Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 04 (e)

MATH152 April 2010

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Question B 04 (e)
The augmented matrix below for a linear system is put in reduced row echelon form with row operations. Write one of the following: the solution if it is unique all solutions if there are more than one state that there are no solutions. (e) $\left[{\begin{array}{cccc}1&0&2&-1\\0&1&2&-2\end{array}}\right.\left|{\begin{array}{c}1\\2\end{array}}\right]$

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
When we have more columns that rows in a matrix, what do we know about the number of free variables?

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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. For the matrix, $\left[{\begin{array}{cccc}1&0&2&-1\\0&1&2&-2\end{array}}\right.\left|{\begin{array}{c}1\\2\end{array}}\right]$ we have two more columns than rows (the matrix is rank deficient) and so we have two free variables. Let x3 and x4 be those variables. Then from row 2 we have, ${\begin{aligned}x_{2}+2x_{3}-2x_{4}&=2\\x_{2}&=2-2x_{3}+2x_{4}\end{aligned}}$ and from row 1 we have, ${\begin{aligned}x_{1}+2x_{3}-x_{4}&=1\\x_{1}&=1-2x_{3}+x_{4}.\end{aligned}}$ Therefore we have two null vectors, ${\begin{bmatrix}-2\\-2\\1\\0\end{bmatrix}}$ and ${\begin{bmatrix}1\\2\\0\\1\end{bmatrix}}$ as well as a particular solution ${\begin{bmatrix}1\\2\\0\\0\end{bmatrix}}.$ Therefore the solution is $\mathbf {x} ={\begin{bmatrix}1\\2\\0\\0\end{bmatrix}}+s{\begin{bmatrix}-2\\-2\\1\\0\end{bmatrix}}+t{\begin{bmatrix}1\\2\\0\\1\end{bmatrix}}.$ There are an infinite number of solutions.