11} u_{13} = 3 \end{array} \begin{array}{l} d_{11} l_{21} = 2 \\ d_{22} + d_{11} l_{21} u_{12} = 6 \\ d_{11} l_{21} u_{13} + d_{22} u_{23} = 7 \end{array} \begin{array}{l} d_{11} l_{31} = 2 \\ d_{22} l_{32} + d_{11} + l_{31} u_{12} = 2 \\ d_{33} + d_{11} l_{31} u_{13} + d_{22} l_{32} u_{23} = 4 \end{array}$

Solving these equations:

$d_{11} = 1 d_{11} u_{12} = 2 i m p l i e s u_{12} = 2 d_{11} u_{13} = 3 i m p l i e s u_{13} = 3 d_{11} I_{21} = 2 i m p l i e s I_{21} = 2 \begin{array}{l} d_{22} + d_{11} l_{21} u_{12} = 6 \\ 鈬� d_{22} + (1) (2) (2) = 6 \end{array} i m p l i e s d_{22} = 2 \begin{array}{l} d_{11} l_{21} u_{13} + d_{22} u_{23} = 7 \\ 鈬� (1) (2) (3) + (2) u_{23} = 7 \end{array} i m p l i e s u_{23} = \frac{1}{2} P r o c e e d i n g l i k e t h i s, I_{32} = - 1 I_{31} = 2 a n d d_{33} = - 1$

Therefore, the matrices are

$L = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 鈭� 1 & 1 \end{matrix}], D = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 鈭� 1 \end{matrix}] a n d U = [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix}] a$

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Short Answer

Step by step solution

Given that

Find lower triangular elementary matrices such that Em...E2E1A=U

Find lower triangular matrices such that A=M1M2...MmU

Find a lower triangular matrix L and upper triangular matrix U such that A = LU

Find a lower triangular matrix L and upper triangular matrix U and diagonal matrix U such that A = LDU

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