Differentiation (Part-38) | S N De

Determinants MCQ Solutions (Questions 15 to 29)

Q 15. If \( \begin{vmatrix} a & b & a-b \\ b & c & b-c \\ 2 & 1 & 0 \end{vmatrix} = 0 \), then \( a, b, \) and \( c \) will be in—

(a) A.P.

(b) G.P.

(c) H.P.

(d) None of these

Ans 15.

Applying column operation \( C_3 \to C_3 - (C_1 - C_2) \):

$$ \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ 2 & 1 & 0 - (2 - 1) \end{vmatrix} = 0 $$

$$ \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ 2 & 1 & -1 \end{vmatrix} = 0 $$

Expanding along column 3 (\( C_3 \)):

$$ (-1) \cdot (ac - b^2) = 0 \implies b^2 - ac = 0 \implies b^2 = ac $$

Since \( b^2 = ac \), \( a, b, \) and \( c \) are in Geometric Progression (G.P.).
Correct Option: (b) G.P.

Q 16. If \( xyz \neq 0 \) and \( \begin{vmatrix} 1+x & 1 & 1 \\ 1 & y+1 & 1 \\ 1 & 1 & z+1 \end{vmatrix} = 0 \), then \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \)

(a) 0

(b) 1

(c) -1

(d) \( \frac{1}{2} \)

Ans 16.

Factor out \( x \), \( y \), and \( z \) from rows 1, 2, and 3 respectively:

$$ xyz \begin{vmatrix} \frac{1}{x} + 1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{vmatrix} = 0 $$

Applying row operation \( R_1 \to R_1 + R_2 + R_3 \):

$$ xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{vmatrix} = 0 $$

Evaluating the remaining determinant gives 1:

$$ xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \cdot 1 = 0 $$

Since \( xyz \neq 0 \):

$$ 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1 $$

Correct Option: (c) -1

Q 17. The value of the determinant \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix} \) is—

(a) \( x + y \)

(b) \( xy \)

(c) \( x + y + 1 \)

(d) \( xy + 1 \)

Ans 17.

Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):

$$ \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix} $$

This is an upper triangular matrix, so its determinant is the product of its diagonal elements:

$$ 1 \cdot x \cdot y = xy $$

Correct Option: (b) \( xy \)

Q 18. \( \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = \)

(a) \( a+b+c \)

(b) \( (a+b+c)^2 \)

(c) 0

(d) None of these

Ans 18.

Applying column operation \( C_1 \to C_1 + C_2 + C_3 \):

$$ \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{vmatrix} $$

Factor out \( 2(a+b+c) \) from column 1:

$$ 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{vmatrix} $$

Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):

$$ 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{vmatrix} $$

Expanding along column 1:

$$ 2(a+b+c) \left[ (b-c)(c-b) - (c-a)(b-a) \right] = 2(a+b+c)(3abc - a^3 - b^3 - c^3) $$

Since this standard cyclic symmetric value does not equal options (a), (b), or (c):
Correct Option: (d) None of these

Q 19. The value of \( \begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} \) is—

(a) \( a^2b^2c^2 \)

(b) \( a^2 + b^2 + c^2 \)

(c) \( 4a^2b^2c^2 \)

(d) \( 9(a^2 + b^2 + c^2) \)

Ans 19.

Factor out \( a, b, c \) from columns 1, 2, and 3 respectively:

$$ abc \begin{vmatrix} -a & a & a \\ b & -b & b \\ c & c & -c \end{vmatrix} $$

Now factor out \( a, b, c \) from rows 1, 2, and 3 respectively:

$$ a^2b^2c^2 \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} $$

Evaluate the \( 3 \times 3 \) numerical determinant:

$$ \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = -1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4 $$

Thus, the value is \( 4a^2b^2c^2 \).
Correct Option: (c) \( 4a^2b^2c^2 \)

Q 20. The value of the matrix determinant \( \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \) is—

(a) 0

(b) 1

(c) 4

(d) -4

Ans 20.

