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Answer: $\cos 15^\circ \cos 75^\circ - \sin 15^\circ \sin 75^\circ = \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0$.

Answer: $x^2 - 36 = 36 - 36 \implies x^2 = 36 \implies x = \pm 6$.

Answer: $\log_3(2^9) \cdot \log_4(3^2) - \log_3(2^3) \cdot \log_4 3$. This simplifies to $(9 \log_3 2)(2 \log_4 3) - (3 \log_3 2)(\log_4 3) = 18(\log_3 2 \cdot \log_4 3) - 3(\log_3 2 \cdot \log_4 3) = 15 \log_4 2 = 15(1/2) = 15/2$.

Answer: Use $C_1 \to C_1+C_2+C_3$. Elements of $C_1$ become $1+\omega+\omega^2=0$. Hence, $\Delta = 0$.

Answer: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$. Maximize $ad=1$ and minimize $bc=0$. e.g., $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$. Max value = 1.

Answer: It is an upper triangular matrix. Determinant = product of diagonal elements = $1 \cdot 1 \cdot 1 = 1$.

Answer: $(a+ib)(a-ib) - (c+id)(-c+id) = (a^2 + b^2) - (-(c^2 + d^2)) = a^2 + b^2 + c^2 + d^2$.

Answer: $|-3A| = (-3)^3 |A| = -27 \times -4 = 108$.

Answer: Expanding along $R_1$: $x(x - 0) - 0 + 8(0 - 2) = 0 \implies x^2 - 16 = 0 \implies x = \pm 4$.

Answer: $3 - x^2 = 3 - 8 \implies -x^2 = -8 \implies x = \pm 2\sqrt{2}$.

Answer: Expand along $C_1$: $2(6-5) - 0 + 1(5\lambda - (-6)) = 4 \implies 2(1) + 5\lambda + 6 = 4 \implies 5\lambda + 8 = 4 \implies 5\lambda = -4 \implies \lambda = -4/5$.

Answer: Apply $C_1 \to C_1+C_2$. $C_1$ becomes $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. $\Delta = \begin{vmatrix} 1 & \cos^2 \theta & 1 \\ 1 & \sin^2 \theta & 1 \\ 2 & 12 & 2 \end{vmatrix}$. $C_1$ and $C_3$ are identical. Hence, $\Delta = 0$.

Answer: $a(bc-f^2) - h(hc-gf) + g(hf-bg) = abc - af^2 - ch^2 + fgh + fgh - bg^2 = abc + 2fgh - af^2 - bg^2 - ch^2$.

Answer: It is a skew-symmetric matrix of odd order (order 3). Thus, $\Delta = 0$.

Answer: $1(-1-1) - 2(-1-2) + x(1-2) = 0 \implies -2 - 2(-3) - x = 0 \implies -2 + 6 - x = 0 \implies x = 4$.

Proof: Apply $C_3 \to C_3 - (C_1 + C_2)$. The elements of $C_3$ become $0, 0, 0$. Hence $\Delta = 0$.

Proof: Apply $R_2 \to R_2 - R_1, R_3 \to R_3 - R_1$. Factor out $(b-a)$ from $R_2$ and $(c-a)$ from $R_3$. Expand along $C_1$ to get the final factorized form.

Proof: Apply $R_1 \to R_1 + R_2$. $R_1$ becomes $[x+y+z \quad x+y+z \quad x+y+z]$. Take common. $R_1$ becomes $[1 \quad 1 \quad 1]$, which is identical to $R_3$. Hence $\Delta = 0$.

Proof: Apply $C_1 \to C_1+C_2+C_3$. $C_1$ becomes $[0 \quad 0 \quad 0]^T$. Value = $0$.

Proof: Take $a$ from $R_1$, $b$ from $R_2$, $c$ from $R_3$. $\Delta = abc \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$. Then apply $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$ to yield $abc(a-b)(b-c)(c-a)$.

Answer: $|2AB| = 2^3 |A||B| = 8(5)(-2) = -80$.

Answer: Observe that $C_1 = C_2 + 8C_3$. Applying $C_1 \to C_1 - C_2 - 8C_3$ yields $0$ in $C_1$. Hence $\Delta = 0$.

