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Let $x$ be the number of chairs produced, $y$ be the number of tables produced, and $z$ be the number of beds produced.
We are given the following information:
So, we have the following system of linear equations:
$$x + y + z = 45$$$$-x + z = 8$$$$x - 2y + z = 0$$We can write the system of equations in matrix form as $AX = B$, where:
$$A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 45 \\ 8 \\ 0 \end{bmatrix}$$$$|A| = \begin{vmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{vmatrix} = 1(0 - (-2)) - 1(-1 - 1) + 1(2 - 0) = 2 + 2 + 2 = 6$$Since $|A| \neq 0$, the system has a unique solution.
The matrix of cofactors is:
$$C = \begin{bmatrix} 2 & 2 & 2 \\ -3 & 0 & 3 \\ 1 & -2 & 1 \end{bmatrix}$$The adjoint of $A$ is the transpose of the cofactor matrix:$$adj(A) = C^T = \begin{bmatrix} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{bmatrix}$$The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} adj(A)$:
$$A^{-1} = \frac{1}{6} \begin{bmatrix} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{bmatrix}$$We have $X = A^{-1}B$:
$$X = \frac{1}{6} \begin{bmatrix} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} 45 \\ 8 \\ 0 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 2(45) - 3(8) + 1(0) \\ 2(45) + 0(8) - 2(0) \\ 2(45) + 3(8) + 1(0) \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 90 - 24 \\ 90 \\ 90 + 24 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 66 \\ 90 \\ 114 \end{bmatrix} = \begin{bmatrix} 11 \\ 15 \\ 19 \end{bmatrix}$$Therefore, $x = 11$, $y = 15$, and $z = 19$.The number of chairs produced is 11, the number of tables produced is 15, and the number of beds produced is 19.
Final Answer: Chairs: 11, Tables: 15, Beds: 19