Expanding along row 1 (\( R_1 \)):

$$ = -1((-1)(-1) - (1)(1)) - 1((1)(-1) - (1)(1)) + 1((1)(1) - (-1)(1)) $$

$$ = -1(1 - 1) - 1(-1 - 1) + 1(1 + 1) $$

$$ = 0 + 2 + 2 = 4 $$

Correct Option: (c) 4

Q 21. The value of the determinant \( \begin{vmatrix} a & 0 & c \\ a & b & 0 \\ 0 & b & c \end{vmatrix} \) will be—

(a) \( abc \)

(b) \( 2abc \)

(c) \( 3abc \)

(d) 0

Ans 21.

Expanding along row 1 (\( R_1 \)):

$$ = a(bc - 0) - 0 + c(ab - 0) $$

$$ = abc + abc = 2abc $$

Correct Option: (b) \( 2abc \)

Q 22. If \( a, b, c \) are in Arithmetic Progression (A.P.), then the value of the determinant \( \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} \) will be—

(a) \( x^3 \)

(b) \( x^2 \)

(c) 0

(d) 1

Ans 22.

Since \( a, b, c \) are in A.P., we have \( a + c = 2b \implies a - 2b + c = 0 \).
Apply row operation \( R_1 \to R_1 - 2R_2 + R_3 \):

$$ \text{First entry of } R_1 = (x+1) - 2(x+2) + (x+3) = 0 $$

$$ \text{Second entry of } R_1 = (x+2) - 2(x+3) + (x+4) = 0 $$

$$ \text{Third entry of } R_1 = (x+a) - 2(x+b) + (x+c) = a - 2b + c = 0 $$

The determinant becomes:

$$ \begin{vmatrix} 0 & 0 & 0 \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} = 0 $$

Since row 1 is entirely zeroes, the value of the determinant is \( 0 \).
Correct Option: (c) 0

Q 23. The value of \( \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sec\theta + \tan\theta & 1 \\ 2 & 3 & \sec\theta - \tan\theta + 2 \end{vmatrix} \) will be—

(a) 0

(b) 1

(c) -1

(d) \( 2(\sec\theta + \tan\theta) \)

Ans 23.

Applying column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):

$$ \begin{vmatrix} 1 & 0 & 0 \\ 0 & \sec\theta + \tan\theta & 1 \\ 2 & 1 & \sec\theta - \tan\theta \end{vmatrix} $$

Expanding along row 1 (\( R_1 \)):

$$ = 1 \cdot \left[ (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) - 1 \cdot 1 \right] $$

Using trigonometric identity \( \sec^2\theta - \tan^2\theta = 1 \):

$$ = (\sec^2\theta - \tan^2\theta) - 1 = 1 - 1 = 0 $$

Correct Option: (a) 0

Q 24. The value of \( \begin{vmatrix} 1 & 0 & 0 \\ 0 & \log_x y & 1 \\ 0 & 1 & \log_y x \end{vmatrix} \) is—

(a) 0

(b) 1

(c) -1

(d) \( \log(xy) \)

Ans 24.

Expanding along row 1 (\( R_1 \)):

$$ = 1 \cdot \left( (\log_x y)(\log_y x) - 1 \cdot 1 \right) $$

Using the change of base rule \( \log_x y \cdot \log_y x = 1 \):

$$ = 1 - 1 = 0 $$

Correct Option: (a) 0

Q 25. The value of \( \begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} \) is—

(a) \( (a^2+b^2) - (c^2+d^2) \)

(b) \( a^2 + b^2 + c^2 + d^2 \)

(c) \( a + b + c + d \)

(d) \( (a+b) - (c+d) \)

Ans 25.

Expand the \( 2 \times 2 \) determinant:

$$ = (a+ib)(a-ib) - (c+id)(-c+id) $$

$$ = (a^2 - i^2b^2) - (i^2d^2 - c^2) $$

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

$$ = (a^2 + b^2) - (-d^2 - c^2) = a^2 + b^2 + c^2 + d^2 $$

Correct Option: (b) \( a^2 + b^2 + c^2 + d^2 \)

Q 26. The value of \( \begin{vmatrix} \beta+\gamma & \alpha & 1 \\ \gamma+\alpha & \beta & 1 \\ \alpha+\beta & \gamma & 1 \end{vmatrix} \) will be—

(a) \( 1 + \alpha + \beta + \gamma \)

(b) \( (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) \)

(c) 0

(d) 1

Ans 26.

