(a) $\Delta=15,\ \Delta_x=15,\ \Delta_y=30,\ \Delta_z=30\Rightarrow x=1,y=2,z=2$. (b) VAM gives an IBFS with total cost $\mathbf{550}$.

Part (a) — Cramer's rule

$x+2y+z=7,\quad 2x-y+2z=4,\quad x+y-2z=-1.$

Step 1 — Main determinant:

$\Delta=\begin{vmatrix}1 & 2 & 1\\ 2 & -1 & 2\\ 1 & 1 & -2\end{vmatrix}=1(2-2)-2(-4-2)+1(2+1)=0+12+3=15.$

Step 2 — Replace columns by the constants $[7,4,-1]$:

$\Delta_x=\begin{vmatrix}7 & 2 & 1\\ 4 & -1 & 2\\ -1 & 1 & -2\end{vmatrix}=7(0)-2(-6)+1(3)=15,$

$\Delta_y=\begin{vmatrix}1 & 7 & 1\\ 2 & 4 & 2\\ 1 & -1 & -2\end{vmatrix}=1(-6)-7(-6)+1(-6)=30,$

$\Delta_z=\begin{vmatrix}1 & 2 & 7\\ 2 & -1 & 4\\ 1 & 1 & -1\end{vmatrix}=1(-3)-2(-6)+7(3)=30.$

Step 3 — Solve:

$x=\frac{\Delta_x}{\Delta}=\frac{15}{15}=1,\quad y=\frac{30}{15}=2,\quad z=\frac{30}{15}=2.$

(Check: $1+2(2)+2=7$ ✓.)

Part (b) — Vogel's Approximation Method (VAM)

Supply $=25+35+40=100=$ demand $30+25+45$ (balanced).

Iteration 1 — penalties (difference of two least costs): rows $S_1{=}3,S_2{=}2,S_3{=}1$; cols $D_1{=}1,D_2{=}2,D_3{=}1$. Largest $=3$ at $S_1$; least cost $5$ in $D_3$ → allocate $\min(25,45)=25$ to $S_1D_3$. $S_1$ exhausted; $D_3$ left $=20$.

Iteration 2 — remaining rows $S_2,S_3$: column $D_3$ penalty $=9-4=5$ (largest); least cost $4$ at $S_2D_3$ → allocate $\min(35,20)=20$. $D_3$ exhausted; $S_2$ left $=15$.

…