Form two simplified simultaneous equations from the given information about production costs in two factories and solve for the cost of producing 1 kg of chocolate bars.

Step 1: Define variables and set up initial relationships.
Let $x$ be the cost of producing 1 kg of chocolate bars in factory A.
Let $y$ be the cost of producing 1 kg of eclairs in factory A.

The cost of 1 kg of chocolate bars in factory B is $\frac{6}{5}$ times the cost in factory A, so it is $\frac{6}{5}x$.
The cost of 1 kg of eclairs in factory B is $\frac{4}{5}$ times the cost in factory A, so it is $\frac{4}{5}y$.

a) Form two simplified simultaneous equations representing this information.

From the problem statement:

1. It costs Kshs 460,000 to produce 1 tonne (1000 kg) of chocolate bars and 800 kg of eclairs in factory A.

$$1000x + 800y = 460,000$$

Divide by 100:

$$10x + 8y = 4600$$

Divide by 2:

$$5x + 4y = 2300 \quad \text{(Equation 1)}$$

2. It costs Kshs 534,000 to produce the same quantities (1000 kg of chocolate bars and 800 kg of eclairs) in factory B. Cost of 1000 kg chocolate bars in factory B: $1000 \times \frac{6}{5}x = 1200x$. Cost of 800 kg eclairs in factory B: $800 \times \frac{4}{5}y = 640y$.

$$1200x + 640y = 534,000$$

Divide by 20:

$$60x + 32y = 26,700$$

Divide by 4:

$$15x + 8y = 6675 \quad \text{(Equation 2)}$$

The two simplified simultaneous equations are:

$$\boxed{\begin{align*} 5x + 4y &= 2300 \\ 15x + 8y &= 6675 \end{align*}}$$

b) Use matrix method to find the cost of producing 1 kg of chocolate bars and 1 kg of eclairs in factory A.

The system of equations from part (a) is:

$$\begin{align*} 5x + 4y &= 2300 \\ 15x + 8y &= 6675 \end{align*}$$

Write this system in matrix form $AX = B$:

$$\begin{bmatrix} 5 & 4 \\ 15 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2300 \\ 6675 \end{bmatrix}$$

Find the determinant of matrix $A$:

$$\det(A) = (5 \times 8) - (4 \times 15) = 40 - 60 = -20$$

Find the inverse of matrix $A$:

$$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 8 & -4 \\ -15 & 5 \end{bmatrix} = \frac{1}{-20} \begin{bmatrix} 8 & -4 \\ -15 & 5 \end{bmatrix}$$

Now, calculate $X = A^{-1}B$:

$$\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{20} \begin{bmatrix} 8 & -4 \\ -15 & 5 \end{bmatrix} \begin{bmatrix} 2300 \\ 6675 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{20} \begin{bmatrix} (8 \times 2300) + (-4 \times 6675) \\ (-15 \times 2300) + (5 \times 6675) \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{20} \begin{bmatrix} 18400 - 26700 \\ -34500 + 33375 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{20} \begin{bmatrix} -8300 \\ -1125 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{-8300}{-20} \\ \frac{-1125}{-20} \end{bmatrix} = \begin{bmatrix} 415 \\ 56.25 \end{bmatrix}$$

So, $x = 415$ and $y = 56.25$.
The cost of producing 1 kg of chocolate bars in factory A is Kshs 415.
The cost of producing 1 kg of eclairs in factory A is Kshs 56.25.

$$\boxed{\text{1 kg chocolate bars: Kshs 415, 1 kg eclairs: Kshs 56.25}}$$

c) Find the cost of producing 100 kg of chocolate bars and 50 kg of eclairs in factory B.

First, find the cost per kg in factory B:
Cost of 1 kg of chocolate bars in factory B = $\frac{6}{5}x = \frac{6}{5} \times 415 = 6 \times 83 = 498$ Kshs.
Cost of 1 kg of eclairs in factory B = $\frac{4}{5}y = \frac{4}{5} \times 56.25 = 4 \times 11.25 = 45$ Kshs.

Now, calculate the total cost for 100 kg of chocolate bars and 50 kg of eclairs in factory B:
Total cost = $(100 \text{ kg} \times 498 \text{ Kshs/kg}) + (50 \text{ kg} \times 45 \text{ Kshs/kg})$
Total cost = $49800 + 2250$
Total cost = 52050 Kshs.

$$\boxed{\text{Kshs 52,050}}$$