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I am having trouble finding the general solution for this question:

Show that $a=(1,1,0)^T$ and $b = (-3,3,-2)^T$ are particular solutions to the system $\begin{bmatrix}2&-1&-5\\ 1&-4&-6\\ 3&2&-4 \end{bmatrix}X=\begin{bmatrix}1\\-3\\5\end{bmatrix}$ and find the general solution.

The book says the general solution is

$\begin{align}X & = a && + t(b-a) \\&= (1,1,0)^T && + t(-4,2,-2)^T\end{align}$

but how did they get that as the general solution?

1 Answers 1

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One way to get the general solution, is to apply elementary row operations on the augmented matrix $\hat{A}=\left[\begin{array}{rrr|r}2&-1&-5&1\\ 1&-4&-6&-3\\ 3&2&-4&5 \end{array}\right]$

After some row operations, the given system is equivalent to the following system $\begin{bmatrix}0&1&1\\ 1&0&-2\\ 0&0&0 \end{bmatrix}X=\begin{bmatrix}1\\1\\0\end{bmatrix}$

So, the general solution is given by $y+z=1, x-2z=1 $ which can be written as $(x,y,z)=(1+2z,1-z,z)=(1,1,0)+z(2,-1,1)$

To match with your book, put $z=-2t$.

Note that $a$ and $b$ corresponds to the particular value of $t=0$ and $t=1$ respectively. Hence they are particular solutions.