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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Exercises 13 and 14, the columns of \(Q\) were obtained by applying the Gram Schmidt process to the columns of \(A\). Find anupper triangular matrix \(R\) such that \(A = QR\). Check your work.

14.\(A = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\), \(Q = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\)

The upper triangular matrix is \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\).

And it is verified that upper triangular matrix is correct.

See the step by step solution

Step by Step Solution

Step 1: \(QR\) factorization of a Matrix

A matrix with order \(m \times n\) can be written as the product of an upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying Gram–Schmidt orthogonalization process to the \({\rm{col}}\left( A \right)\)

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Step 2: Find the matrix \(R\)

The given matrices are:

\(A = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\)and \(Q = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\)

Find the upper triangular matrix by using \({Q^T}A = R\).

\(\begin{aligned}{}R & = {Q^T}A\\ & = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}&{\frac{2}{7}}&{\frac{4}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}&{\frac{{ - 4}}{7}}&{\frac{2}{7}}\end{aligned}} \right)\left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{\frac{{ - 2}}{7} \times \left( { - 2} \right) + \frac{5}{7} \times 5 + \frac{2}{7} \times 2 + \frac{4}{7} \times 4}&{\frac{{ - 2}}{7} \times 3 + \frac{5}{7} \times 7 + \frac{2}{7} \times \left( { - 2} \right) + \frac{4}{7} \times 6}\\{\frac{5}{7} \times \left( { - 2} \right) + \frac{2}{7} \times 5 + \left( {\frac{{ - 4}}{7}} \right) \times 2 + \frac{2}{7} \times 4}&{\frac{5}{7} \times \left( 3 \right) + \frac{2}{7} \times 7 + \left( {\frac{{ - 4}}{7}} \right) \times \left( { - 2} \right) + \frac{2}{7} \times 6}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\end{aligned}\)

Hence, the upper triangular matrix is,\(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\).

Step 3: Checking Whether \(R\) is correct or not

Now, check that whether the obtained matrix \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\) is correct or not. Find the product of \(Q\) and \(R\), if the product is the same as matrix \(A\), then \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\) is correct.

\(\begin{aligned}{}QR & = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\\ & = A\end{aligned}\)

Hence verified, the obtained upper triangular matrix is correct.

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