Science:Math Exam Resources/Courses/MATH152/April 2012/Question 02 (c)/Solution 1

If we multiply the matrix A to the vector [1,1,1] we get

${\begin{bmatrix}2&-1&2\\1&3&-6\\3&2&-4\end{bmatrix}}{\begin{bmatrix}1\\1\\1\end{bmatrix}}={\begin{bmatrix}3\\-2\\1\end{bmatrix}}$

and therefore a=3, b=-2, and c=1. If we set those values permanently then we know that [1,1,1] is a particular solution to the problem. However, from part(b) we have that the solution to the homogeneous problem was

$\mathbf {x} _{h}=t{\begin{bmatrix}0\\2\\1\end{bmatrix}}$

and therefore we can write that the general solution $\mathbf {x}$ is the sum of the particular and homogeneous solution,

$\mathbf {x} ={\begin{bmatrix}1\\1\\1\end{bmatrix}}+t{\begin{bmatrix}0\\2\\1\end{bmatrix}}.$