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

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

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

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

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

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

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

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

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