1. **Problem 1: Show that** $A + (B + C) = (A + B) + C$ for given matrices. 2. **Given:** $$A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix},\quad B = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 3 \\ 1 & -2 \end{bmatrix}$$ 3. **Matrix addition rule:** Add corresponding elements. 4. Calculate $B + C$: $$B + C = \begin{bmatrix} 1+1 & 2+3 \\ -2+1 & 3+(-2) \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ -1 & 1 \end{bmatrix}$$ 5. Calculate $A + (B + C)$: $$A + (B + C) = \begin{bmatrix} 2+2 & 1+5 \\ 4+(-1) & 3+1 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 3 & 4 \end{bmatrix}$$ 6. Calculate $A + B$: $$A + B = \begin{bmatrix} 2+1 & 1+2 \\ 4+(-2) & 3+3 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 2 & 6 \end{bmatrix}$$ 7. Calculate $(A + B) + C$: $$ (A + B) + C = \begin{bmatrix} 3+1 & 3+3 \\ 2+1 & 6+(-2) \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 3 & 4 \end{bmatrix}$$ 8. Since $A + (B + C) = (A + B) + C = \begin{bmatrix} 4 & 6 \\ 3 & 4 \end{bmatrix}$, the equality holds. --- 1. **Problem 2: Show that** $A(BC) = (AB)C$ for given matrices. 2. **Given:** $$A = \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix},\quad B = \begin{bmatrix} 1 & 1 \\ 3 & -2 \end{bmatrix},\quad C = \begin{bmatrix} 5 & 0 \\ -1 & 2 \end{bmatrix}$$ 3. **Matrix multiplication rule:** For matrices $X$ and $Y$, element $(i,j)$ of $XY$ is sum of products of row $i$ of $X$ and column $j$ of $Y$. 4. Calculate $BC$: $$BC = \begin{bmatrix} 1\cdot5 + 1\cdot(-1) & 1\cdot0 + 1\cdot2 \\ 3\cdot5 + (-2)\cdot(-1) & 3\cdot0 + (-2)\cdot2 \end{bmatrix} = \begin{bmatrix} 5 -1 & 0 + 2 \\ 15 + 2 & 0 -4 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 17 & -4 \end{bmatrix}$$ 5. Calculate $A(BC)$: $$A(BC) = \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ 17 & -4 \end{bmatrix} = \begin{bmatrix} 1\cdot4 + (-3)\cdot17 & 1\cdot2 + (-3)\cdot(-4) \\ 2\cdot4 + 4\cdot17 & 2\cdot2 + 4\cdot(-4) \end{bmatrix} = \begin{bmatrix} 4 - 51 & 2 + 12 \\ 8 + 68 & 4 - 16 \end{bmatrix} = \begin{bmatrix} -47 & 14 \\ 76 & -12 \end{bmatrix}$$ 6. Calculate $AB$: $$AB = \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + (-3)\cdot3 & 1\cdot1 + (-3)\cdot(-2) \\ 2\cdot1 + 4\cdot3 & 2\cdot1 + 4\cdot(-2) \end{bmatrix} = \begin{bmatrix} 1 - 9 & 1 + 6 \\ 2 + 12 & 2 - 8 \end{bmatrix} = \begin{bmatrix} -8 & 7 \\ 14 & -6 \end{bmatrix}$$ 7. Calculate $(AB)C$: $$ (AB)C = \begin{bmatrix} -8 & 7 \\ 14 & -6 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (-8)\cdot5 + 7\cdot(-1) & (-8)\cdot0 + 7\cdot2 \\ 14\cdot5 + (-6)\cdot(-1) & 14\cdot0 + (-6)\cdot2 \end{bmatrix} = \begin{bmatrix} -40 -7 & 0 + 14 \\ 70 + 6 & 0 - 12 \end{bmatrix} = \begin{bmatrix} -47 & 14 \\ 76 & -12 \end{bmatrix}$$ 8. Since $A(BC) = (AB)C = \begin{bmatrix} -47 & 14 \\ 76 & -12 \end{bmatrix}$, the equality holds. **Final answers:** - $A + (B + C) = (A + B) + C = \begin{bmatrix} 4 & 6 \\ 3 & 4 \end{bmatrix}$ - $A(BC) = (AB)C = \begin{bmatrix} -47 & 14 \\ 76 & -12 \end{bmatrix}$