QR Factorisation
Find $Q R$ factorisation of the following matrix:
$$
A=\left(\begin{array}{cc}
5 & -4 \
12 & 6
\end{array}\right)
$$

Answer:
$\square$
Note:
Enter your answer as two element Maple list: $[Q, R]$. Each element of the list is Maple matrix:

Question

QR Factorisation
Find $Q R$ factorisation of the following matrix:
$$
A=\left(\begin{array}{cc}
5 & -4 \
12 & 6
\end{array}\right)
$$

Answer:
$\square$
Note:
Enter your answer as two element Maple list: $[Q, R]$. Each element of the list is Maple matrix: < col1 | col2 | . . . . Each col1, col2,... is matrix column in Maple vector syntax: $\langle a, b, \ldots$.

Note 2:
Double check your answer by multiplying $Q R$ before submitting this form. Use preview function to ensure that your answer is typo-free.

Note 3:
if you need to type square root, use sqrt Maple function.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given matrix $ A = \begin{pmatrix} 5 & -4 \ 12 & 6 \end{pmatrix} $, we need to find its QR factorization. ‖ ‖ step 2 ⋮ First, we find the orthonormal basis vectors for the columns of $ A $. Let $ a_1 = \begin{pmatrix} 5 \ 12 \end{pmatrix} $ and $ a_2 = \begin{pmatrix} -4 \ 6 \end{pmatrix} $. ‖ ‖ step 3 ⋮ Normalize $ a_1 $ to get $ q_1 $: $$ q_1 = \frac{a_1}{\|a_1\|} = \frac{1}{13} \begin{pmatrix} 5 \ 12 \end{pmatrix} = \begin{pmatrix} \frac{5}{13} \ \frac{12}{13} \end{pmatrix} $$ ‖ ‖ step 4 ⋮ Project $ a_2 $ onto $ q_1 $ and subtract to get the orthogonal component: $$ \text{proj}_{q_1}(a_2) = \left( \begin{pmatrix} \frac{5}{13} \ \frac{12}{13} \end{pmatrix} \cdot \begin{pmatrix} -4 \ 6 \end{pmatrix} \right) \begin{pmatrix} \frac{5}{13} \ \frac{12}{13} \end{pmatrix} = \frac{20}{13} \begin{pmatrix} \frac{5}{13} \ \frac{12}{13} \end{pmatrix} = \begin{pmatrix} \frac{100}{169} \ \frac{240}{169} \end{pmatrix} $$ ‖ ‖ step 5 ⋮ Subtract the projection from $ a_2 $ to get $ u_2 $: $$ u_2 = a_2 - \text{proj}_{q_1}(a_2) = \begin{pmatrix} -4 \ 6 \end{pmatrix} - \begin{pmatrix} \frac{100}{169} \ \frac{240}{169} \end{pmatrix} = \begin{pmatrix} -\frac{776}{169} \ \frac{774}{169} \end{pmatrix} $$ ‖ ‖ step 6 ⋮ Normalize $ u_2 $ to get $ q_2 $: $$ q_2 = \frac{u_2}{\|u_2\|} = \frac{1}{\sqrt{\left( -\frac{776}{169} \right)^2 + \left( \frac{774}{169} \right)^2}} \begin{pmatrix} -\frac{776}{169} \ \frac{774}{169} \end{pmatrix} = \begin{pmatrix} -\frac{12}{13} \ \frac{5}{13} \end{pmatrix} $$ ‖ ‖ step 7 ⋮ Form the orthogonal matrix $ Q $: $$ Q = \begin{pmatrix} \frac{5}{13} & -\frac{12}{13} \ \frac{12}{13} & \frac{5}{13} \end{pmatrix} $$ ‖ ‖ step 8 ⋮ Compute the upper triangular matrix $ R $ using $ R = Q^T A $: $$ R = \begin{pmatrix} \frac{5}{13} & \frac{12}{13} \ -\frac{12}{13} & \frac{5}{13} \end{pmatrix} \begin{pmatrix} 5 & -4 \ 12 & 6 \end{pmatrix} = \begin{pmatrix} 13 & 4 \ 0 & 6 \end{pmatrix} $$ ‖ ∻Answer∻ ⚹ The QR factorization of the matrix $ A $ is: $$ Q = \begin{pmatrix} \frac{5}{13} & -\frac{12}{13} \ \frac{12}{13} & \frac{5}{13} \end{pmatrix}, \quad R = \begin{pmatrix} 13 & 4 \ 0 & 6 \end{pmatrix} $$ ⚹ ∻Key Concept∻ ⚹QR Factorization⚹ ∻Explanation∻ ⚹QR factorization decomposes a matrix into an orthogonal matrix $ Q $ and an upper triangular matrix $ R $. This is useful for solving linear systems and eigenvalue problems.⚹◊What is the formula for QR factorization of a matrix $A$?⍭ Generate me a similar question◊

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