Differentiation (Part-38) | S N De

Determinants MCQ Solutions (Questions 58 to 68)

Q 58. Which statement is correct regarding the points \( (0, -3), (3, 0) \) and \( (5, 2) \)?

(a) The points form a right-angled triangle of area \( \frac{9}{2} \) sq units

(b) The points are collinear

(c) The points form an equilateral triangle of area \( \frac{9}{2} \) sq units

(d) The points form an isosceles triangle of area \( \frac{9}{2} \) sq units

Ans 58.

Let's calculate the area of the triangle formed by these three points using the determinant method:

$$ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $$

$$ \text{Area} = \frac{1}{2} \begin{vmatrix} 0 & -3 & 1 \\ 3 & 0 & 1 \\ 5 & 2 & 1 \end{vmatrix} $$

Expanding along the first row:

$$ = \frac{1}{2} \left[ 0 - (-3)(3 \cdot 1 - 5 \cdot 1) + 1(3 \cdot 2 - 5 \cdot 0) \right] $$

$$ = \frac{1}{2} \left[ 3(3 - 5) + (6 - 0) \right] = \frac{1}{2} \left[ 3(-2) + 6 \right] = \frac{1}{2} \left[ -6 + 6 \right] = 0 $$

Since the area is 0, the points lie on the same straight line.
Correct Option: (b) The points are collinear

Q 59. If the points \( (a_1, b_1), (a_2, b_2) \) and \( (a_1+a_2, b_1+b_2) \) are collinear, which of the following is correct?

(a) \( a_1a_2 = b_1b_2 \)

(b) \( a_1a_2 + b_1b_2 = 0 \)

(c) \( a_1b_2 = a_2b_1 \)

(d) \( a_1b_2 + a_2b_1 = 0 \)

Ans 59.

For the points to be collinear, the area of the triangle they form must be 0:

$$ \frac{1}{2} \begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \\ a_1+a_2 & b_1+b_2 & 1 \end{vmatrix} = 0 $$

Applying the row operation \( R_3 \to R_3 - R_1 - R_2 \):

$$ \begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \\ 0 & 0 & -1 \end{vmatrix} = 0 $$

Expanding along the third row (\( R_3 \)):

$$ -1(a_1b_2 - a_2b_1) = 0 \implies a_1b_2 - a_2b_1 = 0 \implies a_1b_2 = a_2b_1 $$

Correct Option: (c) \( a_1b_2 = a_2b_1 \)

Q 60. \( \Delta ABC \) is an equilateral triangle with side length '\( a \)'. If the coordinates of A, B, C are \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \) respectively, then the value of \( \begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix}^2 \) will be—

(a) \( a^3 \)

(b) \( 2a^3 \)

(c) \( a^4 \)

(d) \( 3a^4 \)

Ans 60.

The area of an equilateral triangle with side length \( a \) is given by \( \text{Area} = \frac{\sqrt{3}}{4} a^2 \).
Also, the area calculated via coordinates is:

$$ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \implies \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 2 \times \text{Area} $$

The determinant in the question can be factored by taking 2 out of the third column:

$$ D = \begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix} = 2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} $$

Substituting the relation for the Area:

$$ D = 2 \times (2 \times \text{Area}) = 4 \times \text{Area} = 4 \left( \frac{\sqrt{3}}{4} a^2 \right) = \sqrt{3} a^2 $$

Squaring this result gives:

$$ D^2 = (\sqrt{3} a^2)^2 = 3a^4 $$

Correct Option: (d) \( 3a^4 \)

Q 61. For what value of \( k \) will the following system of equations be dependent/not have a unique solution?
\( x + y - 2z = 2, \ 2x - 3y + z = 0, \ x - 5y + kz = 1 \)

(a) \( k = 4 \)

(b) \( k \neq 4 \)

(c) \( k = 0, 4 \)

(d) \( k \neq 0 \)

Ans 61.

For a system of 3 linear equations to lack a unique solution (i.e., dependent or inconsistent), the determinant of the coefficient matrix \( \Delta \) must equal zero.

$$ \Delta = \begin{vmatrix} 1 & 1 & -2 \\ 2 & -3 & 1 \\ 1 & -5 & k \end{vmatrix} = 0 $$

Expanding along the first row:

$$ 1(-3k - (-5)) - 1(2k - 1) - 2(-10 - (-3)) = 0 $$

$$ (-3k + 5) - (2k - 1) - 2(-7) = 0 $$

$$ -5k + 5 + 1 + 14 = 0 \implies -5k + 20 = 0 \implies 5k = 20 \implies k = 4 $$

Correct Option: (a) \( k = 4 \)

Q 62. The condition for the system of equations \( a_i x + b_i y + c_i z = d_i \) (\( i = 1, 2, 3 \)) to have a unique solution is—

(a) \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \)

(b) \( \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} \neq 0 \)

(c) \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \neq 0 \)

(d) \( \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} \neq 0 \)

Ans 62.

