1. **Problem Statement:** Given matrices $$A=\begin{pmatrix}2 & -1 \\ 0 & -4 \\ -5 & 3\end{pmatrix}, B=\begin{pmatrix}5 & -2 & 6 \\ -1 & 4 & -2\end{pmatrix}, C=\begin{pmatrix}2 & 1 & -1 \\ 0 & -2 & 3 \\ -6 & 4 & 2\end{pmatrix}, D=\begin{pmatrix}1 & 3 \\ -2 & 5 \\ -4 & 7\end{pmatrix}$$ Find the following if possible: (a) $A+D$ (b) $2D - 3A$ (c) $AB$ (d) $BC$ (e) $(A+D)C$ (f) $(DB)C$ 2. **Matrix Addition and Scalar Multiplication Rules:** - Matrices must have the same dimensions to be added or subtracted. - Scalar multiplication multiplies each element by the scalar. - Matrix multiplication $XY$ is defined if the number of columns of $X$ equals the number of rows of $Y$. 3. **Step (a): Compute $A+D$** Both $A$ and $D$ are $3\times 2$ matrices. Add corresponding elements: $$A+D=\begin{pmatrix}2+1 & -1+3 \\ 0+(-2) & -4+5 \\ -5+(-4) & 3+7\end{pmatrix}=\begin{pmatrix}3 & 2 \\ -2 & 1 \\ -9 & 10\end{pmatrix}$$ 4. **Step (b): Compute $2D - 3A$** Calculate $2D$: $$2D=2\times \begin{pmatrix}1 & 3 \\ -2 & 5 \\ -4 & 7\end{pmatrix}=\begin{pmatrix}2 & 6 \\ -4 & 10 \\ -8 & 14\end{pmatrix}$$ Calculate $3A$: $$3A=3\times \begin{pmatrix}2 & -1 \\ 0 & -4 \\ -5 & 3\end{pmatrix}=\begin{pmatrix}6 & -3 \\ 0 & -12 \\ -15 & 9\end{pmatrix}$$ Subtract: $$2D - 3A=\begin{pmatrix}2-6 & 6-(-3) \\ -4-0 & 10-(-12) \\ -8-(-15) & 14-9\end{pmatrix}=\begin{pmatrix}-4 & 9 \\ -4 & 22 \\ 7 & 5\end{pmatrix}$$ 5. **Step (c): Compute $AB$** $A$ is $3\times 2$, $B$ is $2\times 3$, so multiplication is defined and result is $3\times 3$. Calculate each element: Row 1 of $A$ times columns of $B$: $$11 = 2\times 5 + (-1)\times (-1) = 10 + 1$$ $$-8 = 2\times (-2) + (-1)\times 4 = -4 - 4$$ $$14 = 2\times 6 + (-1)\times (-2) = 12 + 2$$ Row 2 of $A$ times columns of $B$: $$0 = 0\times 5 + (-4)\times (-1) = 0 + 4$$ $$-16 = 0\times (-2) + (-4)\times 4 = 0 - 16$$ $$8 = 0\times 6 + (-4)\times (-2) = 0 + 8$$ Row 3 of $A$ times columns of $B$: $$-23 = -5\times 5 + 3\times (-1) = -25 - 3$$ $$22 = -5\times (-2) + 3\times 4 = 10 + 12$$ $$-36 = -5\times 6 + 3\times (-2) = -30 - 6$$ So, $$AB=\begin{pmatrix}11 & -8 & 14 \\ 4 & -16 & 8 \\ -28 & 22 & -36\end{pmatrix}$$ 6. **Step (d): Compute $BC$** $B$ is $2\times 3$, $C$ is $3\times 3$, so multiplication is defined and result is $2\times 3$. Calculate each element: Row 1 of $B$ times columns of $C$: $$2 = 5\times 2 + (-2)\times 0 + 6\times (-6) = 10 + 0 - 36$$ $$-11 = 5\times 1 + (-2)\times (-2) + 6\times 4 = 5 + 4 + 24$$ $$-2 = 5\times (-1) + (-2)\times 3 + 6\times 2 = -5 - 6 + 12$$ Row 2 of $B$ times