1. **Problem Statement:** An amount of 5000 is invested in three parts at 6%, 7%, and 8% per annum respectively. The total annual income is 358. The income from the first two investments is 70 more than the income from the third. We need to find the amount invested in each using the matrix method. 2. **Define variables:** Let the amounts invested at 6%, 7%, and 8% be $x$, $y$, and $z$ respectively. 3. **Form equations:** - Total investment: $$x + y + z = 5000$$ - Total income: $$0.06x + 0.07y + 0.08z = 358$$ - Income from first two is 70 more than third: $$0.06x + 0.07y = 0.08z + 70$$ 4. **Rewrite third equation:** $$0.06x + 0.07y - 0.08z = 70$$ 5. **Matrix form:** $$\begin{bmatrix}1 & 1 & 1 \\ 0.06 & 0.07 & 0.08 \\ 0.06 & 0.07 & -0.08\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}5000 \\ 358 \\ 70\end{bmatrix}$$ 6. **Solve the system:** Subtract third equation from second: $$ (0.06x + 0.07y + 0.08z) - (0.06x + 0.07y - 0.08z) = 358 - 70 $$ $$ 0.16z = 288 $$ $$ z = \frac{288}{0.16} = 1800 $$ 7. **Substitute $z=1800$ into first equation:** $$ x + y + 1800 = 5000 $$ $$ x + y = 3200 $$ 8. **Substitute $z=1800$ into second equation:** $$ 0.06x + 0.07y + 0.08 \times 1800 = 358 $$ $$ 0.06x + 0.07y + 144 = 358 $$ $$ 0.06x + 0.07y = 214 $$ 9. **Solve for $x$ and $y$:** From step 7: $$ y = 3200 - x $$ Substitute into step 8: $$ 0.06x + 0.07(3200 - x) = 214 $$ $$ 0.06x + 224 - 0.07x = 214 $$ $$ -0.01x = -10 $$ $$ x = 1000 $$ 10. **Find $y$:** $$ y = 3200 - 1000 = 2200 $$ **Final answer:** - Amount invested at 6%: 1000 - Amount invested at 7%: 2200 - Amount invested at 8%: 1800