1. **Problem 39:** For how many distinct values of $x \in \mathbb{C}$ is the given matrix noninvertible? The matrix is upper triangular: $$ \begin{pmatrix} 1+x & x & 1 \\ 0 & 1+x^{2} & -x \\ 0 & 0 & 1-x \end{pmatrix} $$ The determinant of a triangular matrix is the product of its diagonal entries: $$\det = (1+x)(1+x^{2})(1-x)$$ 2. To find when the matrix is noninvertible, solve: $$ (1+x)(1+x^{2})(1-x) = 0 $$ 3. Solutions: - $1+x=0 \implies x=-1$ - $1+x^{2}=0 \implies x^{2} = -1 \implies x=\pm i$ - $1 - x = 0 \implies x = 1$ There are 4 distinct values: $x = -1, i, -i, 1$. **Answer Choice:** (E) Four --- 4. **Problem 40:** Number of regions formed by 10 lines in the plane with no two parallel and no three concurrent. The formula for the maximum number of regions formed by $n$ such lines is: $$ R_n = \frac{n(n+1)}{2} + 1 $$ 5. Substitute $n=10$: $$ R_{10} = \frac{10 \times 11}{2} + 1 = 55 + 1 = 56 $$ **Answer Choice:** (E) 56 --- 6. **Problem 41:** Statement true for every $3 \times 3$ matrix $M$ with real entries. - (A) $M$ has 3 linearly independent eigenvectors: Not always true (e.g., defective matrices). - (B) $M$ has at most one complex eigenvalue: False, complex eigenvalues come in conjugate pairs. - (C) $M$ has at least one real eigenvalue: True by Fundamental Theorem of Algebra for characteristic polynomial with real coefficients of odd degree. - (D) If $M$ invertible, then 3 distinct eigenvalues: False. - (E) If $M$ has 2 orthogonal eigenvectors, then at least 2 distinct eigenvalues: False (orthogonal eigenvectors can exist for repeated eigenvalues in symmetric matrices). **Answer Choice:** (C) $M$ has at least one real eigenvalue. --- 7. **Problem 42:** Find area of triangle with vertices $(1,3,2)$, $(3,1,2)$, and $(-2,0,4)$ in $\mathbb{R}^3$. 8. Compute vectors: $$ \vec{AB} = (3-1, 1-3, 2-2) = (2, -2, 0) $$ $$ \vec{AC} = (-2-1, 0-3, 4-2) = (-3, -3, 2) $$ 9. Compute cross product: $$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 0 \\ -3 & -3 & 2 \end{vmatrix} = \mathbf{i}((-2)(2)-0(-3)) - \mathbf{j}(2(2)-0(-3)) + \mathbf{k}(2(-3)-(-2)(-3)) $$ $$ = \mathbf{i}(-4) - \mathbf{j}(4) + \mathbf{k}(-6 - 6) = (-4, -4, -12) $$ 10. Compute its norm: $$ ||\vec{AB} \times \vec{AC}|| = \sqrt{(-4)^2 + (-4)^2 + (-12)^2} = \sqrt{16 + 16 + 144} = \sqrt{176} = 4\sqrt{11} $$ 11. Area of triangle: $$ \text{Area} = \frac{1}{2} || \vec{AB} \times \vec{AC} || = \frac{1}{2} \times 4\sqrt{11} = 2\sqrt{11} $$ **Answer Choice:** (A) $2\sqrt{11}$ --- 12. **Problem 43:** Relation $R$ on $\mathbb{R}$ defined by $xRy$ if: $$ (x - y)(xy + 2) = 0 $$ Check properties: I. Reflexivity: $xRx$ means $(x-x)(x\cdot x + 2) = 0$, since $x-x=0$, always true. II. Symmetry: If $xRy$, then $(x-y)(xy+2) = 0$. - If $x=y$, then $y=x$ obviously. - If $xy = -2$, then $yx = -2$, so $yRx$. Thus symmetric. III. Transitivity: If $xRy$ and $yRz$, check if $xRz$ holds: - Cases: - If $x=y$, $y=z$ then $x=z$, trivially true. - Suppose $(x-y)=0$ and $(y-z)(yz+2)=0$. If $x=y$, then $xRz$ holds if $(x-z)(xz+2)=0$. But consider counter example: Take $x=1$, $y=1$, $z=-2$: - $xRy$ since $x=y$ - $yRz$ since $(1 - (-2))=3 \neq 0$, but check $yz + 2 = 1 \times (-2) + 2 = 0$, so true. - Check $xRz$: $(1 - (-2))=3 \neq 0$, and $1 \times (-2) + 2=0$, so yes $xRz$ holds here. Try to check a counterexample: Take $x=1$, $y=-2$, $z=2$: - $xRy$: $(1 - (-2))=3 \neq 0$, so need $xy+2 = 1 \times (-2) + 2= 0$, yes. - $yRz$: $(-2 - 2) = -4 \neq 0$, need $yz + 2 = (-2 \times 2) + 2 = -2 \neq 0$, fails. So $yRz$ false. Try $x=2$, $y=-1$, $z=2$: - $xRy$: $(2 +1) \neq 0$, check $2 \times (-1) + 2 = 0$, true. - $yRz$: $(-1 - 2) \neq 0$, check $(-1) \times 2 + 2 = 0$, true. - $xRz$: $(2-2) = 0$, true. So transitive holds in these examples, but let's try if fails. Another counterexample: $x=2$, $y=-1$, $z=0$ - $xRy$: $(2 - (-1))=3 \neq 0$, and $2 \times (-1) + 2 = 0$, true. - $yRz$: $(-1 - 0) = -1 \neq 0$, check $(-1)\times 0 + 2 = 2 \neq 0$, so false. So if $yRz$ fails, no contradiction. Try $x=1$, $y=1$, $z=y$; trivially transitive. Hence, **III is true**. **Answer Choice:** (E) I, II, and III