Science:Math Exam Resources/Courses/MATH221/April 2010/Question 01 (d)

MATH221 April 2010

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Question 01 (d)
Let $\displaystyle A={\begin{bmatrix}1&2&3&-2&4\\2&5&8&-1&6\\1&4&7&5&2\\1&3&5&1&2\\\end{bmatrix}}$ and $\displaystyle U={\begin{bmatrix}1&2&3&-2&4\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\\\end{bmatrix}}$ Then the matrix U is an echelon form for A (you may assume this, you don't have to do the row reduction again.) Solve the equation $\displaystyle Ax=0$

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Hint
Find the reduced echelon form, that is, use row operations to set the entries above the pivot elements to zero.

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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. In order to solve $Ax=0$ we need to find the reduced echelon form. Starting with the echelon form $U$, we first subtract the second row from the first row twice, $R_{1}=R_{1}-2R_{2}$, to obtain ${\begin{bmatrix}1&2&3&-2&4\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\end{bmatrix}}\rightarrow {\begin{bmatrix}1&0&-1&-8&8\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\end{bmatrix}}$ Now we add the third row eight times to the first row, and subtract the third row three times from the second row, $R_{1}=R_{1}+8R_{3}$, $R_{2}=R_{2}-3R_{3}$, to obtain ${\begin{bmatrix}1&0&-1&-8&8\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\end{bmatrix}}\rightarrow {\begin{bmatrix}1&0&-1&0&24\\0&1&2&0&-8\\0&0&0&1&2\\0&0&0&0&0\end{bmatrix}}$ In plain notation this means that ${\begin{aligned}x_{1}-x_{3}+24x_{5}&=0\\x_{2}+2x_{3}-8x_{5}&=0\\x_{4}+2x_{5}&=0\end{aligned}}$ Fixing the pivot variables $x_{1},x_{2},x_{4}$ and solving the above for the remaining two variables $x_{3},x_{5}$ we find that the solution to $Ax=0$ has the form ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}}={\begin{bmatrix}x_{3}-24x_{5}\\-2x_{3}+8x_{5}\\x_{3}\\-2x_{5}\\x_{5}\end{bmatrix}}=x_{3}{\begin{bmatrix}1\\-2\\1\\0\\0\end{bmatrix}}+x_{5}{\begin{bmatrix}-24\\8\\0\\-2\\1\end{bmatrix}}$

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