Determinants MCQ Solutions (Questions 30 to 41)
Q 30. The value of \( \begin{vmatrix} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^2+c^2 & c^2+a^2 & a^2+b^2 \end{vmatrix} \) is—
Ans 30.
Applying column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
$$ \begin{vmatrix} 1 & 0 & 0 \\ b+c & a-b & a-c \\ b^2+c^2 & a^2-b^2 & a^2-c^2 \end{vmatrix} $$
Factor out \( (a-b) \) from column 2 and \( (a-c) \) from column 3:
$$ (a-b)(a-c) \begin{vmatrix} 1 & 0 & 0 \\ b+c & 1 & 1 \\ b^2+c^2 & a+b & a+c \end{vmatrix} $$
Applying column operation \( C_3 \to C_3 - C_2 \):
$$ (a-b)(a-c) \begin{vmatrix} 1 & 0 & 0 \\ b+c & 1 & 0 \\ b^2+c^2 & a+b & c-b \end{vmatrix} $$
Expanding along row 1:
$$ = (a-b)(a-c)(1)(1(c-b) - 0) = (a-b)(a-c)(c-b) $$
Rewriting to standard cyclic form:
$$ = (b-c)(c-a)(a-b) $$
Correct Option: (b) \( (b-c)(c-a)(a-b) \)
Q 31. The value of \( \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^3 & \beta^3 & \gamma^3 \end{vmatrix} \) is—
Ans 31.
Applying column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
$$ \begin{vmatrix} 1 & 0 & 0 \\ \alpha & \beta-\alpha & \gamma-\alpha \\ \alpha^3 & \beta^3-\alpha^3 & \gamma^3-\alpha^3 \end{vmatrix} $$
Factor out \( (\beta-\alpha) \) from column 2 and \( (\gamma-\alpha) \) from column 3:
$$ = (\beta-\alpha)(\gamma-\alpha) \begin{vmatrix} 1 & 0 & 0 \\ \alpha & 1 & 1 \\ \alpha^3 & \beta^2+\alpha\beta+\alpha^2 & \gamma^2+\alpha\gamma+\alpha^2 \end{vmatrix} $$
Applying column operation \( C_3 \to C_3 - C_2 \):
$$ = (\beta-\alpha)(\gamma-\alpha) \begin{vmatrix} 1 & 0 & 0 \\ \alpha & 1 & 0 \\ \alpha^3 & \beta^2+\alpha\beta+\alpha^2 & \gamma^2-\beta^2+\alpha(\gamma-\beta) \end{vmatrix} $$
Note that \( \gamma^2-\beta^2+\alpha(\gamma-\beta) = (\gamma-\beta)(\gamma+\beta+\alpha) \). Factoring out \( (\gamma-\beta) \):
$$ = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) $$
Correct Option: (b) \( (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) \)
Q 32. The value of \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \) is—
Ans 32.
Applying column operation \( C_1 \to C_1 + C_2 + C_3 \):
$$ \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} $$
Factor out \( (a+b+c) \) from column 1:
$$ (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} $$
Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
$$ (a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} $$
Expanding along column 1:
$$ = (a+b+c) \left[ (c-b)(b-c) - (a-c)(a-b) \right] $$
$$ = - (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) = -(a^3 + b^3 + c^3 - 3abc) $$
Correct Option: (b) \( -(a^3 + b^3 + c^3 - 3abc) \)
Q 33. If \( A + B + C = \pi \), then the value of \( \begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix} \) will be—
Ans 33.
Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
$$ \begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B - \sin^2 A & \cot B - \cot A & 0 \\ \sin^2 C - \sin^2 A & \cot C - \cot A & 0 \end{vmatrix} $$
Expanding along column 3:
$$ = 1 \cdot \left[ (\sin^2 B - \sin^2 A)(\cot C - \cot A) - (\sin^2 C - \sin^2 A)(\cot B - \cot A) \right] $$
Using trigonometric simplifications given \( A+B+C = \pi \), the cross terms cancel out and the determinant evaluates identically to zero.
Correct Option: (d) 0
Q 34. The value of \( \begin{vmatrix} b^2+c^2 & ab & ac \\ ab & c^2+a^2 & bc \\ ac & bc & a^2+b^2 \end{vmatrix} \) is—
Ans 34.
Multiply rows 1, 2, and 3 by \( a, b, c \) respectively, and divide the determinant by \( abc \):
$$ = \frac{1}{abc} \begin{vmatrix} a(b^2+c^2) & a^2b & a^2c \\ ab^2 & b(c^2+a^2) & b^2c \\ ac^2 & bc^2 & c(a^2+b^2) \end{vmatrix} $$
Take out \( a, b, c \) common from columns 1, 2, and 3 respectively:
$$ = \frac{abc}{abc} \begin{vmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{vmatrix} $$
Applying column operation \( C_1 \to C_1 + C_2 + C_3 \):
$$ = \begin{vmatrix} 2(b^2+c^2) & a^2 & a^2 \\ 2(c^2+a^2) & c^2+a^2 & b^2 \\ 2(a^2+b^2) & c^2 & a^2+b^2 \end{vmatrix} $$
Further reduction yields:
$$ 4a^2b^2c^2 $$
Correct Option: (b) \( 4a^2b^2c^2 \)
Q 35. The value of \( \begin{vmatrix} \cos x & 1 & 0 \\ 1 & 2\cos x & 1 \\ 0 & 1 & 2\cos x \end{vmatrix} \) is—
Ans 35.
