1. Problem: Determine which statement about matrices is ALWAYS TRUE. 2. Analyze each option: (A) If $A \neq 0$ is a square matrix, then $A$ is invertible. - This is false because a nonzero matrix can be singular. (B) If $A$ and $B$ are invertible, then $A + B$ is invertible. - False; sum of invertible matrices need not be invertible. (C) If $A$ is square, then $A + A^T$ is skew-symmetric. - False; $A + A^T$ is symmetric, not skew-symmetric. (D) If $A$ and $B$ are invertible, then $(A^{-1}B)^T$ is invertible. - True; transpose and product of invertible matrices are invertible. (E) If the $(3,1)$-entry of $A$ is 5, then the $(1,3)$-entry of $A^T$ is $-5$. - False; transpose entry equals original entry, not negative. Answer: (D) --- 3. Problem: For which values of $a,b$ does the system have more than one solution? Given system: $$\begin{cases}-x + 3y + 2z = -8 \\ x + z = 2 \\ 2x + 2y + az = b \end{cases}$$ 4. Use row operations and analyze the augmented matrix rank. From the given row operations, the system has more than one solution if the coefficient matrix is singular and the augmented matrix is consistent. The key condition is when $a = -4$ and $b = 0$ for infinite solutions. Answer: (D) --- 5. Problem: Let $A$ and $B$ be symmetric matrices of the same order. Which statement is ALWAYS TRUE? (A) $AB$ never symmetric. - False. (B) $AB$ is skew-symmetric. - False. (C) $AB$ is invertible. - False. (D) $AB$ is symmetric if and only if $AB = BA$. - True; product of symmetric matrices is symmetric iff they commute. (E) $AB$ is always symmetric. - False. Answer: (D) --- 6. Problem: Main diagonal of $A^{-1}$ for $$A = \begin{bmatrix}1 & -1 & 2 \\ -5 & 7 & -11 \\ -2 & 3 & -5 \end{bmatrix}$$ Calculate $A^{-1}$ or use known results. Answer: (D) $-2, 1, -2$ --- 7. Problem: Given a non-homogeneous system of 5 equations and 7 unknowns, answer: - Can it have no solution? Yes. - Can it have infinitely many solutions? Yes. - Can it have a unique solution? No (since variables > equations). Answer: (E) Yes, Yes, No --- 8. Problem: Find $(a,b,c)$ so that $$\begin{bmatrix}a & 1 & b & b & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & c \end{bmatrix}$$ is in RREF. RREF requires leading 1s and zeros below and above pivots. Answer: (A) $(0,0,1)$ --- 9. Problem: Convert complex number $$\frac{(16 + 13i)(1 + 2i)}{10 + 5i}$$ to $a + bi$ form. Calculate numerator: $$(16 + 13i)(1 + 2i) = 16 + 32i + 13i + 26i^2 = 16 + 45i - 26 = -10 + 45i$$ Multiply numerator and denominator by conjugate of denominator: $$\frac{-10 + 45i}{10 + 5i} \times \frac{10 - 5i}{10 - 5i} = \frac{-100 - 225i^2 + 50i + 450i}{100 - 25i^2} = \frac{-100 + 225 + 500i}{100 + 25} = \frac{125 + 500i}{125} = 1 + 4i$$ Answer: (A) $1 + 4i$ --- 10. Problem: Polar form of $$\frac{3\sqrt{3} - 3i}{\sqrt{2} + i\sqrt{2}}$$ Calculate magnitude and argument: Magnitude numerator: $\sqrt{(3\sqrt{3})^2 + (-3)^2} = \sqrt{27 + 9} = 6$ Magnitude denominator: $\sqrt{2 + 2} = 2$ Magnitude ratio: $6/2 = 3$ Argument numerator: $\tan^{-1}(-3 / (3\sqrt{3})) = \tan^{-1}(-1/\sqrt{3}) = -\pi/6$ Argument denominator: $\tan^{-1}(\sqrt{2}/\sqrt{2}) = \pi/4$ Total argument: $-\pi/6 - \pi/4 = -5\pi/12$ Answer: (B) $3 (\cos(-5\pi/12) + i \sin(-5\pi/12))$ --- 11. Problem: Find Im$(z)$ if $$z = \frac{1}{(1 - i)(3 - 2i)}$$ Calculate denominator: $$(1 - i)(3 - 2i) = 3 - 2i - 3i + 2i^2 = 3 - 5i - 2 = 1 - 5i$$ So, $$z = \frac{1}{1 - 5i} = \frac{1}{1 - 5i} \times \frac{1 + 5i}{1 + 5i} = \frac{1 + 5i}{1 + 25} = \frac{1 + 5i}{26} = \frac{1}{26} + \frac{5}{26}i$$ Imaginary part is $5/26$ Answer: (D) $5/26$ --- 12. Problem: Rank of matrix $$A = \begin{bmatrix}1 & 4 & 5 & 6 \\ 3 & -2 & 1 & 4 \\ -1 & 0 & -1 & -2 \end{bmatrix}$$ Row operations show three nonzero rows. Answer: (C) 3 --- 13. Problem: Find (2,3)-entry of product of matrices $$\begin{bmatrix}1 & 2 & 4 \\ 2 & 6 & 0 \end{bmatrix} \times \begin{bmatrix}4 & 1 & 4 & 3 \\ 0 & -1 & 3 & 1 \\ 2 & 7 & 5 & 2 \end{bmatrix}$$ Calculate entry (2,3): $$2 \times 4 + 6 \times 3 + 0 \times 5 = 8 + 18 + 0 = 26$$ Answer: (A) 26 --- 14. Problem: Given $$(X^{T} Y Z^{-1})^{-1} W^{T} Y^{-1} = Z$$ Find $W$. Rearranging and using properties of inverses and transposes, Answer: (D) $W = Y^{T} (Z^{2})^{T} Y^{T} X$ --- 15. Problem: Solve for symmetric matrix $A$ given $$A^{T} + \begin{bmatrix}2 & 1 \\ 5 & 1 \end{bmatrix} A = \begin{bmatrix}1 & -3 \\ 1 & -4 \end{bmatrix}$$ Solution yields $$A = \begin{bmatrix}3 & -2 \\ -2 & 1 \end{bmatrix}$$ Answer: (A) --- 16. Problem: Which matrices are involutory ($A = A^{-1}$)? Check each: $B = \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$ is involutory. $D = \begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$ is involutory. Answer: (D) A and B --- Summary of correct answers: 1: (D) 2: (D) 3: (D) 4: (D) 5: (D) 6: (E) 7: (A) 8: (B) 9: (D) 10: (A) 11: (C) 12: (A) 13: (D) 14: (A) 15: (B) Total questions: 15