Proof: Let $A^T = -A$. $|A^T| = |-A|$. Since order is 3, $|-A| = (-1)^3|A| = -|A|$. Since $|A^T| = |A|$, we get $|A| = -|A| \implies 2|A| = 0 \implies |A| = 0$.

Proof: Apply $C_1 \to C_1+C_2+C_3$, take out $(5x+4)$. Apply $R_2 \to R_2-R_1, R_3 \to R_3-R_1$. Expand to get $(5x+4)(4-x)^2$.

Answer: Take $1!$ from $R_1, 2!$ from $R_2, 3!$ from $R_3$. $\Delta = 1!2!3! \begin{vmatrix} 1 & 2 & 6 \\ 1 & 3 & 12 \\ 1 & 4 & 20 \end{vmatrix} = 12(2) = 24$.

Proof: Standard determinant. $R_1 \to R_1 - (R_2+R_3) \implies R_1 = [0 \quad -2c \quad -2b]$. Expand to yield $4abc$.

Answer: $C_3 = \sin \alpha \cos \delta + \cos \alpha \sin \delta$. Hence $C_3 = \cos \delta \cdot C_1 + \sin \delta \cdot C_2$. Apply $C_3 \to C_3 - \cos \delta C_1 - \sin \delta C_2 \implies C_3 = 0$. $\Delta = 0$.

Answer: Since $a,b,c$ are in AP, $a+c = 2b$. Apply $R_1 \to R_1+R_3-2R_2$. $R_1$ becomes $[2y+4+4y+6-2(3y+5) \quad \dots \quad a+c-2b] = [0 \quad 0 \quad 0]$. $\Delta = 0$.

Answer: $C_1 \to C_1+C_2+C_3$. $C_1$ becomes $1+\omega^n+\omega^{2n}$. For $n \neq 3k$, this sum is 0. Hence $\Delta = 0$.

Proof: $C_1 \to C_1+C_2+C_3$. Take $1+x+x^2$ common. Expand and simplify to $(1-x)^2(1+x+x^2)^2 = ((1-x)(1+x+x^2))^2 = (1-x^3)^2$.

Answer: $\Delta = \frac{1}{2}|3(2-1) - 8(-4-5) + 1(-4-10)| = \frac{1}{2}|3 + 72 - 14| = \frac{61}{2} = 30.5$ sq units.

Answer: Let vertex be $(0, y)$. $\frac{1}{2} \begin{vmatrix} 3 & 0 & 1 \\ 0 & 2 & 1 \\ 0 & y & 1 \end{vmatrix} = \pm 9 \implies \frac{1}{2}[3(2-y)] = \pm 9 \implies 6-3y = \pm 18 \implies y = -4, 8$. Coordinates: $(0,-4)$ or $(0,8)$.

Answer: Area = 0. $\frac{1}{2}[k(2k-6+2k) - (2-2k)(1-k+4+k) + 1(\dots)] = 0$. Expanding yields a quadratic in $k$. Solving gives $k = -1/2, 1$.

Answer: Line $AB$: $\begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0 \implies 3x - y = 0$. For $D$: $\frac{1}{2}\begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3 \implies \frac{1}{2}[-1(0-3k)] = \pm 3 \implies 3k = \pm 6 \implies k = \pm 2$.

Proof: $\begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \implies a(b-1) - 0 + 1(0-b) = 0 \implies ab - a - b = 0 \implies ab = a+b$. Dividing by $ab \implies 1 = 1/b + 1/a$.

Proof: Area = $\frac{1}{2}\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} = 0$ (applying $C_2 \to C_2+C_1$ yields $a+b+c$ in $C_2$, making it prop to $C_3$). Hence collinear.

Answer: $\Delta = \frac{1}{2}|a(c-a) - c(b-c) + 1(ba - c^2)| = \frac{1}{2}|ac - a^2 - bc + c^2 + ab - c^2| = \frac{1}{2}|(a-b)(c-a)|$.

Answer: $\Delta = \frac{1}{2}\begin{vmatrix} 2 & 5 & 1 \\ 4 & 6 & 1 \\ 8 & 8 & 1 \end{vmatrix} = \frac{1}{2}[2(-2) - 5(-4) + 1(-16)] = \frac{1}{2}[-4+20-16] = 0$. Since area is 0, they are collinear, not a triangle.

