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Concept:

Sum of Binomial Coefficients and Greatest Binomial Coefficient:

• The sum of binomial coefficients in the expansion of $(x + y)^n$ is calculated by substituting x = 1 and y = 1. The result is $2^n$.
• To find the greatest binomial coefficient, we analyze the coefficients $C(n, r)$ where r is the term index in the expansion. The greatest coefficient occurs near the middle term(s).
• Key Formulae:
• Sum of binomial coefficients: $\text{Sum} = 2^n$
• Binomial coefficient: $C(n, r) = \frac{n!}{r!(n-r)!}$
• Greatest binomial coefficient: For even n, it occurs at r = n/2. For odd n, it occurs at r = (n-1)/2 and r = (n+1)/2.

Calculation:

Given,

Sum of binomial coefficients = $2^n = 256$

We calculate n:

$2^n = 256$

⇒ $2^8 = 256$

Greatest Binomial Coefficient:

For $n = 8$ (even), the greatest binomial coefficient occurs at $r = n/2 = 8/2 = 4$.

⇒ The term index is r = 4, which corresponds to the 5^th term (since indexing starts from 0).

∴ The greatest binomial coefficient occurs in the 5^th term.

Hence, the correct answer is Option 3.

Calculation:

Given,

The inequality is: $k < (√{2} + 1)^3 < k + 2$, where k is a natural number.

We need to compute $(√{2} + 1)^3$

$(√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7$

Approximating √ 2≈ 1.414 , we find:

$5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07$

The inequality becomes:

$k < 14.07 < k + 2$

This implies that k > 12.07 \), so the smallest integer value of k  is:

$k = 13$

∴ The value of k  is 13.

Hence, the correct answer is Option 2.

Calculation:

Multiply Matrix 1 and Matrix 2:

$\begin{bmatrix} x & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} x+4+7 & 2x+5+8 & 3x+6+9 \end{bmatrix}$

⇒ $\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix}$

Multiply the resulting matrix with Matrix 3:

$\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix} \times \begin{bmatrix} 1 \ 1 \ x \end{bmatrix}$

$= (x+11) \cdot 1 + (2x+13) \cdot 1 + (3x+15) \cdot x$

⇒ $(x+11) + (2x+13) + (3x^2+15x)$

⇒ $(3x^2 + 18x + 24)$

Equate the result to 45:

$(3x^2 + 18x + 24 = 45)$

⇒ $(3x^2 + 18x - 21 = 0)$

$(x^2 + 6x - 7 = 0)$

$(x^2 + 7x - x - 7 = 0)$

⇒ $(x = 1 \text{ or } x = -7)$

Step 5: Verify:

For $x = 1$, substitute back:

$(3(1)^2 + 18(1) + 24 = 45)$

45 =45

∴ The correct value of x is 1.

Hence, the correct answer is Option 4.

Calculation:

Given,

The matrix A is:

$A = \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix}$

Since A  is an orthogonal matrix, we know that:

$A^T = A^{-1} \quad \Rightarrow \quad A^T A = I$

This property tells us that A  is orthogonal, and it implies that $A^T A$ (the product of A's transpose and A is equal to the identity matrix I , which is:

$A^T A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$

Now, let’s calculate $A^T A$ step by step. The transpose of matrix A , denoted $A^T$ is:

$A^T = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix}$

Now, we perform matrix multiplication between $A^T$ and A:

$A^T A = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix} \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix}$

Performing this multiplication, we get the following matrix:

$A^T A = \begin{bmatrix} y^2 + z^2 + x^2 & xy + zx + xy & xz + yz + x^2 \ xy + zx + xy & x^2 + x^2 + y^2 & xy + xz + yz \ xz + yz + x^2 & xy + xz + yz & x^2 + y^2 + z^2 \end{bmatrix}$

This matrix must be equal to the identity matrix I , which is:

$\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$

By comparing the elements of the matrices, we get the following system of equations:

1. $y^2 + z^2 + x^2 = 1$ 2. $xy + zx + xy = 0$ 3. $xz + yz + x^2 = 1$

Thus, the key result from the orthogonality condition is:

$x^2 + y^2 + z^2 = 1$

Hence, the correct answer is Option 2.

