Differentiation (Part-38) | S N De

Determinants MCQ Solutions (Questions 42 to 57)

Q 42. If \( \det(A) = 25 \), then what is \( \det(A^T) \)?

(a) \( \frac{1}{25} \)

(b) 25

(c) 5

(d) \( \frac{1}{5} \)

Ans 42.

According to the properties of determinants, the determinant of a matrix and its transpose are always equal: \( \det(A^T) = \det(A) \).
Given \( \det(A) = 25 \), we have:

$$ \det(A^T) = 25 $$

Correct Option: (b) 25

Q 43. If \( A \) is a matrix of order \( 3 \times 3 \) and \( \det(A) = 3 \), then what is \( \det(3A^T) \)?

(a) 27

(b) 3

(c) \( \frac{1}{3} \)

(d) 81

Ans 43.

For a scalar \( k \) and an \( n \times n \) matrix \( A \), \( \det(kA) = k^n \det(A) \).
Here \( k = 3 \) and \( n = 3 \):

$$ \det(3A^T) = 3^3 \cdot \det(A^T) = 27 \cdot \det(A) $$

$$ = 27 \times 3 = 81 $$

Correct Option: (d) 81

Q 44. Interchange of any two rows of a square matrix \( A \) results in a determinant value that is—

(a) twice of \( |A| \)

(b) equal to \( |A| \)

(c) equal to \( -|A| \)

(d) equal to \( 0 \)

Ans 44.

A fundamental property of determinants states that if any two parallel lines (rows or columns) of a determinant are interchanged, the sign of the determinant changes, but its magnitude remains the same.
Correct Option: (c) equal to \( -|A| \)

Q 45. If \( \begin{vmatrix} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{vmatrix} = k(a+b)(b+c)(c+a) \), then the value of \( k \) is—

(a) 4

(b) 2

(c) 0

(d) 1

Ans 45.

Apply row operation \( R_1 \to R_1 + R_2 + R_3 \), which allows factoring out \( 2(a+b+c) \) and systematically expanding the reduced determinant. The final simplified form equates to \( 2(a+b)(b+c)(c+a) \), making the constant \( k = 2 \).
Correct Option: (b) 2

Q 46. If \( \begin{vmatrix} x & b & c \\ a & y & c \\ a & b & z \end{vmatrix} = p\left(\frac{x}{x-a} + \frac{y}{y-b} + \frac{z}{z-c} - 2\right) \), then the value of \( p \) is—

(a) \( xyz \)

(b) \( abc \)

(c) \( (x-a)(y-b)(z-c) \)

(d) \( (x+a)(y+b)(z+c) \)

Ans 46.

Subtracting row 2 from row 1 and row 3 from row 2 creates factors of \( (x-a) \), \( (y-b) \), and \( (z-c) \). Factoring these out from the columns gives the remaining expression. The scaling coefficient \( p \) corresponds to the product of these differences.
Correct Option: (c) \( (x-a)(y-b)(z-c) \)

Q 47. If \( \begin{vmatrix} 1 & 1 & 1 \\ \sin\alpha & \sin\beta & \sin\gamma \\ \cos\alpha & \cos\beta & \cos\gamma \end{vmatrix} = \lambda \sin\frac{\alpha-\beta}{2} \sin\frac{\beta-\gamma}{2} \sin\frac{\gamma-\alpha}{2} \), then the value of \( \lambda \) is—

(a) 2

(b) -2

(c) 4

(d) -4

Ans 47.

Apply column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \). Expand using trigonometric identity \( \sin X - \sin Y = 2\sin\frac{X-Y}{2}\cos\frac{X+Y}{2} \). The resulting coefficient after full simplification is \( -4 \).
Correct Option: (d) -4

Q 48. If \( \begin{vmatrix} y+z & z+x & x+y \\ z+x & x+y & y+z \\ x+y & y+z & z+x \end{vmatrix} = \lambda \begin{vmatrix} x & y & z \\ y & z & x \\ z & x & y \end{vmatrix} \), then the value of \( \lambda \) is—

(a) 2

(b) -2

(c) 4

(d) -4

Ans 48.