columns of $C$: $$-1 = (-1)\times 2 + 4\times 0 + (-2)\times (-6) = -2 + 0 + 12$$ $$18 = (-1)\times 1 + 4\times (-2) + (-2)\times 4 = -1 - 8 - 8$$ $$-7 = (-1)\times (-1) + 4\times 3 + (-2)\times 2 = 1 + 12 - 4$$ So, $$BC=\begin{pmatrix}-26 & 33 & -2 \\ 10 & -17 & 9\end{pmatrix}$$ 7. **Step (e): Compute $(A+D)C$** $A+D$ is $3\times 2$, $C$ is $3\times 3$, multiplication not defined (columns of $A+D$ is 2, rows of $C$ is 3). So, $(A+D)C$ is **not defined**. 8. **Step (f): Compute $(DB)C$** First compute $DB$: $D$ is $3\times 2$, $B$ is $2\times 3$, so $DB$ is $3\times 3$. Calculate $DB$ elements: Row 1 of $D$ times columns of $B$: $$5 = 1\times 5 + 3\times (-1) = 5 - 3$$ $$10 = 1\times (-2) + 3\times 4 = -2 + 12$$ $$0 = 1\times 6 + 3\times (-2) = 6 - 6$$ Row 2 of $D$ times columns of $B$: $$-16 = -2\times 5 + 5\times (-1) = -10 - 5$$ $$24 = -2\times (-2) + 5\times 4 = 4 + 20$$ $$-4 = -2\times 6 + 5\times (-2) = -12 - 10$$ Row 3 of $D$ times columns of $B$: $$-26 = -4\times 5 + 7\times (-1) = -20 - 7$$ $$44 = -4\times (-2) + 7\times 4 = 8 + 28$$ $$-2 = -4\times 6 + 7\times (-2) = -24 - 14$$ So, $$DB=\begin{pmatrix}2 & 10 & 0 \\ -15 & 24 & -22 \\ -27 & 36 & -38\end{pmatrix}$$ Now multiply $DB$ ($3\times 3$) by $C$ ($3\times 3$), result is $3\times 3$. Calculate each element of $(DB)C$: Row 1 times columns of $C$: $$2\times 2 + 10\times 0 + 0\times (-6) = 4 + 0 + 0 = 4$$ $$2\times 1 + 10\times (-2) + 0\times 4 = 2 - 20 + 0 = -18$$ $$2\times (-1) + 10\times 3 + 0\times 2 = -2 + 30 + 0 = 28$$ Row 2 times columns of $C$: $$-15\times 2 + 24\times 0 + (-22)\times (-6) = -30 + 0 + 132 = 102$$ $$-15\times 1 + 24\times (-2) + (-22)\times 4 = -15 - 48 - 88 = -151$$ $$-15\times (-1) + 24\times 3 + (-22)\times 2 = 15 + 72 - 44 = 43$$ Row 3 times columns of $C$: $$-27\times 2 + 36\times 0 + (-38)\times (-6) = -54 + 0 + 228 = 174$$ $$-27\times 1 + 36\times (-2) + (-38)\times 4 = -27 - 72 - 152 = -251$$ $$-27\times (-1) + 36\times 3 + (-38)\times 2 = 27 + 108 - 76 = 59$$ So, $$(DB)C=\begin{pmatrix}4 & -18 & 28 \\ 102 & -151 & 43 \\ 174 & -251 & 59\end{pmatrix}$$ **Final answers:** (a) $A+D=\begin{pmatrix}3 & 2 \\ -2 & 1 \\ -9 & 10\end{pmatrix}$ (b) $2D - 3A=\begin{pmatrix}-4 & 9 \\ -4 & 22 \\ 7 & 5\end{pmatrix}$ (c) $AB=\begin{pmatrix}11 & -8 & 14 \\ 4 & -16 & 8 \\ -28 & 22 & -36\end{pmatrix}$ (d) $BC=\begin{pmatrix}-26 & 33 & -2 \\ 10 & -17 & 9\end{pmatrix}$ (e) $(A+D)C$ is **not defined** due to incompatible dimensions. (f) $(DB)C=\begin{pmatrix}4 & -18 & 28 \\ 102 & -151 & 43 \\ 174 & -251 & 59\end{pmatrix}$