Science:Math Exam Resources/Courses/MATH221/December 2009/Question 01

MATH221 December 2009

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[hide] Question 01
Consider the system of equations: ${\begin{matrix}x_{1}+&x_{2}-&-x_{3}&=2\\x_{1}+&2x_{2}+&x_{3}&=3\\x_{1}+&x_{2}+&(c^{2}-5)x_{3}&=c\end{matrix}}$ Find all values of c such that the system has: (a) no solutions (b) a unique solution (c) infinitely many solutions In case c. write the general solution in the parametric vector form.

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?

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[show] Hint

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[show] Solution

Step 1: Write the system of equations in matrix form $A{\vec {x}}={\vec {b}}$ In this case $A={\begin{bmatrix}1&1&-1\\1&2&1\\1&1&c^{2}-5\end{bmatrix}}\quad {\vec {b}}={\begin{bmatrix}2\\3\\c\end{bmatrix}}\quad {\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$ Step 2: Write A and b as augmented matrix. Augmented form : $\left[{\begin{array}{ccc|ccc}1&1&-1&2\\1&2&1&3\\1&1&c^{2}-5&c\end{array}}\right]$ Step 3: Reduce the augmented matrix in row echelon form. After performing the row operations row2 = row2 - row1 and row3 = row3 - row1, we get the following reduced row echelon matrix $\left[{\begin{array}{ccc|ccc}1&1&-1&2\\0&1&2&1\\0&0&c^{2}-4&c-2\end{array}}\right]$ For part (a), we can see that in order for the system of equations to have no solution, $c^{2}-4=0$ while $c-2\neq 0$ and thus for $c=-2$ there is no solution. With this value of $c$ our augment matrix looks like $\left[{\begin{array}{ccc|ccc}1&1&-1&2\\0&1&2&1\\0&0&0&-4\end{array}}\right]$ indiacting no solution For part (b), that is, for the system of equations to have unique solution, we do not want any row of our matrix to be zero entries. Thus we want $c^{2}-4\neq 0$ and $c-2\neq 0$ Thus for all values of $c$ except $c=2$ and $c=-2$, we have unique solution. For part (c), For the system of equation to have an infinitely many solutions, we want at least one free variable, i.e we want at least one row to be all zero entries. This can be achieved if $c-2=0$ and $c^{2}-4=0$. Thus with $c=2$, we have the third row as all zero entries. Thus our augmented matrix for $c=2$ is $\left[{\begin{array}{ccc|ccc}1&1&-1&2\\0&1&2&1\\0&0&0&0\end{array}}\right]$ We know that the first column represents entries of $x_{1}$, the second column represents entries of $x_{2}$ and the third column represents entries of $x_{3}$. Thus looking at our augmented matrix, it can be said column 1 and 2 are pivots columns and so only $x_{3}$ is a free variable. We solve for general solution as follows: $x_{2}+2x_{3}=1\Rightarrow x_{2}=1-2x_{3}$ $x_{1}=2-x_{2}+x_{3}\Rightarrow x_{1}=2-1+2x_{3}+x_{3}\Rightarrow x_{1}=1+3x_{3}$ ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$ $\Rightarrow$ ${\vec {x}}={\begin{bmatrix}1+3x_{3}\\1-2x_{3}\\x_{3}\end{bmatrix}}$ $\Rightarrow$ ${\begin{bmatrix}1\\1\\0\end{bmatrix}}+t{\begin{bmatrix}3\\-2\\1\end{bmatrix}},\;t\in \mathbb {R}$