Science:Math Exam Resources/Courses/MATH307/April 2006/Question 01 (b)

MATH307 April 2006

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Question 01 (b)
Let PA = LU be ${\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\2&1&8&8\\0&1&4&2\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&1&0\\2&1&1\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\0&1&4&2\\0&0&0&0\end{bmatrix}}$ and $b={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}$ (b) Find the condition on b1, b2, and b3 so that Ax = b has at least one solution.

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. To find the condition on $b_{1},b_{2},b_{3}$, let us first solve the matrix problem $Ax=b$. ${\begin{bmatrix}1&0&2&3&|&b_{1}\\2&1&8&8&|&b_{2}\\0&1&4&2&|&b_{3}\end{bmatrix}}$ $P={\begin{bmatrix}1&0&0\\{\color {Red}{0}}&{\color {Red}{0}}&{\color {Red}{1}}\\{\color {Red}{0}}&{\color {Red}{1}}&{\color {Red}{0}}\end{bmatrix}}$ $L={\begin{bmatrix}1&0&0\\0&1&0\\{\color {Green}{2}}&{\color {Purple}{1}}&1\end{bmatrix}}$ Since we are given the LU decomposition and partial pivoting of matrix A, we can simply follow the steps described by the matrix P (swapping the second and third row which is highlighted in ${\color {Red}{red}}$) and L (subtracting twice the first row (${\color {Green}{green}}$) from the third row and subtracting once the second row from the third row(${\color {Purple}{purple}}$)), we would arrive at ${\begin{bmatrix}1&0&2&3&|&b_{1}\\0&1&4&2&|&b_{3}\\0&0&0&0&|&b_{2}-b_{3}-2b_{1}\end{bmatrix}}$ In order for $Ax=b$ to have at least solution, the above matrix must be consistent, therefore the last row must contain all zeros. The constraint on $b_{1},b_{2},b_{3}$ is the equation $-2b_{1}+b_{2}-b_{3}=0$.