1. **State the problem:** A company produces three products daily with total production 90 kg. The first product's production exceeds the second by 20 kg. The sum of the first and second products equals twice the third product's production. Find the production levels of each product using matrix algebra. 2. **Define variables:** Let $x$, $y$, and $z$ be the production (in kg) of the first, second, and third products respectively. 3. **Write equations from the problem:** - Total production: $$x + y + z = 90$$ - First exceeds second by 20: $$x = y + 20$$ - Sum of first and second is twice third: $$x + y = 2z$$ 4. **Rewrite equations in standard form:** From $x = y + 20$, we get $$x - y = 20$$ From $x + y = 2z$, rewrite as $$x + y - 2z = 0$$ Total production is $$x + y + z = 90$$ 5. **Matrix form:** $$\begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -2\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}90 \\ 20 \\ 0\end{bmatrix}$$ 6. **Solve the system:** From second equation: $$x = y + 20$$ Substitute into first and third: - First: $$(y + 20) + y + z = 90 \Rightarrow 2y + z = 70$$ - Third: $$(y + 20) + y - 2z = 0 \Rightarrow 2y - 2z = -20 \Rightarrow y - z = -10$$ 7. **From $y - z = -10$, express $y = z - 10$** Substitute into $2y + z = 70$: $$2(z - 10) + z = 70 \Rightarrow 2z - 20 + z = 70 \Rightarrow 3z = 90 \Rightarrow z = 30$$ 8. **Find $y$ and $x$:** $$y = 30 - 10 = 20$$ $$x = y + 20 = 20 + 20 = 40$$ 9. **Final answer:** The production levels are: - First product: $40$ kg - Second product: $20$ kg - Third product: $30$ kg These satisfy all given conditions.