Applying column operation \( C_1 \to C_1 + C_2 \):

$$ \begin{vmatrix} \alpha+\beta+\gamma & \alpha & 1 \\ \alpha+\beta+\gamma & \beta & 1 \\ \alpha+\beta+\gamma & \gamma & 1 \end{vmatrix} $$

Factor out \( \alpha+\beta+\gamma \) from column 1:

$$ (\alpha+\beta+\gamma) \begin{vmatrix} 1 & \alpha & 1 \\ 1 & \beta & 1 \\ 1 & \gamma & 1 \end{vmatrix} $$

Since column 1 and column 3 are identical, the determinant equals \( 0 \):

$$ (\alpha+\beta+\gamma) \cdot 0 = 0 $$

Correct Option: (c) 0

Q 27. If \( a + b + c = 0 \), then the value of \( \begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix} \) is—

(a) \( -(a-b)(b-c)(c-a) \)

(b) \( abc(a-b)(b-c)(c-a) \)

(c) 0

(d) \( 8abc \)

Ans 27.

Since \( a+b+c = 0 \), we have:

$$ b+c = -a, \quad c+a = -b, \quad a+b = -c $$

Substitute these expressions into column 2:

$$ \begin{vmatrix} a & -a & a^2 \\ b & -b & b^2 \\ c & -c & c^2 \end{vmatrix} $$

Taking out factor \( -1 \) from column 2:

$$ -1 \begin{vmatrix} a & a & a^2 \\ b & b & b^2 \\ c & c & c^2 \end{vmatrix} $$

Since column 1 and column 2 are identical, the value of the determinant is \( 0 \):

$$ -1 \times 0 = 0 $$

Correct Option: (c) 0

Q 28. The value of \( \begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} \) is—

(a) 2

(b) 0

(c) -1

(d) \( \log_{xyz} xyz \)

Ans 28.

Change base using natural logarithm: \( \log_a b = \frac{\ln b}{\ln a} \).
The determinant becomes:

$$ \begin{vmatrix} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\[6pt] \frac{\ln x}{\ln y} & 1 & \frac{\ln z}{\ln y} \\[6pt] \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 1 \end{vmatrix} $$

Factor out \( \frac{1}{\ln x} \), \( \frac{1}{\ln y} \), and \( \frac{1}{\ln z} \) from rows 1, 2, and 3:

$$ \frac{1}{\ln x \cdot \ln y \cdot \ln z} \begin{vmatrix} \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \end{vmatrix} $$

Since all three rows are identical, the determinant is \( 0 \).
Correct Option: (b) 0

Q 29. The value of \( \begin{vmatrix} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{vmatrix} \) is—

(a) \( -(a^3 + b^3)^2 \)

(b) \( (a + b)^2 \)

(c) \( -(a + b)^2 \)

(d) \( (a^3 + b^3)^2 \)

Ans 29.

Applying column operation \( C_1 \to C_1 + C_2 + C_3 \):

$$ \text{First column elements} = 2ab + a^2 + b^2 = (a+b)^2 $$

Factor out \( (a+b)^2 \) from column 1:

$$ (a+b)^2 \begin{vmatrix} 1 & a^2 & b^2 \\ 1 & b^2 & 2ab \\ 1 & 2ab & a^2 \end{vmatrix} $$

Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):

$$ (a+b)^2 \begin{vmatrix} 1 & a^2 & b^2 \\ 0 & b^2-a^2 & 2ab-b^2 \\ 0 & 2ab-a^2 & a^2-b^2 \end{vmatrix} $$

Expanding along column 1 gives:

$$ (a+b)^2 \left[ (b^2-a^2)(a^2-b^2) - (2ab-b^2)(2ab-a^2) \right] $$

$$ = (a+b)^2 \left[ -(a^2-b^2)^2 - ab(2b-a)(2a-b) \right] = -(a^3 + b^3)^2 $$

Correct Option: (a) \( -(a^3 + b^3)^2 \)