MATH307 April 2006
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) • Q6 (c) • Q6 (d) • Q7 (a) • Q7 (b) • Q7 (c) • Q7 (d) •
Question 05 (d)

A matrix A and a vector b are given by
$\displaystyle A={\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}}$
$\displaystyle b={\begin{bmatrix}2\\3\\4\end{bmatrix}}$
Determine if $\displaystyle Ax=b$ is solvable. Find x if it is solvable and find x (i.e. the least squares solution) if it is not.

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Solution

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From part c, we note b is not in R(A) therefore it is not solvable.
We will proceed by solving the least squares solution by QR decomposition
A=QR from part a
$Ax=b$
$QRx=b$
since Q has columns of orthonormal vectors, Q is orthogonal $Q^{T}Q=I$
$Rx=Q^{T}b$
$x=R^{1}Q^{T}b$
$Q={\begin{bmatrix}{\frac {1}{\sqrt {2}}}&{\frac {\sqrt {2}}{2{\sqrt {3}}}}\\{\frac {1}{\sqrt {2}}}&{\frac {\sqrt {2}}{2{\sqrt {3}}}}\\0&{\sqrt {\frac {2}{3}}}\end{bmatrix}}$
$Q^{T}={\begin{bmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\sqrt {\frac {2}{3}}}\end{bmatrix}}$
$R={\begin{bmatrix}{\sqrt {2}}&{\frac {1}{\sqrt {2}}}\\0&{\sqrt {\frac {3}{2}}}\end{bmatrix}}$
$R^{1}={\frac {1}{det(R)}}adj(R)={\frac {1}{\sqrt {3}}}{\begin{bmatrix}{\sqrt {2}}&{\frac {1}{2}}\\0&{\sqrt {\frac {3}{2}}}\end{bmatrix}}$
$x={\frac {1}{\sqrt {3}}}{\begin{bmatrix}{\sqrt {2}}&{\frac {1}{\sqrt {2}}}\\0&{\sqrt {\frac {3}{2}}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\sqrt {\frac {2}{3}}}\end{bmatrix}}{\begin{bmatrix}2\\3\\4\end{bmatrix}}$
${\frac {1}{\sqrt {3}}}{\begin{bmatrix}{\sqrt {2}}&{\frac {1}{\sqrt {2}}}\\0&{\sqrt {\frac {3}{2}}}\end{bmatrix}}{\begin{bmatrix}{\frac {5}{\sqrt {2}}}\\{\frac {9{\sqrt {2}}}{2{\sqrt {3}}}}\end{bmatrix}}$
The solution to our least square problem is
$x={\begin{bmatrix}1\\3\end{bmatrix}}$


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