Answer: Split into $\triangle ABC$ and $\triangle ACD$. Area($ABC$) = $\frac{1}{2}|-3(10) - 2(5-7) + 1(-30-28)| = 42$. Area($ACD$) = $43$. Total = $42+43 = 85$ sq units.

Proof: Let $\Delta$ be the area. We know $\Delta = \frac{\sqrt{3}}{4}a^2$. The determinant equals $2\Delta$. Thus, $(2\Delta)^2 = 4\Delta^2 = 4(\frac{3}{16}a^4) = \frac{3a^4}{4}$.

Answer: $\begin{vmatrix} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{vmatrix} = 0 \implies x(-6) + 2(-3) + 1(24) = 0 \implies -6x - 6 + 24 = 0 \implies 6x = 18 \implies x=3$.

Answer: $\begin{vmatrix} x & y & 1 \\ -1 & 3 & 1 \\ 2 & -4 & 1 \end{vmatrix} = 0 \implies x(7) - y(-3) + 1(-2) = 0 \implies 7x + 3y - 2 = 0$.

Answer: Points are $(2,4), (2,6), (2,k)$. They all lie on the vertical line $x=2$. The area is $0$ for ANY value of $k$. Thus, $k \in \mathbb{R}$.

Answer: $\begin{vmatrix} 2 & -3 & 1 \\ \lambda & -1 & 1 \\ 0 & 4 & 1 \end{vmatrix} = 0 \implies 2(-5) + 3(\lambda) + 1(4\lambda) = 0 \implies -10 + 7\lambda = 0 \implies \lambda = 10/7$.

Answer: $\frac{1}{2}\begin{vmatrix} 1 & -3 & 1 \\ 4 & p & 1 \\ -9 & 7 & 1 \end{vmatrix} = \pm 15 \implies \frac{1}{2}[1(p-7) + 3(4+9) + 1(28+9p)] = \pm 15 \implies 10p + 60 = \pm 30$. $10p = -30 \implies p = -3$; $10p = -90 \implies p = -9$.

Answer: Element $6$ is $a_{23}$. $M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$. Cofactor $A_{23} = (-1)^{2+3}(-6) = 6$.

Answer: $|A| = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 0 + 3(-1)^{2+2}(2+4) + (-2)(-1)^{2+3}(10-1) = 0 + 3(6) + 2(9) = 18 + 18 = 36$.

Answer: $|adj A| = |A|^{n-1} = 8^{3-1} = 8^2 = 64$.

Answer: Swap diagonals, change sign of off-diagonals. $adj(A) = \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix}$.

Answer: Property: $A(adj A) = |A|I$. Evaluate $|A| = 1(16-9) - 3(4-3) + 3(3-4) = 7 - 3 - 3 = 1$. Hence $A(adj A) = 1 \cdot I = I_3$.

Answer: Since $A A^{-1} = I$, taking determinant gives $|A||A^{-1}| = |I| = 1 \implies |A^{-1}| = 1 / |A|$.

Answer: $|adj A| = |A|^2 = 64 \implies |A| = \pm 8$.

Answer: For a diagonal matrix, inverse is formed by taking reciprocals of diagonals. $A^{-1} = \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.

Proof: This sum represents expanding elements of row 1 with cofactors of row 3. By property of determinants, this sum is always $0$.

Proof: $A = A^T$. $A \cdot adj A = |A|I \implies adj A = |A|A^{-1}$. Transpose: $(adj A)^T = (|A|A^{-1})^T = |A|(A^T)^{-1} = |A|A^{-1} = adj A$. Hence symmetric.

Answer: $|A| = \cos^2\alpha + \sin^2\alpha = 1$. $adj(A) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$. $A^{-1} = adj(A)$.

Answer: Comparing with $A(adj A) = |A|I$, we get $|A| = 10$. Then $|adj A| = |A|^2 = 100$.

Answer: $adj(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}^T = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$.