Add columns \( C_2 \) and \( C_3 \) to \( C_1 \) to get \( 2(x+y+z) \) in the first column. Factoring out 2 and simplifying yields exactly the right-hand determinant.
Correct Option: (a) 2

Q 49. If \( \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix} \), then the value of \( x \) is—

(a) \( \pm 3 \)

(b) \( \pm 6 \)

(c) \( \pm 4 \)

(d) \( \pm 5 \)

Ans 49.

Expand the determinants on both sides:

$$ (2x)(x) - (5)(8) = (6)(3) - (-2)(7) $$

$$ 2x^2 - 40 = 18 - (-14) $$

$$ 2x^2 - 40 = 32 \implies 2x^2 = 72 $$

$$ x^2 = 36 \implies x = \pm 6 $$

Correct Option: (b) \( \pm 6 \)

Q 50. If \( \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} = x + iy \), then the value of \( (x, y) \) is—

(a) \( (1, 0) \)

(b) \( (0, 1) \)

(c) \( (0, 0) \)

(d) \( (1, 1) \)

Ans 50.

Expanding the determinant along the first row:

$$ \Delta = 6i(3i^2 - (-3)) - (-3i)(4i - (-20)) + 1(12 - 60i) $$

Substitute \( i^2 = -1 \):

$$ = 6i(-3 + 3) + 3i(4i + 20) + 12 - 60i $$

$$ = 0 + 12i^2 + 60i + 12 - 60i $$

$$ = 12(-1) + 12 = 0 $$

Since \( x + iy = 0 \), comparing real and imaginary parts gives \( x = 0 \) and \( y = 0 \).
Correct Option: (c) \( (0, 0) \)

Q 51. If \( \begin{vmatrix} 1-x & 2 & 3 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = 0 \), then the values of \( x \) will be—

(a) \( 1, -1 \)

(b) \( 0, 1 \)

(c) \( 0, 2 \)

(d) \( 0, -1 \)

Ans 51.

The matrix is upper triangular. The determinant of an upper triangular matrix is the product of its diagonal elements:

$$ (1-x)(x)(x) = 0 \implies x^2(1-x) = 0 $$

This gives \( x = 0 \) and \( 1-x = 0 \implies x = 1 \).
Correct Option: (b) \( 0, 1 \)

Q 52. The solutions of the equation \( \begin{vmatrix} 2-x & 2 & 3 \\ 1 & 3-x & 3 \\ 2 & 1 & 1 \end{vmatrix} = 0 \) will be—

(a) \( 2, 3 \)

(b) \( 2, -3 \)

(c) \( 1, -5 \)

(d) \( -1, 5 \)

Ans 52.

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

$$ \begin{vmatrix} 1-x & -(1-x) & 0 \\ 1 & 3-x & 3 \\ 2 & 1 & 1 \end{vmatrix} = 0 $$

Factor out \( (1-x) \) from \( R_1 \):

$$ (1-x) \begin{vmatrix} 1 & -1 & 0 \\ 1 & 3-x & 3 \\ 2 & 1 & 1 \end{vmatrix} = 0 $$

Expanding the remaining determinant:

$$ (1-x)[1((3-x)(1) - 3(1)) - (-1)(1(1) - 3(2))] = 0 $$

$$ (1-x)[(3-x - 3) + (1 - 6)] = 0 $$

$$ (1-x)(-x - 5) = 0 $$

This yields roots \( x = 1 \) and \( x = -5 \).
Correct Option: (c) \( 1, -5 \)

Q 53. In the determinant \( A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \), the cofactor of \( a_{22} \) is—

(a) \( \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} \)

(b) \( -\begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix} \)

(c) \( \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} \)

(d) \( -\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} \)

Ans 53.

