1. **Problem statement:** Given matrices $$A=\begin{pmatrix}1 & -2 & 3 \\ 4 & 0 & -2\end{pmatrix}, B=\begin{pmatrix}3 & 0 & 2 \\ 4 & 1 & 5\end{pmatrix}, C=\begin{pmatrix}1 & 3 \\ 2 & -1\end{pmatrix}, D=\begin{pmatrix}1 \\ -1\end{pmatrix}, E=\begin{pmatrix}2 & 1 & 1 \\ 0 & -1 & 2 \\ 1 & 2 & 1\end{pmatrix}, F=\begin{pmatrix}1 & 1 \\ 3 & 2 \\ -1 & 0\end{pmatrix}$$ Find: 1) $AB^T - 3I$ 2) $|2E^3| + 2 \operatorname{trace}(C)$ 3) $E^{-1}$ 4) $(F^T D)^T$ --- 2. **Step 1: Calculate $AB^T - 3I$** - Compute $B^T$ (transpose of $B$): $$B^T = \begin{pmatrix}3 & 4 \\ 0 & 1 \\ 2 & 5\end{pmatrix}$$ - Multiply $A$ (2x3) by $B^T$ (3x2): $$AB^T = \begin{pmatrix}1 & -2 & 3 \\ 4 & 0 & -2\end{pmatrix} \begin{pmatrix}3 & 4 \\ 0 & 1 \\ 2 & 5\end{pmatrix} = \begin{pmatrix} (1)(3)+(-2)(0)+(3)(2) & (1)(4)+(-2)(1)+(3)(5) \\ (4)(3)+(0)(0)+(-2)(2) & (4)(4)+(0)(1)+(-2)(5) \end{pmatrix}$$ $$= \begin{pmatrix}3 + 0 + 6 & 4 - 2 + 15 \\ 12 + 0 - 4 & 16 + 0 - 10 \end{pmatrix} = \begin{pmatrix}9 & 17 \\ 8 & 6\end{pmatrix}$$ - $I$ is the $2 \times 2$ identity matrix: $$I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$ - Calculate $AB^T - 3I$: $$\begin{pmatrix}9 & 17 \\ 8 & 6\end{pmatrix} - 3 \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}9 - 3 & 17 - 0 \\ 8 - 0 & 6 - 3\end{pmatrix} = \begin{pmatrix}6 & 17 \\ 8 & 3\end{pmatrix}$$ --- 3. **Step 2: Calculate $|2E^3| + 2 \operatorname{trace}(C)$** - First find $|E|$ (determinant of $E$): $$E = \begin{pmatrix}2 & 1 & 1 \\ 0 & -1 & 2 \\ 1 & 2 & 1\end{pmatrix}$$ Calculate determinant by expansion: $$|E| = 2 \begin{vmatrix}-1 & 2 \\ 2 & 1\end{vmatrix} - 1 \begin{vmatrix}0 & 2 \\ 1 & 1\end{vmatrix} + 1 \begin{vmatrix}0 & -1 \\ 1 & 2\end{vmatrix}$$ $$= 2((-1)(1) - (2)(2)) - 1(0 \cdot 1 - 2 \cdot 1) + 1(0 \cdot 2 - (-1) \cdot 1)$$ $$= 2(-1 - 4) - 1(0 - 2) + 1(0 + 1) = 2(-5) + 2 + 1 = -10 + 2 + 1 = -7$$ - Since $|E^3| = |E|^3 = (-7)^3 = -343$ - Then $|2E^3| = 2^3 |E^3| = 8 \times (-343) = -2744$ - Calculate $\operatorname{trace}(C)$ (sum of diagonal elements of $C$): $$\operatorname{trace}(C) = 1 + (-1) = 0$$ - So, $$|2E^3| + 2 \operatorname{trace}(C) = -2744 + 2 \times 0 = -2744$$ --- 4. **Step 3: Calculate $E^{-1}$** - Find inverse of $E$ using formula $E^{-1} = \frac{1}{|E|} \operatorname{adj}(E)$ - We already have $|E| = -7$ - Calculate cofactors matrix and then adjugate (transpose of cofactor matrix): Cofactors: $$C_{11} = \begin{vmatrix}-1 & 2 \\ 2 & 1\end{vmatrix} = (-1)(1) - (2)(2) = -1 - 4 = -5$$ $$C_{12} = - \begin{vmatrix}0 & 2 \\ 1 & 1\end{vmatrix} = - (0 \cdot 1 - 2 \cdot 1) = - (0 - 2) = 2$$ $$C_{13} = \begin{vmatrix}0 & -1 \\ 1 & 2\end{vmatrix} = 0 \cdot 2 - (-1) \cdot 1 = 1$$ $$C_{21} = - \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - (1 \cdot 1 - 1 \cdot 2) = - (1 - 2) = 1$$ $$C_{22} = \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1$$ $$C_{23} = - \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = - (2 \cdot 2 - 1 \cdot 1) = - (4 - 1) = -3$$ $$C_{31} = \begin{vmatrix}1 & 1 \\ -1 & 2\end{vmatrix} = 1 \cdot 2 - 1 \cdot (-1) = 2 + 1 = 3$$ $$C_{32} = - \begin{vmatrix}2 & 1 \\ 0 & 2\end{vmatrix} = - (2 \cdot 2 - 1 \cdot 0) = -4$$ $$C_{33} = \begin{vmatrix}2 & 1 \\ 0 & -1\end{vmatrix} = 2 \cdot (-1) - 1 \cdot 0 = -2$$ Cofactor matrix: $$\begin{pmatrix}-5 & 2 & 1 \\ 1 & 1 & -3 \\ 3 & -4 & -2\end{pmatrix}$$ Adjugate (transpose): $$\operatorname{adj}(E) = \begin{pmatrix}-5 & 1 & 3 \\ 2 & 1 & -4 \\ 1 & -3 & -2\end{pmatrix}$$ Inverse: $$E^{-1} = \frac{1}{-7} \begin{pmatrix}-5 & 1 & 3 \\ 2 & 1 & -4 \\ 1 & -3 & -2\end{pmatrix} = \begin{pmatrix}5/7 & -1/7 & -3/7 \\ -2/7 & -1/7 & 4/7 \\ -1/7 & 3/7 & 2/7\end{pmatrix}$$ --- 5. **Step 4: Calculate $(F^T D)^T$** - Compute $F^T$ (transpose of $F$): $$F^T = \begin{pmatrix}1 & 3 & -1 \\ 1 & 2 & 0\end{pmatrix}$$ - Multiply $F^T$ (2x3) by $D$ (3x1): $$D = \begin{pmatrix}1 \\ -1\end{pmatrix}$$ Note: $D$ is $2 \times 1$ matrix, but given as $\begin{pmatrix}1 \\ -1\end{pmatrix}$ which is $2 \times 1$, so dimensions match for $F^T D$: $$F^T D = \begin{pmatrix}1 & 3 & -1 \\ 1 & 2 & 0\end{pmatrix} \begin{pmatrix}1 \\ -1\end{pmatrix}$$ This is invalid because $F^T$ is $2 \times 3$ and $D$ is $2 \times 1$, so multiplication is not defined. Check original $D$ matrix: User gave $D = \begin{pmatrix}1 \\ -1\end{pmatrix}$ which is $2 \times 1$. $F^T$ is $2 \times 3$, so $F^T D$ is not defined. Check if $D$ should be $3 \times 1$ to multiply with $F^T$: No, $D$ is $2 \times 1$, $F^T$ is $2 \times 3$, so multiplication $F^T D$ is undefined. Check if user meant $F^T$ times $D$ transpose or $F^T$ times $D$ with different dimensions. Alternatively, multiply $F^T$ (2x3) by $D$ (2x1) is invalid. Try $F^T$ (2x3) times $D$ transpose (1x2): $D^T = (1, -1)$ (1x2) $F^T D^T$ is $2x3$ times $1x2$ invalid. Try $F^T$ (2x3) times $D$ (3x1) if $D$ is $3x1$: User gave $D$ as $2x1$, so no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (3x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. Try $F^T$ (2x3) times $D$ (2x1) no. **Conclusion:** Multiplication $F^T D$ is not defined due to incompatible dimensions. Hence, $(F^T D)^T$ is not defined. --- **Final answers:** 1) $AB^T - 3I = \begin{pmatrix}6 & 17 \\ 8 & 3\end{pmatrix}$ 2) $|2E^3| + 2 \operatorname{trace}(C) = -2744$ 3) $$E^{-1} = \begin{pmatrix}5/7 & -1/7 & -3/7 \\ -2/7 & -1/7 & 4/7 \\ -1/7 & 3/7 & 2/7\end{pmatrix}$$ 4) $(F^T D)^T$ is not defined due to incompatible matrix dimensions.