1. Problem 16: Given matrices $$A = \begin{pmatrix}2 & -2 \\ -1 & 3\end{pmatrix}, \quad AB = \begin{pmatrix}4 & -2 \\ 0 & 7\end{pmatrix}$$ Find matrix $B$. Step 1: We know $AB = \text{given}$ and $A$ is invertible, so multiply both sides by $A^{-1}$: $$B = A^{-1}AB = A^{-1}(AB) = B$$ Step 2: Find $A^{-1}$. The determinant of $A$ is $$|A| = (2)(3) - (-2)(-1) = 6 - 2 = 4$$ Step 3: The inverse matrix $A^{-1} = \frac{1}{|A|} \begin{pmatrix}3 & 2 \\ 1 & 2\end{pmatrix}$, switching and negating elements accordingly: $$A^{-1} = \frac{1}{4} \begin{pmatrix}3 & 2 \\ 1 & 2\end{pmatrix}$$ Step 4: Calculate $B = A^{-1} AB = A^{-1} (AB)$: $$B = \frac{1}{4} \begin{pmatrix}3 & 2 \\ 1 & 2\end{pmatrix} \begin{pmatrix}4 & -2 \\ 0 & 7\end{pmatrix} = \frac{1}{4} \begin{pmatrix}3 \times 4 + 2 \times 0 & 3 \times (-2) + 2 \times 7 \\ 1 \times 4 + 2 \times 0 & 1 \times (-2) + 2 \times 7\end{pmatrix} = \frac{1}{4} \begin{pmatrix}12 & 8 \\ 4 & 12\end{pmatrix} = \begin{pmatrix}3 & 2 \\ 1 & 3\end{pmatrix}$$ 2. Problem 17: Given $$P = \begin{pmatrix}1 & 2 \\ 1 & 3\end{pmatrix}, R = \begin{pmatrix}1 & 1 & 0 \\ 2 & -1 & 1\end{pmatrix}, S = \begin{pmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{pmatrix}$$ Find matrix $X$ such that $$|P|X^T - 2P^{-1} + RS + 2\operatorname{diag}(1, -2) - 5I = 0$$ Step 1: Calculate determinant of $P$: $$|P| = 1 \times 3 - 2 \times 1 = 3 - 2 = 1$$ Step 2: Find $P^{-1}$: $$P^{-1} = \frac{1}{1} \begin{pmatrix}3 & -2 \\ -1 & 1\end{pmatrix} = \begin{pmatrix}3 & -2 \\ -1 & 1\end{pmatrix}$$ Step 3: Calculate $RS$: Multiplying $2 \times 3$ matrix $R$ with $3 \times 2$ matrix $S$ results in $2 \times 2$: $$RS = \begin{pmatrix}1 & 1 & 0 \\ 2 & -1 & 1\end{pmatrix} \begin{pmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1\times1+1\times1+0\times0 & 1\times0+1\times1+0\times1 \\ 2\times1 + (-1)\times1 + 1\times0 & 2\times0 + (-1)\times1 + 1\times1 \end{pmatrix} = \begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}$$ Step 4: Define $D = 2\operatorname{diag}(1, -2) = \begin{pmatrix}2 & 0 \\ 0 & -4\end{pmatrix}$ and $I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ scalar multiplied by 5: $$5I = \begin{pmatrix}5 & 0 \\ 0 & 5\end{pmatrix}$$ Step 5: Solve for $X^T$: $$1 \times X^T - 2P^{-1} + RS + D - 5I = 0$$ $$X^T = 2P^{-1} - RS - D + 5I$$ Calculate right side: $$2P^{-1} = 2 \times \begin{pmatrix}3 & -2 \\ -1 & 1\end{pmatrix} = \begin{pmatrix}6 & -4 \\ -2 & 2\end{pmatrix}$$ $$X^T = \begin{pmatrix}6 & -4 \\ -2 & 2\end{pmatrix} - \begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix} - \begin{pmatrix}2 & 0 \\ 0 & -4\end{pmatrix} + \begin{pmatrix}5 & 0 \\ 0 & 5\end{pmatrix} = \begin{pmatrix}6-2-2+5 & -4-1-0+0 \\ -2-1-0+0 & 2-0+4+5\end{pmatrix} = \begin{pmatrix}7 & -5 \\ -3 & 11\end{pmatrix}$$ Step 6: Transpose both sides to find $X$: $$X = (X^T)^T = \begin{pmatrix}7 & -3 \\ -5 & 11\end{pmatrix}$$ 3. Problem 18: Given $$A = \begin{pmatrix}x & 1 \\ y & 2\end{pmatrix}$$ Find $x + y$ if $$A = 3A^{-1}$$ Step 1: Multiply both sides by $A$: $$A^2 = 3I$$ Step 2: Calculate $A^2$: $$A^2 = \begin{pmatrix}x & 1 \\ y & 2\end{pmatrix} \begin{pmatrix}x & 1 \\ y & 2\end{pmatrix} = \begin{pmatrix}x^2 + y & x + 2 \\ yx + 2y & y + 4\end{pmatrix}$$ Step 3: Equate to $3I = \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}$: $$\begin{cases} x^2 + y = 3 \\ x + 2 = 0 \\ yx + 2y = 0 \\ y + 4 = 3 \end{cases}$$ Step 4: From second equation: $$x = -2$$ Step 5: From fourth equation: $$y = -1$$ Step 6: Verify first equation: $$(-2)^2 + (-1) = 4 -1 = 3$$ valid Step 7: Find $x + y$: $$x + y = -2 + (-1) = -3$$ 4. Problem 19: Given $$A=\begin{pmatrix}2 & 1 \\ 3 & 2\end{pmatrix}, B=\begin{pmatrix}1 & -1 \\ 0 & 2\end{pmatrix}, C=\begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix}$$ Find $X$ such that $$|A|X - \operatorname{tr}(B) A^{-1} = C^T B$$ Step 1: Calculate $|A|$: $$|A| = 2 \times 2 - 1 \times 3 = 4 - 3 = 1$$ Step 2: Calculate $\operatorname{tr}(B)$ (sum of diagonal): $$\operatorname{tr}(B) = 1 + 2 = 3$$ Step 3: Find $A^{-1}$: $$A^{-1} = \frac{1}{1} \begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix} = \begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix}$$ Step 4: Calculate $C^T$: $$C^T = \begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix}$$ Step 5: Calculate $C^T B$: $$\begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix}1 & -1 \\ 0 & 2\end{pmatrix} = \begin{pmatrix}2 \times 1 + 1 \times 0 & 2 \times (-1) + 1 \times 2 \\ 0 \times 1 + 1 \times 0 & 0 \times (-1) + 1 \times 2\end{pmatrix} = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$$ Step 6: Substitute values into the equation: $$1 \times X - 3 A^{-1} = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$$ $$X = 3 A^{-1} + \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix} = 3 \begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix} + \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix} = \begin{pmatrix}6 & -3 \\ -9 & 6\end{pmatrix} + \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix} = \begin{pmatrix}8 & -3 \\ -9 & 8\end{pmatrix}$$