Given matrix A:

a) Find the value of the constant $k$ for which the matrix A is singular.
A matrix is singular if its determinant is zero.
Given matrix A:

$$A = \begin{pmatrix} 2 & 3 & 4 \\ 1 & -1 & 2 \\ 5 & 0 & k \end{pmatrix}$$

Step 1: Calculate the determinant of A.

$$\det(A) = 2\begin{vmatrix} -1 & 2 \\ 0 & k \end{vmatrix} - 3\begin{vmatrix} 1 & 2 \\ 5 & k \end{vmatrix} + 4\begin{vmatrix} 1 & -1 \\ 5 & 0 \end{vmatrix}$$ $$\det(A) = 2((-1)(k) - (2)(0)) - 3((1)(k) - (2)(5)) + 4((1)(0) - (-1)(5))$$ $$\det(A) = 2(-k - 0) - 3(k - 10) + 4(0 - (-5))$$ $$\det(A) = -2k - 3k + 30 + 20$$ $$\det(A) = -5k + 50$$

Step 2: Set the determinant to zero and solve for $k$.

$$-5k + 50 = 0$$ $$-5k = -50$$ $$k = \frac{-50}{-5}$$ $$k = 10$$ $$\boxed{k=10}$$

b) Find the inverse of matrix B, if $B = \begin{pmatrix} 2 & 3 & 4 \\ 1 & -1 & 2 \\ 5 & 0 & 2 \end{pmatrix}$.
Hence solve the system of equations:
$2x + 3y + 4z = 1$
$x - y + 2z = 2$
$5x + 2z = 5$

Step 1: Calculate the determinant of B.

$$\det(B) = 2\begin{vmatrix} -1 & 2 \\ 0 & 2 \end{vmatrix} - 3\begin{vmatrix} 1 & 2 \\ 5 & 2 \end{vmatrix} + 4\begin{vmatrix} 1 & -1 \\ 5 & 0 \end{vmatrix}$$ $$\det(B) = 2((-1)(2) - (2)(0)) - 3((1)(2) - (2)(5)) + 4((1)(0) - (-1)(5))$$ $$\det(B) = 2(-2 - 0) - 3(2 - 10) + 4(0 - (-5))$$ $$\det(B) = 2(-2) - 3(-8) + 4(5)$$ $$\det(B) = -4 + 24 + 20$$ $$\det(B) = 40$$

Step 2: Calculate the matrix of cofactors, $C$.

$$C_{11} = \begin{vmatrix} -1 & 2 \\ 0 & 2 \end{vmatrix} = -2$$ $$C_{12} = -\begin{vmatrix} 1 & 2 \\ 5 & 2 \end{vmatrix} = -(2 - 10) = 8$$ $$C_{13} = \begin{vmatrix} 1 & -1 \\ 5 & 0 \end{vmatrix} = 0 - (-5) = 5$$ $$C_{21} = -\begin{vmatrix} 3 & 4 \\ 0 & 2 \end{vmatrix} = -(6 - 0) = -6$$ $$C_{22} = \begin{vmatrix} 2 & 4 \\ 5 & 2 \end{vmatrix} = 4 - 20 = -16$$ $$C_{23} = -\begin{vmatrix} 2 & 3 \\ 5 & 0 \end{vmatrix} = -(0 - 15) = 15$$ $$$$