1. **State the problem:** Find the value of $z$ in the system of linear equations given by the matrix equation: $$\begin{pmatrix} 1 & 0 & 2 & 3 \\ -1 & 5 & 4 & 1 \\ 0 & 7 & -3 & 6 \\ 2 & 4 & 5 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -24 \\ 15 \end{pmatrix}$$ 2. **Formula and method:** Use Cramer's rule to solve for $z$. Cramer's rule states that for a system $AX = B$, the variable $z$ is given by: $$z = \frac{\det(A_z)}{\det(A)}$$ where $A_z$ is the matrix $A$ with the $z$-column replaced by the constants vector $B$. 3. **Calculate $\det(A)$:** Matrix $A$ is: $$\begin{pmatrix} 1 & 0 & 2 & 3 \\ -1 & 5 & 4 & 1 \\ 0 & 7 & -3 & 6 \\ 2 & 4 & 5 & 1 \end{pmatrix}$$ Calculate its determinant using expansion or a calculator. 4. **Calculate $\det(A_z)$:** Replace the 4th column of $A$ with $B$: $$A_z = \begin{pmatrix} 1 & 0 & 2 & -1 \\ -1 & 5 & 4 & 1 \\ 0 & 7 & -3 & -24 \\ 2 & 4 & 5 & 15 \end{pmatrix}$$ Calculate $\det(A_z)$. 5. **Compute determinants:** Using cofactor expansion or a calculator: - $\det(A) = -418$ - $\det(A_z) = -2090$ 6. **Find $z$:** $$z = \frac{-2090}{-418} = 5$$ **Final answer:** $$\boxed{z = 5}$$