1. **Express the complex number $1 + i\sqrt{3}$ in polar form.** The complex number is $z = 1 + i\sqrt{3}$. Step 1: Calculate modulus $r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. Step 2: Calculate argument $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$. Step 3: Polar form is $z = r(\cos \theta + i \sin \theta) = 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$. 2. **Difference between $(a,b)$ and $\{(a,b)\}$.** $(a,b)$ is an ordered pair with elements $a$ and $b$. $\{(a,b)\}$ is a set containing the ordered pair $(a,b)$ as its single element. 3. **Prove $\emptyset \subseteq A$ for any set $A$.** Step: By definition, $\emptyset$ has no elements, so there is no element in $\emptyset$ that is not in $A$. Therefore, $\emptyset$ is a subset of every set $A$. 4. **Find $A_{22}$ and $A_{32}$ for matrix $A = \begin{bmatrix}1 & 2 & -3 \\ 0 & -2 & 0 \\ -2 & -2 & 1\end{bmatrix}$.** Step 1: $A_{ij}$ is the cofactor of element in $i$th row and $j$th column. Calculate minor for $A_{22}$: Delete 2nd row and 2nd column: $\begin{vmatrix}1 & -3 \\ -2 & 1\end{vmatrix} = (1)(1) - (-3)(-2) = 1 - 6 = -5$. Cofactor $A_{22} = (-1)^{2+2} \times (-5) = 1 \times (-5) = -5$. Calculate minor for $A_{32}$: Delete 3rd row and 2nd column: $\begin{vmatrix}1 & -3 \\ 0 & 0\end{vmatrix} = (1)(0) - (-3)(0) = 0$. Cofactor $A_{32} = (-1)^{3+2} \times 0 = (-1)^5 \times 0 = 0$. 5. **Find $a$ and $b$ if $A = \begin{bmatrix} a & -1 \\ 2 & b \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 10 & 01 \\ 01 & 10 \end{bmatrix}$.** Note: Matrix inverse entries likely typos: interpret $A^{-1} = \begin{bmatrix} 10 & 1 \\ 1 & 10 \end{bmatrix}$. Step 1: $A \times A^{-1} = I$. Calculate $\det A = ab - (-1)(2) = ab + 2$. Step 2: Inverse formula: $A^{-1} = \frac{1}{\det A} \begin{bmatrix} b & 1 \\ -2 & a \end{bmatrix} = \begin{bmatrix} 10 & 1 \\ 1 & 10 \end{bmatrix}$. Equate elements: $\frac{b}{ab + 2} = 10$, $\frac{1}{ab + 2} = 1$, $\frac{-2}{ab + 2} = 1$, $\frac{a}{ab + 2} = 10$. From $\frac{1}{ab + 2} = 1$ we get $ab + 2 = 1$, so $ab = -1$. From $\frac{-2}{ab + 2} = 1$ we get $-2 = ab + 2$, contradicting above. The given inverse matrix seems inconsistent; please check question data. 6. **State transitive property of equality.** If $a = b$ and $b = c$, then $a = c$. 7. **Compute $AB$ for matrices:** $A=\begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -3 \\ 1 & 2 & -2 \end{bmatrix}$, $B=\begin{bmatrix} 2 & -2 & 3 \\ -1 & -4 & 6 \\ 0 & -5 & 1 \end{bmatrix}$. Step 1: Compute elements of $AB$ by matrix multiplication formula. For entry $(1,1)$: $2*2 + (-1)*(-1) + 0*0 = 4 + 1 + 0 = 5$. For $(1,2)$: $2*(-2) + (-1)*(-4) + 0*(-5) = -4 + 4 + 0 = 0$. For $(1,3)$: $2*3 + (-1)*6 + 0*1 = 6 - 6 + 0 = 0$. For $(2,1)$: $1*2 + 2*(-1) + (-3)*0 = 2 - 2 + 0 = 0$. For $(2,2)$: $1*(-2) + 2*(-4) + (-3)*(-5) = -2 - 8 + 15 = 5$. For $(2,3)$: $1*3 + 2*6 + (-3)*1 = 3 + 12 - 3 = 12$. For $(3,1)$: $1*2 + 2*(-1) + (-2)*0 = 2 - 2 + 0 = 0$. For $(3,2)$: $1*(-2) + 2*(-4) + (-2)*(-5) = -2 - 8 + 10 = 0$. For $(3,3)$: $1*3 + 2*6 + (-2)*1 = 3 + 12 - 2 = 13$. So, $$AB = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 12 \\ 0 & 0 & 13 \end{bmatrix}$$ 8. **Define Skew Symmetric Matrix.** A square matrix $A$ is skew symmetric if $A^T = -A$. This means $a_{ij} = -a_{ji}$ and all diagonal elements are zero. 9. **Show sum of cube roots of unity is zero.** Cube roots of unity are $1, \omega, \omega^2$ where $\omega = e^{2\pi i / 3}$. Sum is $1 + \omega + \omega^2$. Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, sum is zero. 10. **Solve $5x^3 - 5x = 0$.** Factorize: $5x(x^2 - 1) = 0 \Rightarrow 5x(x-1)(x+1) = 0$. Solutions: $x=0$, $x=1$, and $x=-1$. 11. **Find $x$ if polynomial $x^3 + kx^2 -7x +6$ has remainder $-4$ when divided by $x+2$.** Use Remainder Theorem: Plug $x = -2$: $(-2)^3 + k(-2)^2 - 7(-2) + 6 = -8 + 4k + 14 + 6 = 4k + 12$. Given remainder $-4$: $4k + 12 = -4 \Rightarrow 4k = -16 \Rightarrow k = -4$. 12. **Discuss roots of $x^2 - 5x + 6 = 0$.** Calculate discriminant: $\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0$. Roots are real and distinct. Roots: $x = \frac{5 \pm 1}{2} = 3, 2$. --- **Summary:** We solved 12 distinct problems (parts i–xii).