This is what my lecturer says on standard and canonical forms: "Two particular forms of the linear programming problem are of interest for m equations in n unknowns. These are called the standard form, i.e. minimise $\underline{c}^T\underline{x}$ subject to $A\underline{x} ≥ \underline{b}$, $\underline{x} ≥ \underline{0}$ and the canonical form, i.e. minimise $\underline{c}^T\underline{x}$ subject to $A\underline{x} = \underline{b}$, $x ≥ 0$."
The question says: Put the following LP into canonical form (do not try to solve)
Minimise $z=x_1+2x_2+3x_3$ subject to $$3x_1+4x_3\leq5$$ $$5x_1+x_2+6x_3 \leq 7$$ $$8x_1+9x_3\geq2$$
With all vairables non-negative.
So I started off by trying to convert it to standard form and adding in the slack and surplus variables so I have
$$3x_1+4x_3+s_1=5$$ $$5x_1+x_2+6x_3+s_2= 7$$ $$8x_1+9x_3-s_3=2$$
but I'm confused about what the vectors $\underline{x}, \underline{c}, \underline{b}$ and the matrix $A$ is.
I read another a question where the answer suggested that $$\underline{x}=\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix},
\underline{c}= \begin{pmatrix}
1\\
2\\
3
\end{pmatrix},
\underline{b}=\begin{pmatrix}
5\\
7\\
2
\end{pmatrix}, A= \begin{pmatrix}
3 & 0 & 4 \\
5 & 1 & 6 \\
8 & 0 & 9
\end{pmatrix}$$
but I don't see where the slack and surplus variables come into play.
Best Answer
Your $\underline{b}$ is correct. The rest should be: $$\underline{x} = \begin{pmatrix}x_1\\x_2\\x_3\\s_1\\s_2\\s_3 \end{pmatrix},\ \underline{c} = \begin{pmatrix}1\\2\\3\\0\\0\\0 \end{pmatrix},\ A = \begin{pmatrix}3 &0 &4 &1 &0 &0\\ 5 &1 &6 &0 &1 &0\\ 8 &0 &9 &0 &0 &-1 \end{pmatrix}$$