Science:Math Exam Resources/Courses/MATH307/April 2006/Question 05 (d)

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MATH307 April 2006

• 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) •

Other MATH307 Exams

• December 2012 • April 2013 • April 2012 • April 2006 • December 2008 • December 2010 •

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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Hint
Science:Math Exam Resources/Courses/MATH307/April 2006/Question 05 (d)/Hint 1

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Solution
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 $x={\begin{bmatrix}1\\3\end{bmatrix}}$