Find the QR factorization of the matrix \(\begin{bmatrix}2 & 1\\ 3 & -1\end{bmatrix}\)

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LEt the columns of A is x1, x2. So, x1 = (2, 3) and x2 = (1, -1)

Let v1 = x1 = (2, 3)

v2 = x2 – \(\frac{x_2 . v_1}{v_1 . v_2}\) . v1

= (1, -1) – \(\frac{(1, -1) . (2, 3)}{(2, 3) (2, 3)}\) (2, 3)

= (1, -1) – \(\frac{-1}{13}\) (2, 3)

= \(\left ( 1 + \frac{2}{13}, -1 + \frac{3}{13}\right )\)

= \(\left ( \frac{15}{13}, \frac{-10}{13} \right )\)

= \(\frac{5}{13}\) (3, -2)

set v2 = (3, -2)

Thus, {v1, v2 } be an orthogonal basis. Let {u1, u2} be normalize of the orthogonal basis.

so, u1 = \(\frac{v_1}{||v_1||}\) = \(\frac{(2, 3)}{13}\)

u2 = \(\frac{v_2}{||v_2||}\) = \(\frac{(3, -2)}{13}\)

Let ! be a matrix whose columns are u1 , u2

Q = \(\begin{bmatrix}\frac{2}{\sqrt{13}} & \frac{3}{\sqrt{13}}\\ \frac{3}{\sqrt{13}} & \frac{-2}{\sqrt{13}}\end{bmatrix}\)

since, we have

A = QR (by QR – Factorization theorem)

then

QTA = QT(QR)

QTA = R

R = QTA

= \(\begin{bmatrix}\frac{2}{\sqrt{13}} & \frac{3}{\sqrt{13}}\\ \frac{3}{\sqrt{13}} & \frac{-2}{\sqrt{13}}\end{bmatrix}\) \(\begin{bmatrix}2 & 1\\ 3 & -1\end{bmatrix}\)

=\(\begin{bmatrix}2 & 1\\ 3 & -1\end{bmatrix}\)

= \(\begin{bmatrix}\frac{4}{\sqrt{13}} + \frac{4}{\sqrt{13}} & \frac{2}{\sqrt{13}} -\frac{3}{\sqrt{13}}\\ \frac{6}{\sqrt{13}} – \frac{6}{\sqrt{13}} & \frac{3}{\sqrt{13}} + \frac{2}{\sqrt{13}}\end{bmatrix}\)

= \(\begin{bmatrix}\frac{13}{\sqrt{13}} & \frac{-1}{\sqrt{13}}\\ 0 & \frac{5}{\sqrt{13}}\end{bmatrix}\)

= \(\begin{bmatrix}\sqrt{13} & \frac{-1}{\sqrt{13}}\\ 0 & \frac{5}{\sqrt{13}}\end{bmatrix}\)

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