Three computer keyboards and 2 chargers cost Kshs. 5,100. Four keyboards and 3 chargers cost Kshs. 7,050. Find the cost of one keyboard and one charger.

Let $k$ be the cost of one keyboard and $c$ be the cost of one charger.
From the first statement: "Three computer key boards and 2 charger cost Kshs. 5,100"

$$3k + 2c = 5100 \quad (1)$$

From the second statement: "4 similar keyboards and 3 similar chargers cost Kshs. 7,050"

$$4k + 3c = 7050 \quad (2)$$

$$A = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} k \\ c \end{bmatrix}, \quad B = \begin{bmatrix} 5100 \\ 7050 \end{bmatrix}$$

So, the matrix equation is:

$$\begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} k \\ c \end{bmatrix} = \begin{bmatrix} 5100 \\ 7050 \end{bmatrix}$$

The determinant of $A$ is $\det(A) = (3)(3) - (2)(4) = 9 - 8 = 1$.
The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix}$$

$$X = A^{-1}B$$ $$\begin{bmatrix} k \\ c \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix} \begin{bmatrix} 5100 \\ 7050 \end{bmatrix}$$

Perform the matrix multiplication:

$$k = (3)(5100) + (-2)(7050)$$ $$k = 15300 - 14100$$ $$k = 1200$$ $$c = (-4)(5100) + (3)(7050)$$ $$c = -20400 + 21150$$ $$c = 750$$

The cost of each keyboard is Kshs. 1200 and the cost of each charger is Kshs. 750.

$$\boxed{\text{Cost of keyboard} = \text{Kshs. } 1200, \quad \text{Cost of charger} = \text{Kshs. } 750}$$