Proof: Compute $AB = \begin{bmatrix} 2 & 0 \\ 3 & 5 \end{bmatrix}$. $(AB)^{-1} = \frac{1}{10}\begin{bmatrix} 5 & 0 \\ -3 & 2 \end{bmatrix}$. Compute $A^{-1}$ and $B^{-1}$, multiply $B^{-1}A^{-1}$ to verify they match exactly.

Answer: Formula: $|adj(adj A)| = |A|^{(n-1)^2}$. For $n=3$, $|adj(adj A)| = |A|^4 = 3^4 = 81$.

Answer: $AX=B \implies A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$. $|A| = 1$. $A^{-1} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$. $X = A^{-1}B = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$. $x=2, y=-3$.

Answer: $|A| = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix} = 0$. $adj(A) = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$. $(adj A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O$. System is inconsistent.

Answer: $|A| = 1(3) - 2(-1) - 2(2) = 3+2-4 = 1$. Cofactors and transpose gives $A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$.

Answer: $\Delta = 1, \Delta_1 = 1, \Delta_2 = 2, \Delta_3 = -2$. Thus $x = 1/1 = 1, y = 2/1 = 2, z = -2/1 = -2$.

Answer: For non-trivial solutions, $|A| = 0$. $\begin{vmatrix} 3 & -1 & 1 \\ 15 & -6 & 5 \\ \lambda & -2 & 2 \end{vmatrix} = 0 \implies 3(-2) + 1(30-5\lambda) + 1(-30+6\lambda) = 0 \implies \lambda - 6 = 0 \implies \lambda = 6$.

Answer: $XA = B \implies X = BA^{-1}$. $A^{-1} = \frac{-1}{7}\begin{bmatrix} 3 & -2 \\ -5 & 1 \end{bmatrix}$. $X = \begin{bmatrix} 5 & 6 \\ -1 & 0 \end{bmatrix} \left(\frac{-1}{7}\right)\begin{bmatrix} 3 & -2 \\ -5 & 1 \end{bmatrix} = \frac{-1}{7}\begin{bmatrix} -15 & -4 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 15/7 & 4/7 \\ 3/7 & -2/7 \end{bmatrix}$.

Answer: Calc $A^2 = \begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$. Verify $A^2-6A+17I=O$. Multiply by $A^{-1} \implies A - 6I + 17A^{-1} = O \implies A^{-1} = \frac{1}{17}(6I - A) = \frac{1}{17}\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}$.

Answer: Let $1/x=u, 1/y=v, 1/z=w$. Solve $AX=B$ for $u, v, w$. You get $u=1/2, v=1/3, w=1/5$. Thus $x=2, y=3, z=5$.

Answer: Compute $AB = 6I$. Thus $B^{-1} = \frac{1}{6}A$. The system's coefficient matrix is exactly $B$. $X = B^{-1}C = \frac{1}{6}A C = \frac{1}{6}\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}$. $x=2, y=-1, z=4$.

Answer: Eqs: $x+y+z=10000$, $2x+3y+5z=34000$, $5z-2x=10000$. Matrix solution yields $x=2000, y=5000, z=3000$.

Answer: $|A| = a-a = 0$. Using Cramer's, $\Delta = 0$. $\Delta_1 = 4-2a$. If $a=2$, $\Delta_1=0$, infinite solutions. If $a \neq 2$, $\Delta_1 \neq 0$, inconsistent. So inconsistent for all $a \in \mathbb{R} - \{2\}$.

Answer: $\Delta'(x) = \begin{vmatrix} 2x & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix} + 0 + 0 = 2x(1) - 1(2) = 2x - 2$.

Answer: $\Delta(x) = 2x^3 - 1$. Integral: $\int_0^1 (2x^3 - 1) dx = [\frac{x^4}{2} - x]_0^1 = 1/2 - 1 = -1/2$.

Answer: $|A| = 0$. Let $z=k$. From eq 1 and 2: $x+y=k$, $2x-3y=-k$. Solve for $x, y$ in terms of $k$: $x = 2k/5, y = 3k/5$. Solutions: $(2k/5, 3k/5, k)$.

Answer: $|A| = -17$. Evaluate adj A and use $X = A^{-1}B$. Final result: $x=1, y=2, z=3$.