According to Cramer's Rule, a system of linear equations possesses a unique solution if and only if the determinant of the coefficient matrix is non-zero (\( \Delta \neq 0 \)).

$$ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \neq 0 $$

Correct Option: (c) \( \Delta \neq 0 \)

Q 63. The solution of the equations \( 2x + y + z = 2, \ -x + y + 2z = 0, \ 2x + 3y - z = -2 \) is—

(a) \( x = 1, y = 1, z = -1 \)

(b) \( x = 1, y = 0, z = 0 \)

(c) \( x = 1, y = -1, z = -1 \)

(d) \( x = 1, y = -1, z = 1 \)

Ans 63.

We can quickly verify by substituting the options into the original equations.
Testing Option (d): \( x = 1, y = -1, z = 1 \)

Eq 1: \( 2(1) + (-1) + 1 = 2 - 1 + 1 = 2 \) (True)

Eq 2: \( -(1) + (-1) + 2(1) = -1 - 1 + 2 = 0 \) (True)

Eq 3: \( 2(1) + 3(-1) - 1 = 2 - 3 - 1 = -2 \) (True)

Since these values satisfy all three equations, it is the correct solution.
Correct Option: (d) \( x = 1, y = -1, z = 1 \)

Q 64. The solution of the equations \( \frac{1}{x} + \frac{2}{y} + \frac{1}{z} = \frac{1}{2}, \ \frac{4}{x} + \frac{2}{y} - \frac{3}{z} = \frac{2}{3}, \ \frac{3}{x} - \frac{4}{y} + \frac{4}{z} = \frac{1}{3} \) is—

(a) \( x = 1, y = 1, z = 1 \)

(b) \( x = 6, y = 8, z = 12 \)

(c) \( x = 6, y = 1, z = 12 \)

(d) \( x = -6, y = 8, z = 12 \)

Ans 64.

Testing Option (b) where \( x = 6, y = 8, z = 12 \):

Eq 1: \( \frac{1}{6} + \frac{2}{8} + \frac{1}{12} = \frac{2}{12} + \frac{3}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \) (True)

Eq 2: \( \frac{4}{6} + \frac{2}{8} - \frac{3}{12} = \frac{8}{12} + \frac{3}{12} - \frac{3}{12} = \frac{8}{12} = \frac{2}{3} \) (True)

Eq 3: \( \frac{3}{6} - \frac{4}{8} + \frac{4}{12} = \frac{6}{12} - \frac{6}{12} + \frac{4}{12} = \frac{4}{12} = \frac{1}{3} \) (True)

Correct Option: (b) \( x = 6, y = 8, z = 12 \)

Q 65. If \( A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix} \), then the solutions of the equation \( \det(A - xI) = 0 \) will be—

(a) 1, 1, 5

(b) 1, 5, 5

(c) 2, 3, 2

(d) 2, 3, -2

Ans 65.

The equation \( \det(A - xI) = 0 \) finds the eigenvalues of the matrix. We can use trace and determinant properties:
1. Sum of eigenvalues = Trace(A) (sum of main diagonal elements).

$$ \text{Trace(A)} = 2 + 3 + 2 = 7 $$

2. Product of eigenvalues = \(\det(A)\).

$$ \det(A) = 2(6-2) - 2(2-1) + 1(2-3) = 2(4) - 2(1) - 1 = 8 - 2 - 1 = 5 $$

Checking the options for Sum = 7 and Product = 5:
Option (a) \( 1+1+5 = 7 \) and \( 1 \times 1 \times 5 = 5 \). This is the only matching set.
Correct Option: (a) 1, 1, 5

Q 66. The solutions of the equations \( x + y + z = 1, \ \alpha x + \beta y + \gamma z = k, \ \alpha^2 x + \beta^2 y + \gamma^2 z = k^2 \) where \( [\alpha \neq \beta, \beta \neq \gamma, \gamma \neq \alpha] \) are—

(a) \( x = 1, y = 1, z = 1 \)

(b) \( x = \alpha - \beta, y = \beta - \gamma, z = \gamma - \alpha \)

(c) \( x = - \frac{(\beta-k)(\gamma-k)}{(\alpha-\beta)(\gamma-\alpha)}, \ y = - \frac{(\gamma-k)(\alpha-k)}{(\alpha-\beta)(\beta-\gamma)}, \ z = - \frac{(\alpha-k)(\beta-k)}{(\beta-\gamma)(\gamma-\alpha)} \)

(d) \( x = \frac{(\beta-k)(\gamma-k)}{(\alpha-\beta)(\gamma-\alpha)}, \ y = \dots \)

Ans 66.

