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Q14E

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Found in: Page 331

### 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

# 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 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.