1.Answer: $\Delta = (4 \times 3) - (5 \times -2) = 12 - (-10) = 22$.
2.Answer: $\cos^2 60^\circ - (-\sin^2 60^\circ) = \cos^2 60^\circ + \sin^2 60^\circ = 1$.
3.Answer: $2x^2 - 40 = 18 - 40 \implies 2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.
4.Answer: $(a+b)(a-b) - ab = a^2 - b^2 - ab$.
5.Answer: Expand along Row 2 (or Col 3): $-(-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 1(-15 - (-3)) = -12$.
6.Answer: Expand along Col 3: $4 \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} = 4(-1 - 12) = 4(-13) = -52$.
7.Answer: $|A| = (1)(3) - (-2)(1) = 3 + 2 = 5$. Since order $n=2$, $|2A| = 2^2|A| = 4(5) = 20$.
8.Answer: $2x(x-2) - (-3)(3x) = 3 \implies 2x^2 - 4x + 9x = 3 \implies 2x^2 + 5x - 3 = 0$. Factoring: $(2x-1)(x+3) = 0 \implies x = 1/2, -3$.
9.Answer: The determinant of a triangular matrix is the product of its diagonal elements: $2 \times (-3) \times 5 = -30$.
10.Answer: $0(0) - a(-a) = 0 + a^2 = a^2$.
11.Answer: $0$, because Row 1 and Row 3 are identical.
12.Answer: $-10$. Row 2 and Row 3 are interchanged, which multiplies the determinant by $-1$.
13.Answer: $|3A| = 3^3 |A| = 27 \times 7 = 189$.
14.Proof: Apply $C_3 \to C_3 - 9C_2$. The new $C_3$ becomes $\begin{bmatrix} 65-63 \\ 75-72 \\ 86-81 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}$, which is identical to $C_1$. Hence, $\Delta = 0$.
15.Answer: Apply $R_3 \to R_3 - (R_1 + R_2)$. $R_3$ becomes $\begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. Therefore, $\Delta = 0$.
16.Answer: $0$. (Property: Determinant of a skew-symmetric matrix of odd order is always zero).
17.Answer: $-4$. (Property: $|A| = |A^T|$).
18.Answer: Take $6$ common from $R_1$: $6 \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$. Since $R_1$ and $R_3$ are identical, value is $0$.
19.Answer: The value remains unchanged. $\Delta_{new} = \Delta$.
20.Answer: Multiplying all elements by $-1$ is equivalent to taking $-1$ common from each of the 3 rows. The new value is $(-1)^3 \Delta = -\Delta$.
21.Answer: $\Delta = \frac{1}{2} |2(1-8) - 7(1-10) + 1(8-10)| = \frac{1}{2} |-14 + 63 - 2| = \frac{47}{2} = 23.5$ sq. units.
22.Answer: $\Delta = \frac{1}{2} |-2(2+8) + 3(3+1) + 1(24+2)| = \frac{1}{2} |-20 + 12 + 26| = \frac{18}{2} = 9$ sq. units.
23.Proof: $\Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$. Apply $C_2 \to C_2 + C_1$, $C_2$ becomes $\begin{bmatrix} a+b+c \\ a+b+c \\ a+b+c \end{bmatrix}$. Take $(a+b+c)$ common from $C_2$. $C_2$ and $C_3$ become identical ($1$'s). So $\Delta = 0$, points are collinear.
24.Answer: $\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \implies x(2-6) - y(1-3) + 1(6-6) = 0 \implies -4x + 2y = 0 \implies y = 2x$.
25.Answer: $\begin{vmatrix} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{vmatrix} = 0 \implies 2(-1-4) + 3(k-0) + 1(4k-0) = 0 \implies -10 + 3k + 4k = 0 \implies 7k = 10 \implies k = 10/7$.
26.Answer: $\frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4 \implies \frac{1}{2} [k(-2) - 0 + 1(8)] = \pm 4 \implies -2k + 8 = \pm 8$. If $8$: $-2k=0 \implies k=0$. If $-8$: $-2k=-16 \implies k=8$. So $k = 0, 8$.
27.Answer: $\frac{1}{2} [2(0) + 6(5-k) + 1(20-4k)] = \pm 35 \implies 30 - 6k + 20 - 4k = \pm 70 \implies 50 - 10k = \pm 70$. If $70$: $k = -2$. If $-70$: $k = 12$. $k = 12, -2$.
28.Answer: The determinant value can be negative depending on vertex order, but Area is a physical quantity, so we take the absolute magnitude ($|\Delta|$) for the final answer.
29.Answer: $\frac{1}{2} |0 - 0 + 1(0 - 20)| = \frac{1}{2} |-20| = 10$ sq. units.
30.Answer: $\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0 \implies x(1-3) - y(3-9) + 1(9-9) = 0 \implies -2x + 6y = 0 \implies x - 3y = 0$.
31.Answer: Delete row 2, col 1. $M_{21} = -2$. Cofactor $A_{21} = (-1)^{2+1}M_{21} = -(-2) = 2$.
32.Answer: Cofactors: $A_{11}=4, A_{12}=-3, A_{21}=-(-1)=1, A_{22}=2$. Matrix: $\begin{bmatrix} 4 & -3 \\ 1 & 2 \end{bmatrix}$.
33.Answer: Swap diagonals, change signs of off-diagonals. $adj(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$.
34.Answer: Delete row 2, col 2. $M_{22} = \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = 9 - 21 = -12$.
35.Answer: $M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$. $A_{23} = (-1)^{2+3}(-6) = 6$.
36.Answer: $|adj A| = |A|^{n-1} = 6^{3-1} = 6^2 = 36$.
37.Answer: By property, $A(adj A) = |A|I = 4I = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$.
38.Answer: $|A| = 4 - 6 = -2$. $adj(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$. $A \cdot adj A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} = -2I$. Verified.
39.Answer: The sum is equal to the determinant value, $\Delta$.
40.Answer: $adj(A) = \begin{bmatrix} -2 & 2 \\ -3 & 5 \end{bmatrix}$.
41.Answer: $|A| = 6 - (-8) = 14$. $adj(A) = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$. $A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
42.Answer: $|A| = (6)(2) - (4)(3) = 12 - 12 = 0$. Since $|A| = 0$, it is a singular matrix, hence NOT invertible.
43.Answer: $A^{-1} = I^{-1} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
44.Answer: Singular implies $|A| = 0$. $(3-x)(1) - 8 = 0 \implies 3 - x - 8 = 0 \implies -5 - x = 0 \implies x = -5$.
45.Answer: $\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$.
46.Answer: $|A| = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3 - 4 = -1 \neq 0$. Consistent with a unique solution.
47.Answer: $|A| = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix} = 6 - 6 = 0$. Check $(adj A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O$. Inconsistent.
49.Answer: $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$. $|A| = 15 - 14 = 1$. $adj(A) = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$. $X = A^{-1}B = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$. $x=2, y=-3$.
50.Answer: $X = A^{-1}B = \begin{bmatrix} 3 & -1 \\ -5/2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3(1) - 1(2) \\ -5/2(1) + 1(2) \end{bmatrix} = \begin{bmatrix} 1 \\ -1/2 \end{bmatrix}$. $x=1, y=-1/2$.
