Science:Math Exam Resources/Courses/MATH152/April 2012/Question 06 (a)

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MATH152 April 2012

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Question 06 (a)
This question requires little or no calculations. For what value(s) of a are the following vectors $\mathbf {v} _{1}=[1,2,-1]$, $\mathbf {v} _{2}=[0,1,3]$, $\mathbf {v} _{3}=[a,4,5]$, linearly dependent?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A set of three vectors is linearly independent if $c_{1}\mathbf {v} _{1}+c_{2}\mathbf {v} _{2}+c_{3}\mathbf {v} _{3}=\mathbf {0}$ is only satisfied if $\displaystyle c_{1}=c_{2}=c_{3}=0$.

Hint 2
For an alternative solution: What is true about the columns of a matrix when its determinant is equal to zero?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

A set of vectors is linearly dependent if $c_{1}\mathbf {v} _{1}+c_{2}\mathbf {v} _{2}+c_{3}\mathbf {v} _{3}=\mathbf {0}$ has more solutions than just the trivial solution $\displaystyle c_{1}=c_{2}=c_{3}=0$. To determine which values of $a$ will cause the vectors to be linearly dependent, we write the equation $c_{1}\mathbf {v} _{1}+c_{2}\mathbf {v} _{2}+c_{3}\mathbf {v} _{3}=\mathbf {0}$ in its matrix-vector form $M\mathbf {c} =\mathbf {0} ,\quad {\mbox{where}}\quad M=\left[{\begin{array}{ccc}1&0&a\\2&1&4\\-1&3&5\end{array}}\right],\quad \mathbf {c} =\left[{\begin{array}{c}c_{1}\\c_{2}\\c_{3}\end{array}}\right].$ If the row-reduced echelon form of $\displaystyle M$ has a row of zeros, there are non-zero values of $\displaystyle \mathbf {c}$ that will satisfy $c_{1}\mathbf {v} _{1}+c_{2}\mathbf {v} _{2}+c_{3}\mathbf {v} _{3}=\mathbf {0}$. Row-reducing $\displaystyle M$ as follows: ${\begin{aligned}M&=\left[{\begin{array}{ccc}1&0&a\\2&1&4\\-1&3&5\end{array}}\right]\quad \rightarrow \quad \left[{\begin{array}{ccc}1&0&a\\0&1&4-2a\\0&3&5+a\end{array}}\right]\quad \rightarrow \quad \left[{\begin{array}{ccc}1&0&a\\0&1&4-2a\\0&0&-7+7a\end{array}}\right],\end{aligned}}$ we can see that the last row will be a row of zeros if $\displaystyle a=1$. Therefore, the vectors ${\color {blue}\mathbf {v} _{1},\,\mathbf {v} _{2},\,\mathbf {v} _{3}}$ are linearly dependent if ${\color {blue}a=1}$.

Solution 2

As a quick alternative to the first solution, recall that the columns of a matrix, $\displaystyle M$, are linearly dependent if the determinant of $\displaystyle M$ is equal to zero. Hence, we define $M=\left[\mathbf {v} _{1},\mathbf {v} _{2},\mathbf {v} _{3}\right]=\left[{\begin{array}{ccc}1&0&a\\2&1&4\\-1&3&5\end{array}}\right],$ which has the determinant $\operatorname {det} (M)=-7+7a.$ The determinant equal to zero when $\displaystyle a=1$. Therefore, the vectors ${\color {blue}\mathbf {v} _{1},\,\mathbf {v} _{2},\,\mathbf {v} _{3}}$ are linearly dependent if ${\color {blue}a=1}$.

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