The cofactor \( C_{ij} \) of an element \( a_{ij} \) is \( (-1)^{i+j} M_{ij} \). For \( a_{22} \), we remove row 2 and column 2:

$$ C_{22} = (-1)^{2+2} \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} = \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} $$

Correct Option: (c) \( \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} \)

Q 54. In the determinant \( \Delta = \begin{vmatrix} 2 & 5 & -1 \\ -7 & 3 & 0 \\ 2 & 1 & 5 \end{vmatrix} \), the cofactor of '0' is—

(a) 8

(b) -8

(c) 10

(d) -10

Ans 54.

The element '0' is located at position \( a_{23} \) (row 2, column 3). Its cofactor is:

$$ C_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} $$

$$ = -1 (2(1) - 5(2)) = -1 (2 - 10) = -(-8) = 8 $$

Correct Option: (a) 8

Q 55. The area of the triangle with vertices \( (-2, -7), (0, 9), (1, 3) \) will be—

(a) 16 sq units

(b) 14 sq units

(c) -14 sq units

(d) 0 sq units

Ans 55.

The area of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:

$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$

$$ \text{Area} = \frac{1}{2} \left| -2(9 - 3) + 0(3 - (-7)) + 1(-7 - 9) \right| $$

$$ = \frac{1}{2} \left| -2(6) + 0 + 1(-16) \right| = \frac{1}{2} |-12 - 16| $$

$$ = \frac{1}{2} |-28| = 14 $$

Correct Option: (b) 14 sq units

Q 56. If the coordinates of the vertices of a quadrilateral are \( (1, 1), (0, 5), (-1, -3) \) and \( (7, 0) \), the area of the quadrilateral will be—

(a) 19 sq units

(b) 14 sq units

(c) 20 sq units

(d) 5 sq units

Ans 56.

Applying the Shoelace Formula for area:

$$ \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| $$

$$ = \frac{1}{2} | (1(5) + 0(-3) + (-1)(0) + 7(1)) - (1(0) + 5(-1) + (-3)(7) + 0(1)) | $$

$$ = \frac{1}{2} | (5 + 0 + 0 + 7) - (0 - 5 - 21 + 0) | $$

$$ = \frac{1}{2} | 12 - (-26) | = \frac{1}{2} (38) = 19 $$

Correct Option: (a) 19 sq units

Q 57. The area of the triangle formed by the lines \( y = m_1x + c_1 \), \( y = m_2x + c_2 \), and \( x = 0 \) will be—

(a) \( |m_1m_2| \) sq units

(b) \( \left|\frac{(c_1 - c_2)^2}{m_1 - m_2}\right| \) sq units

(c) \( \frac{1}{2} \left|\frac{(c_1 - c_2)^2}{(m_1 - m_2)}\right| \) sq units

(d) \( \frac{1}{2} \left|\frac{(c_1 - c_2)}{(m_1 - m_2)}\right| \) sq units

Ans 57.

The points of intersection are the vertices of the triangle.
1. Intersection of \( x=0 \) and \( y=m_1x+c_1 \) is \( (0, c_1) \).
2. Intersection of \( x=0 \) and \( y=m_2x+c_2 \) is \( (0, c_2) \).
3. Intersection of \( y=m_1x+c_1 \) and \( y=m_2x+c_2 \):

$$ m_1x + c_1 = m_2x + c_2 \implies x = \frac{c_2 - c_1}{m_1 - m_2} $$

Treating the segment on the y-axis (where \( x=0 \)) as the base, the length of the base is \( |c_1 - c_2| \).
The height is the x-coordinate of the third vertex: \( \left| \frac{c_2 - c_1}{m_1 - m_2} \right| \).

$$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} |c_1 - c_2| \left| \frac{c_1 - c_2}{m_1 - m_2} \right| = \frac{1}{2} \left| \frac{(c_1 - c_2)^2}{m_1 - m_2} \right| $$

Correct Option: (c) \( \frac{1}{2} \left|\frac{(c_1 - c_2)^2}{(m_1 - m_2)}\right| \) sq units