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

MATH307 April 2006

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[hide] 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.

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[show] Hint

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[show] Solution

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 $-2b_{1}+b_{2}-b_{3}=0$.