📌 Step-by-Step Answer:

Concept: For a 2×2 matrix, \(\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc\).

Also, we use the identity: \(\cos(A+B) = \cos A \cos B – \sin A \sin B\).

\[
\Delta = \cos75^\circ \cos15^\circ – \sin75^\circ \sin15^\circ
\]

\[
= \cos(75^\circ + 15^\circ) = \cos 90^\circ = 0
\]

✅ Correct Option: (B) zero

📌 Step-by-Step Answer:

Concept: If a square matrix \(A\) is of order \(n\), then \(|kA| = k^n |A|\).

Here, matrix \(A\) has order \(2\) (since it is 2×2). So, \(n = 2\).

\[
|3A| = 3^2 |A| = 9|A|
\]

Given that \(|3A| = k|A|\).

Comparing both sides, we get \(k = 9\).

✅ Correct Option: (B) 9

📌 Step-by-Step Answer:

Concept 1: For any two square matrices, \(|AB| = |A| \cdot |B|\).

Concept 2: For a matrix of order \(n\), \(|kM| = k^n |M|\).

Here, the order is 3 (given). So, \(n = 3\).

\[
|2AB| = 2^3 |AB| = 8 |AB|
\]

\[
|AB| = |A| \times |B| = (-3) \times 2 = -6
\]

\[
|2AB| = 8 \times (-6) = -48
\]

✅ Correct Option: (B) – 48

📌 Step-by-Step Answer:

Step 1: Expand the Left Hand Side (LHS).

\[
\text{LHS} = (x+1)(x+2) – (x-1)(x-3)
\]

\[
= (x^2 + 3x + 2) – (x^2 – 4x + 3)
\]

\[
= x^2 + 3x + 2 – x^2 + 4x – 3 = 7x – 1
\]

Step 2: Expand the Right Hand Side (RHS).

\[
\text{RHS} = (5)(7) – (5)(3) = 35 – 15 = 20
\]

Step 3: Equate LHS and RHS to find \(x\).

\[
7x – 1 = 20 \Rightarrow 7x = 21 \Rightarrow x = 3
\]

✅ Correct Option: (B) 3

📌 Step-by-Step Answer:

Concept: We know the fundamental identity \(A(\text{adj } A) = |A| I\).

Given that \(A(\text{adj } A) = 7 I\).

Comparing the two, we get \(|A| = 7\).

Formula: For a square matrix of order \(n\), \(|\text{adj } A| = |A|^{n-1}\).

Here, the order is 3. So, \(n = 3\).

\[
|\text{adj } A| = |A|^{3-1} = |A|^2 = 7^2 = 49
\]

✅ Correct Option: (C) 49

📌 Step-by-Step Answer:

Formula: Area of triangle = \(\frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |\).

Assign the points: \((x_1, y_1) = (3, 0)\), \((x_2, y_2) = (0, k)\), \((x_3, y_3) = (-3, 0)\).

Substitute these values into the formula.

\[
\text{Area} = \frac{1}{2} | 3(k-0) + 0(0-0) + (-3)(0-k) |
\]

\[
= \frac{1}{2} | 3k + 0 + 3k | = \frac{1}{2} | 6k | = 3|k|
\]

Given that the area is 9 square units.

\[
3|k| = 9 \Rightarrow |k| = 3 \Rightarrow k = \pm 3
\]

Among the given options, \(k = 3\) is present.

✅ Correct Option: (C) 3

📌 Step-by-Step Answer:

Given that \(X^2 = I\).

This means \(X \cdot X = I\).

We want to find \(X^{-1}\).

Multiply both sides of \(X \cdot X = I\) on the left by \(X^{-1}\).

\[
X^{-1} (X \cdot X) = X^{-1} I
\]

Using the associative property: \((X^{-1} X) X = X^{-1}\).

Since \(X^{-1} X = I\), we get \(I \cdot X = X^{-1}\).

Therefore, \(X = X^{-1}\).

✅ Correct Option: (A) \(X\)

📌 Step-by-Step Answer:

Concept: Cofactor \(C_{ij} = (-1)^{i+j} \cdot M_{ij}\), where \(M_{ij}\) is the minor.

