Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 02 (c)

MATH152 April 2016

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Question B 02 (c)
In the system below, $x$, $y$, and $z$ are variables, and $a$ and $b$ are constants. ${\begin{array}{ccccccc}x&+&y&+&z&=&5\\x&~&~&+&z&=&1\\ax&~&~&+&z&=&b\end{array}}$ (c) For what value or values of $a$ and $b$ (if any) does the system have no solutions?

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
A system of linear equations has no solution if a row echelon form of the augmented matrix has a row consisting of all zeros on the left hand side together with a nonzero entry on the right hand side.

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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. By (b), a row echelon form of the augmented matrix is $\left({\begin{array}{ccc|c}1&1&1&5\\0&-1&0&-4\\0&0&1-a&b-a\end{array}}\right).$ The system of linear equations has no solution if a row echelon form of the augmented matrix has a row with all zeros on the left side together with a nonzero entry on the right hand side. The first two rows of this augmented matrix are not zero rows. However, if $a=1$, then the third row would consist of all zeros on the left side. In order to make the right hand side nonzero, we would need $b-a\neq 0,$ i.e., $b-1\neq 0,$ or $b\neq 1.$ Therefore, the answer is $\color {blue}a=1,b\neq 1$.