Applying Cramer's Rule, the system determinant is a standard Vandermonde determinant:

$$ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \end{vmatrix} = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) $$

For \( x \), replace the first column with constants \( 1, k, k^2 \):

$$ \Delta_x = \begin{vmatrix} 1 & 1 & 1 \\ k & \beta & \gamma \\ k^2 & \beta^2 & \gamma^2 \end{vmatrix} = (k-\beta)(\beta-\gamma)(\gamma-k) $$

Solving for \( x \):

$$ x = \frac{\Delta_x}{\Delta} = \frac{(k-\beta)(\beta-\gamma)(\gamma-k)}{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)} = \frac{(k-\beta)(\gamma-k)}{(\alpha-\beta)(\gamma-\alpha)} $$

To match Option (c)'s formatting exactly, we can pull out a negative sign:

$$ x = - \frac{(\beta-k)(\gamma-k)}{(\alpha-\beta)(\gamma-\alpha)} $$

Similar logic applies for \( y \) and \( z \) respecting the cyclic symmetries.
Correct Option: (c)

Q 67. If \( \Delta_r = \begin{vmatrix} 2 & 3 & 4 \\ 2^r & 2 \cdot 3^r & 3 \cdot 4^r \\ 2(2^n-1) & 3(3^n-1) & 4(4^n-1) \end{vmatrix} \), then the value of \( \sum_{r=1}^n \Delta_r \) will be—

(a) \( 24(2^n-1)(3^n-1)(4^n-1) \)

(b) 0

(c) 1

(d) \( 2^n 3^n 4^n (2^n-1)(3^n-1)(4^n-1) \)

Ans 67.

The summation \( \sum_{r=1}^n \) only applies to the second row because only \( R_2 \) depends on \( r \).
Evaluate the sum for each element in \( R_2 \):

$$ \sum_{r=1}^n 2^r = 2 \left( \frac{2^n - 1}{2 - 1} \right) = 2(2^n - 1) $$

$$ \sum_{r=1}^n 2 \cdot 3^r = 2 \cdot 3 \left( \frac{3^n - 1}{3 - 1} \right) = 3(3^n - 1) $$

$$ \sum_{r=1}^n 3 \cdot 4^r = 3 \cdot 4 \left( \frac{4^n - 1}{4 - 1} \right) = 4(4^n - 1) $$

After summation, the second row becomes identical to the third row:

$$ \sum_{r=1}^n \Delta_r = \begin{vmatrix} 2 & 3 & 4 \\ 2(2^n-1) & 3(3^n-1) & 4(4^n-1) \\ 2(2^n-1) & 3(3^n-1) & 4(4^n-1) \end{vmatrix} $$

Since \( R_2 \) and \( R_3 \) are identical, the value of the determinant is 0.
Correct Option: (b) 0

Q 68. The value of \( \begin{vmatrix} \cos(x+y) & \sin(x+y) & -\cos(x+y) \\ \sin(x-y) & \cos(x-y) & \sin(x-y) \\ \sin 2x & 0 & \sin 2y \end{vmatrix} \) will be—

(a) \( \cos 2(x+y) \)

(b) \( \cos 3(x+y) \)

(c) \( \sin 2(x+y) \)

(d) \( \sin 3(x+y) \)

Ans 68.

Expanding the determinant along the third row (\( R_3 \)):

$$ \Delta = \sin 2x \begin{vmatrix} \sin(x+y) & -\cos(x+y) \\ \cos(x-y) & \sin(x-y) \end{vmatrix} + \sin 2y \begin{vmatrix} \cos(x+y) & \sin(x+y) \\ \sin(x-y) & \cos(x-y) \end{vmatrix} $$

Evaluate the \( 2 \times 2 \) determinants:

First Det: \( \sin(x+y)\sin(x-y) - (-\cos(x+y)\cos(x-y)) = \cos(x+y-x+y) = \cos(2y) \)

Second Det: \( \cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos(x+y+x-y) = \cos(2x) \)

Substitute these back into the expansion:

$$ \Delta = \sin 2x \cos 2y + \sin 2y \cos 2x $$

Apply the compound angle formula \( \sin A \cos B + \cos A \sin B = \sin(A+B) \):

$$ \Delta = \sin(2x + 2y) = \sin 2(x+y) $$

Correct Option: (c) \( \sin 2(x+y) \)