We need the cofactor of \(a_{32}\).

Here, \(i=3\) and \(j=2\).

The sign is \((-1)^{3+2} = (-1)^5 = -1\).

So, \(C_{32} = – M_{32}\).

Find the minor \(M_{32}\) by deleting Row 3 and Column 2.

Removing Row 3 \((3, 1, 2)\) and Column 2 \((2, 3, 1)\), we get the 2×2 determinant:

\[
M_{32} = \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}
\]

\[
M_{32} = (1)(1) – (3)(2) = 1 – 6 = -5
\]

Now, \(C_{32} = -(-5) = 5\).

✅ Correct Option: (C) \(5\)

📌 Step-by-Step Answer:

Check Assertion (A):

For a skew-symmetric matrix, \(A’ = -A\).

Taking determinant on both sides: \(|A’| = |-A|\).

Using property of transpose, \(|A’| = |A|\).

Using scalar property for order 3, \(|-A| = (-1)^3 |A| = -|A|\).

So, \(|A| = -|A| \Rightarrow 2|A| = 0 \Rightarrow |A| = 0\).

Thus, Assertion (A) is TRUE.

Check Reason (R):

For any square matrix, the determinant of the matrix equals the determinant of its transpose.

So, \(|A| = |A’|\) is a standard and true property.

Thus, Reason (R) is TRUE.

Does R explain A?

Yes, we used R (the equality \(|A| = |A’|\)) along with the skew-symmetric definition to prove A.

✅ Correct Option: Both (A) and (R) are true and (R) is the correct explanation of (A).

📌 Step-by-Step Solution:

Part 1: Finding \(A^{-1}\)

Step 1: We are given \(|A| = 62\).

Step 2: Find the cofactor matrix.

Calculate \(C_{11} = -3\), \(C_{12} = 26\), \(C_{13} = 19\).

Calculate \(C_{21} = 9\), \(C_{22} = -16\), \(C_{23} = 5\).

Calculate \(C_{31} = 5\), \(C_{32} = -2\), \(C_{33} = -11\).

So, Cofactor Matrix = \(\begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix}\).

Step 3: Find the Adjugate matrix.

Adj(\(A\)) = Transpose of the cofactor matrix.

\[
\text{adj}(A) = \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}
\]

Step 4: Find the inverse using the formula \(A^{-1} = \frac{1}{|A|} \text{adj}(A)\).

\[
A^{-1} = \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}
\]

Part 2: Solving the System of Equations

The system is: \(3x + 4y + 7z = 14\), \(2x – y + 3z = 4\), \(x + 2y – 3z = 0\).

Write it in matrix form \(C X = B\).

\[
C = \begin{bmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix}
\]

Notice that \(C\) is exactly the transpose of the given matrix \(A\).

So, \(C = A’\). The equation becomes \(A’ X = B\).

Multiply both sides by \((A’)^{-1}\) to isolate \(X\).

\[
X = (A’)^{-1} B
\]

Using the property \((A’)^{-1} = (A^{-1})’\), we get:

\[
X = (A^{-1})’ B
\]

Find the transpose of \(A^{-1}\):

\[
(A^{-1})’ = \frac{1}{62} \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix}
\]

Now, multiply by \(B\) to find \(x, y, z\).

\[
x = \frac{1}{62} [(-3)(14) + (26)(4) + (19)(0)]
\]

\[
= \frac{-42 + 104 + 0}{62} = \frac{62}{62} = 1
\]

\[
y = \frac{1}{62} [(9)(14) + (-16)(4) + (5)(0)]
\]

\[
= \frac{126 – 64 + 0}{62} = \frac{62}{62} = 1
\]

\[
z = \frac{1}{62} [(5)(14) + (-2)(4) + (-11)(0)]
\]

\[
= \frac{70 – 8 + 0}{62} = \frac{62}{62} = 1
\]

✅ Final Answer: \(x = 1\), \(y = 1\), \(z = 1\).