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Q14E

Expert-verifiedFound in: Page 331

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.

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 \({Q^T}A = R\).

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,\(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\).

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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