# Science:Math Exam Resources/Courses/MATH221/April 2009/Question 01/Solution 1

From the system we are able to get the following matrix. ${\displaystyle {\begin{bmatrix}1&0&4&-2&|&1\\-1&1&-7&7&|&2\\2&3&-1&c&|&11\end{bmatrix}}}$ By doing Gaussian Elimination, we are able to get the matrix below ${\displaystyle {\begin{bmatrix}1&0&4&-2&|&1\\0&1&-3&5&|&3\\0&0&0&c-11&|&0\end{bmatrix}}}$

If we look at the 3rd row of this matrix, we have ${\displaystyle \displaystyle (c-11)x_{4}=0}$.

In the case of ${\displaystyle c=11}$, the matrix becomes ${\displaystyle {\begin{bmatrix}1&0&4&-2&|&1\\0&1&-3&5&|&3\\0&0&0&0&|&0\end{bmatrix}}.}$

This implies that ${\displaystyle x_{3}}$ and ${\displaystyle x_{4}}$ are free variables and the general solution is

{\displaystyle {\begin{aligned}{\begin{bmatrix}x1\\x2\\x3\\x4\end{bmatrix}}={\begin{bmatrix}1\\3\\0\\0\end{bmatrix}}+{\begin{bmatrix}-4\\3\\1\\0\end{bmatrix}}x_{3}+{\begin{bmatrix}2\\-5\\0\\1\end{bmatrix}}x_{4}.\end{aligned}}}

On the other hand, if ${\displaystyle c\neq 11}$, we can easily see that ${\displaystyle \displaystyle x_{4}=0}$ .

Therefore, the matrix becomes ${\displaystyle {\begin{bmatrix}1&0&4&|&1\\0&1&-3&|&3\\0&0&0&|&0\end{bmatrix}}}$ which can be solved easily. The general solution is

{\displaystyle {\begin{aligned}{\begin{bmatrix}x1\\x2\\x3\\x4\end{bmatrix}}={\begin{bmatrix}1\\3\\0\\0\end{bmatrix}}+{\begin{bmatrix}-4\\3\\1\\0\end{bmatrix}}x_{3}.\end{aligned}}}