Calculation:

Statement I

$|M^2| = |M \times M| = |M| \cdot |M| = |M|^2$

⇒ Statement I is correct.

Statement II

For a non-singular matrix, $M \times M^{-1} = I$, where $I$ is the identity matrix.

$|M| \cdot |M^{-1}| = |I| = 1$

⇒ Statement II is incorrect unless $|M| = \pm 1$.

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

$|M| = |M^T|$

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Hence, the correct answer is Option 3.

Concept:

Rotation Matrix:

• A rotation matrix is used to perform a rotation in a Euclidean space. It is a square matrix that describes the rotation of a vector space.
• For a 2D rotation, the matrix is given by: $f(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix}$
• Here, θ is the angle of rotation in radians.
• cos θ: Represents the cosine of the rotation angle.
• sin θ: Represents the sine of the rotation angle.
• Key property of a rotation matrix:
• The transpose of the matrix is equal to its inverse.
• The determinant of the matrix is always equal to 1.
• When θ = π, the rotation matrix becomes: $f(\pi) = \begin{bmatrix} \cos \pi & \sin \pi \ -\sin \pi & \cos \pi \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$

Calculation:

Given,

Rotation matrix at θ = π:

$f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$

To find (f(π))², multiply the matrix by itself:

$f(\pi) × f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} × \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$

Using matrix multiplication:

Top-left element: (-1)(-1) + (0)(0) = 1

Top-right element: (-1)(0) + (0)(-1) = 0

Bottom-left element: (0)(-1) + (-1)(0) = 0

Bottom-right element: (0)(0) + (-1)(-1) = 1

Resulting matrix:

$f(\pi)^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

∴ (f(π))² is equal to the identity matrix, which is $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.

Hence, the correct answer is Option 4.

Calculation:

Given,

$A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$

$\Rightarrow A^2 = A \times A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$

$\Rightarrow A^2 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}$

Now 4A

$\Rightarrow 4A = 4 \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$

$\Rightarrow 4A = \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}$

Also A² - 4A

$\Rightarrow A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}$

$\Rightarrow A^2 - 4A = \begin{bmatrix} 9-4 & 8-8 & 8-8 \ 8-8 & 9-4 & 8-8 \ 8-8 & 8-8 & 9-4 \end{bmatrix}$

$\Rightarrow A^2 - 4A = \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$

Relate the result to the identity matrix I3I3I3

$\Rightarrow A^2 - 4A = 5I_3$

Hence, the correct answer is option 4.

Calculation:

Given,

The number of selections of r things from 5n things is equal to the number of selections of n+r things from 5n things.

This gives the equation:

$\binom{5n}{r} = \binom{5n}{n+r}$

Using the symmetry property of binomial coefficients, we know that:

$\binom{5n}{r} = \binom{5n}{5n-r}$

Thus, comparing the two sides:

$r = 5n - (n + r)$

Simplifying the equation:

$r = 5n - n - r$

$2r = 4n$

$r = 2n$

∴ The value of r is 2n.

Hence, the correct answer is Option 2.

Calculation:

Given:

Total number of items, n = 6

We need to calculate the total number of selections for r = 0, 1, 2, and 3:

For r = 0

$^6C_0 = \frac{6!}{0!(6-0)!} = \frac{1}{1} = 1$

⇒ Number of ways to select 0 items = 1

For r = 1

$^6C_1 = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6$

⇒ Number of ways to select 1 item = 6

For r = 2

$^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$

⇒ Number of ways to select 2 items = 15

For r = 3

$^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

⇒ Number of ways to select 3 items = 20

Total number of selections:

$^6C_0 + ^6C_1 + ^6C_2 + ^6C_3 = 1 + 6 + 15 + 20 = 42$

∴ The total number of selections of at most 3 things from 6 different things is 42.

Hence, the correct answer is Option 4.

Calculation:

Given,

Matrix A is defined as:

$A = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}$

It is mentioned that A is an orthogonal matrix. Therefore,

$A^T A = I$

Now, we calculate A²:

$A^2 = A \cdot A$

$A^2 = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix} \cdot \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}$

Since A is orthogonal, $A^T A = I$, and hence:

$A^2 = I$

∴ A² = Identity matrix (I).

Hence, the Correct answer is Option 2.