1. **Stating the problem:** We have two matrices: $$A = \begin{bmatrix} 3 & 5 & 12 \\ 10 & 7 & 8 \\ 4 & 3 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 & 5 & -3 \\ 7 & 4 & 5 & 5 \\ 8 & 2 & 9 & 7 \end{bmatrix}$$ We are asked to: a) Compute $A + B$ b) Compute the inverses of $A$ and $B$ c) Compute the transposes of $A$ and $B$ d) Compute the product $A \times B$ 2. **Matrix addition rule:** Matrices can be added only if they have the same dimensions. Here, $A$ is $3 \times 3$ and $B$ is $3 \times 4$, so addition is **not defined**. 3. **Matrix inverse rule:** Only square matrices can have inverses. $A$ is $3 \times 3$ (square), so $A^{-1}$ may exist. $B$ is $3 \times 4$ (not square), so $B^{-1}$ does not exist. 4. **Transpose rule:** Transpose of a matrix flips rows and columns. If $A$ is $m \times n$, then $A^T$ is $n \times m$. 5. **Matrix multiplication rule:** Product $AB$ is defined if number of columns of $A$ equals number of rows of $B$. $A$ is $3 \times 3$, $B$ is $3 \times 4$, so $AB$ is defined and will be $3 \times 4$. --- **a) Addition $A + B$:** Not possible because dimensions differ. **b) Inverse of $A$:** Calculate determinant of $A$: $$\det(A) = 3(7 \times 5 - 8 \times 3) - 5(10 \times 5 - 8 \times 4) + 12(10 \times 3 - 7 \times 4)$$ $$= 3(35 - 24) - 5(50 - 32) + 12(30 - 28) = 3(11) - 5(18) + 12(2) = 33 - 90 + 24 = -33$$ Since $\det(A) \neq 0$, $A$ is invertible. Calculate $A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$ where $\mathrm{adj}(A)$ is adjugate matrix. Compute cofactors: $$C_{11} = +(7 \times 5 - 8 \times 3) = 35 - 24 = 11$$ $$C_{12} = -(10 \times 5 - 8 \times 4) = -(50 - 32) = -18$$ $$C_{13} = +(10 \times 3 - 7 \times 4) = 30 - 28 = 2$$ $$C_{21} = -(5 \times 5 - 12 \times 3) = -(25 - 36) = 11$$ $$C_{22} = +(3 \times 5 - 12 \times 4) = 15 - 48 = -33$$ $$C_{23} = -(3 \times 3 - 5 \times 4) = -(9 - 20) = 11$$ $$C_{31} = +(5 \times 8 - 7 \times 12) = 40 - 84 = -44$$ $$C_{32} = -(3 \times 8 - 10 \times 12) = -(24 - 120) = 96$$ $$C_{33} = +(3 \times 7 - 10 \times 5) = 21 - 50 = -29$$ Adjugate is transpose of cofactor matrix: $$\mathrm{adj}(A) = \begin{bmatrix} 11 & 11 & -44 \\ -18 & -33 & 96 \\ 2 & 11 & -29 \end{bmatrix}^T = \begin{bmatrix} 11 & -18 & 2 \\ 11 & -33 & 11 \\ -44 & 96 & -29 \end{bmatrix}$$ Therefore, $$A^{-1} = \frac{1}{-33} \begin{bmatrix} 11 & -18 & 2 \\ 11 & -33 & 11 \\ -44 & 96 & -29 \end{bmatrix} = \begin{bmatrix} -\frac{11}{33} & \frac{18}{33} & -\frac{2}{33} \\ -\frac{11}{33} & 1 & -\frac{11}{33} \\ \frac{44}{33} & -\frac{96}{33} & \frac{29}{33} \end{bmatrix}$$ Simplify fractions: $$A^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{6}{11} & -\frac{2}{33} \\ -\frac{1}{3} & 1 & -\frac{1}{3} \\ \frac{4}{3} & -\frac{32}{11} & \frac{29}{33} \end{bmatrix}$$ **Inverse of $B$:** Not defined (non-square). **c) Transpose of $A$ and $B$:** $$A^T = \begin{bmatrix} 3 & 10 & 4 \\ 5 & 7 & 3 \\ 12 & 8 & 5 \end{bmatrix}$$ $$B^T = \begin{bmatrix} 1 & 7 & 8 \\ 2 & 4 & 2 \\ 5 & 5 & 9 \\ -3 & 5 & 7 \end{bmatrix}$$ **d) Multiplication $AB$:** Calculate each element of $AB$ by dot product of rows of $A$ with columns of $B$. First row: $$ (3)(1)+(5)(7)+(12)(8) = 3 + 35 + 96 = 134 $$ $$ (3)(2)+(5)(4)+(12)(2) = 6 + 20 + 24 = 50 $$ $$ (3)(5)+(5)(5)+(12)(9) = 15 + 25 + 108 = 148 $$ $$ (3)(-3)+(5)(5)+(12)(7) = -9 + 25 + 84 = 100 $$ Second row: $$ (10)(1)+(7)(7)+(8)(8) = 10 + 49 + 64 = 123 $$ $$ (10)(2)+(7)(4)+(8)(2) = 20 + 28 + 16 = 64 $$ $$ (10)(5)+(7)(5)+(8)(9) = 50 + 35 + 72 = 157 $$ $$ (10)(-3)+(7)(5)+(8)(7) = -30 + 35 + 56 = 61 $$ Third row: $$ (4)(1)+(3)(7)+(5)(8) = 4 + 21 + 40 = 65 $$ $$ (4)(2)+(3)(4)+(5)(2) = 8 + 12 + 10 = 30 $$ $$ (4)(5)+(3)(5)+(5)(9) = 20 + 15 + 45 = 80 $$ $$ (4)(-3)+(3)(5)+(5)(7) = -12 + 15 + 35 = 38 $$ So, $$AB = \begin{bmatrix} 134 & 50 & 148 & 100 \\ 123 & 64 & 157 & 61 \\ 65 & 30 & 80 & 38 \end{bmatrix}$$