Expanding along row 1:
$$ = \cos x \begin{vmatrix} 2\cos x & 1 \\ 1 & 2\cos x \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 0 & 2\cos x \end{vmatrix} + 0 $$
$$ = \cos x (4\cos^2 x - 1) - 1(2\cos x) $$
$$ = 4\cos^3 x - \cos x - 2\cos x = 4\cos^3 x - 3\cos x $$
Using triple angle identity for cosine:
$$ = \cos 3x $$
Correct Option: (d) \( \cos 3x \)
Q 36. The value of \( \begin{vmatrix} 3! & 4! & 5! \\ 4! & 5! & 6! \\ 5! & 6! & 7! \end{vmatrix} \) is—
Ans 36.
Factor out \( 3!, 4!, 5! \) from columns 1, 2, and 3 respectively:
$$ = 3! \cdot 4! \cdot 5! \begin{vmatrix} 1 & 4 & 5 \cdot 4 \\ 4 & 5 \cdot 4 & 6 \cdot 5 \cdot 4 \\ 5 \cdot 4 & 6 \cdot 5 \cdot 4 & 7 \cdot 6 \cdot 5 \cdot 4 \end{vmatrix} $$
Simplifying the determinant evaluates to:
$$ 2 \times 3! \times 4! \times 5! $$
Correct Option: (a) \( 2 \times 3! \times 4! \times 5! \)
Q 37. If in a G.P., \( t_p = a, t_q = b, t_r = c \), then the value of \( \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} \) is—
Ans 37.
Let the first term be \( A \) and common ratio be \( R \). Then:
$$ a = A R^{p-1}, \quad b = A R^{q-1}, \quad c = A R^{r-1} $$
Taking logarithms:
$$ \log a = \log A + (p-1)\log R $$
Substituting into the determinant and performing row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \), we get proportional rows leading to a determinant value of:
$$ 0 $$
Correct Option: (b) 0
Q 38. If \( a \neq p, b \neq q, c \neq r \) and \( \begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} = 0 \), which of the following is true?
Ans 38.
Applying row operations and expanding the given determinant equation yields the linear relation among fractions, which simplifies to:
$$ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 2 $$
Correct Option: (c) \( \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 2 \)
Q 39. The value of \( \begin{vmatrix} a^2 & a^3 & a^4 \\ a^3 & a^4 & a^5 \\ a^4 & a^5 & a^6 \end{vmatrix} \) is—
Ans 39.
Factor out \( a^2 \) from row 1, \( a^3 \) from row 2, and \( a^4 \) from row 3:
$$ = a^9 \begin{vmatrix} 1 & a & a^2 \\ 1 & a & a^2 \\ 1 & a & a^2 \end{vmatrix} $$
Since all three rows are identical, the determinant equals \( 0 \):
$$ = a^9 \cdot 0 = 0 $$
Correct Option: (c) 0
Q 40. The value of \( \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ca \\ \frac{1}{c} & c^2 & ab \end{vmatrix} \) is—
Ans 40.
Multiply row 1 by \( a \), row 2 by \( b \), and row 3 by \( c \), dividing the whole determinant by \( abc \):
Correct Option: (d) 0
$$ = \frac{1}{abc} \begin{vmatrix} 1 & a^3 & abc \\ 1 & b^3 & abc \\ 1 & c^3 & abc \end{vmatrix} $$
Factor out \( abc \) from column 3:
$$ = \frac{abc}{abc} \begin{vmatrix} 1 & a^3 & 1 \\ 1 & b^3 & 1 \\ 1 & c^3 & 1 \end{vmatrix} $$
Since column 1 and column 3 are identical, the determinant equals \( 0 \).Correct Option: (d) 0
Q 41. If \( a, b, c \) are the three roots of the equation \( x^3 + px + q = 0 \), then the value of \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \) is—
Ans 41.
From equation \( x^3 + px + q = 0 \), we know the sum of roots \( a+b+c = 0 \), sum of product of two roots \( ab+bc+ca = p \), and product of roots \( abc = -q \).
From Question 32, the value of the determinant is:
From Question 32, the value of the determinant is:
$$ -(a^3 + b^3 + c^3 - 3abc) $$
Since \( a+b+c = 0 \), \( a^3+b^3+c^3 = 3abc \). Substituting this in:
$$ -(3abc - 3abc) = 0 $$
Correct Option: (a) 0
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