Science:Math Exam Resources/Courses/MATH307/April 2009/Question 01 (d)

MATH307 April 2009

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Question 01 (d)
In trying to solve the equation Ax = b, where A is the matrix in part (c), $A={\begin{bmatrix}-10&0&0\\0&-3.5&0\\0&0&0.5\end{bmatrix}}$ you are able to measure b with an accuracy of ${\frac {\|\Delta b\|}{\|b\|}}\leq 0.1$. What is the largest possible relative error ${\frac {\|\Delta x\|}{\|x\|}}$ in your solution for x?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. METHODOLOGY Suppose the measured data is given by: ${\vec {b'}}={\vec {b}}+{\vec {\Delta b}}$ where ${\vec {b}}$ is the true data, ${\vec {\Delta b}}$ is the error in the data. If a vector ${\vec {x}}$ was the true solution then $A{\vec {x}}={\vec {b}}$. Using ${\vec {b'}}$ we can get an approximate solution for ${\vec {x}}$ given by: ${\vec {x'}}=A^{-1}{\vec {b'}}$ Substituting in ${\vec {b'}}={\vec {b}}+{\vec {\Delta b}}$, we get: ${\vec {x'}}=A^{-1}({\vec {b}}+{\vec {\Delta b}})$ ${\vec {x'}}=A^{-1}{\vec {b}}+A^{-1}{\vec {\Delta b}}$ ${\vec {x'}}=A^{-1}{\vec {b}}+A^{-1}{\vec {\Delta b}}={\vec {x}}+{\vec {\Delta x}}$ This gives us: $A^{-1}{\vec {\Delta b}}={\vec {\Delta x}}$ which we can use to find the upper bound of the relative error ${\frac {\|\Delta x\|}{\|x\|}}$. To do this we will start with: $\|{\vec {\Delta x}}\|=\|A^{-1}{\vec {\Delta b}}\|$ ...and then multiply both sides by $\|{\vec {b}}\|$ to get: $\|{\vec {\Delta x}}\|\|{\vec {b}}\|=\|A^{-1}{\vec {\Delta b}}\|\|{\vec {b}}\|$ We know that $\|{\vec {b}}\|=\|A{\vec {x}}\|$ and we can substitute this into the above equation to get: $\|{\vec {\Delta x}}\|\|{\vec {b}}\|=\|A^{-1}{\vec {\Delta b}}\|\|{\vec {b}}\|=\|A^{-1}{\vec {\Delta b}}\|\|A{\vec {x}}\|$ It can be proven that that for any matrix A and any vector x, the equality $\|A{\vec {x}}\|\leq \|A\|\|{\vec {x}}\|$ will always hold true. We can use this in solving for the relative error. Using this equality, we get: $\|{\vec {\Delta x}}\|\|{\vec {b}}\|\leq \|A^{-1}{\vec {\Delta b}}\|\|A{\vec {x}}\|$ Divide both sides by $\|{\vec {x}}\|\|{\vec {b}}\|$: ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq \|A\|\|A^{-1}\|{\frac {\|\Delta {\vec {b}}\|}{\|{\vec {b}}\|}}$ Since $\|A\|\|A^{-1}\|=cond(A)$: ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq cond(A){\frac {\|\Delta {\vec {b}}\|}{\|{\vec {b}}\|}}$ We have finished finding an equation which tells us an upper bound on the relative error ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}$. This says the larger the condition number, the less control we have on the relative error. ANSWER To solve this question we have to substitute in ${\frac {\|\Delta b\|}{\|b\|}}$ and $cond(A)$. The question gives ${\frac {\|\Delta b\|}{\|b\|}}\leq 0.1$, but we want the largest possible relative error ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}$. Since ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}$ is proportional to ${\frac {\|\Delta b\|}{\|b\|}}$, we want the largest value of ${\frac {\|\Delta b\|}{\|b\|}}$ which is 0.1, so we will use ${\frac {\|\Delta b\|}{\|b\|}}=0.1$. From the part (c) of question 1, we got that $cond(A)=20$, and we have both values needed to solve for the relative error. ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq cond(A){\frac {\|\Delta {\vec {b}}\|}{\|{\vec {b}}\|}}$ ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq 20\cdot 0.1$ ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq 2$ FINAL ANSWER ${\frac {\|\Delta {\vec {x}}\|}{\|{\vec {x}}\|}}\leq 2$