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

MATH221 April 2010

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Question 01 (h)

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.) Find all solutions to the system of equations ${\begin{aligned}1x+2y+3z-2w=4\\2x+5y+8z-1w=6\\1x+4y+7z+5w=2\\1x+3y+5z+1w=2\\\end{aligned}}$ (Hint: What is the augmented matrix for this system?)

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. 1.) Transform the system of equations into an augmented matrix. x,y,z,&w all represent one column each and the number of system of equations represent the rows of the matrix. The answers to each equation will be on the right-most of the matrix. 2.) ${\begin{vmatrix}1&2&3&-2&4\\2&5&8&-1&6\\1&4&7&5&2\\1&3&5&1&2\end{vmatrix}}$ 3.) Let’s reduce it to row echelon form by following these steps: 1. $R_{2}$ = -2$R_{1}$ + $R_{2}$ 2. $R_{3}$ = -1$R_{1}$ + $R_{3}$ 3. $R_{4}$ = -1$R_{1}$ + $R_{4}$ 4. $R_{3}$ = -2$R_{2}$ + $R_{3}$ 5. $R_{4}$ = -1$R_{2}$ + $R_{4}$ 6. $R_{2}$ = -3$R_{3}$ + $R_{2}$ 7. $R_{1}$ = 2$R_{3}$ + $R_{1}$ 8. $R_{1}$ = -2$R_{2}$ + $R_{1}$ and we will end up with ${\begin{vmatrix}1&0&-1&0&24\\0&1&2&0&-8\\0&0&0&1&2\\0&0&0&0&0\end{vmatrix}}$ The right most values in column 5 are the answers. 4.) Therefore, the system of equations correspond to: $x_{1}$= $x_{3}$ + 24 $x_{2}$ = -2$x_{3}$ - 8 $x_{3}$= $x_{3}$ $x_{4}$= 2 All solutions are represented by: ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\\end{bmatrix}}$ = $x_{3}{\begin{bmatrix}1\\-2\\1\\0\end{bmatrix}}+{\begin{bmatrix}24\\-8\\0\\2\end{bmatrix}}$

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