JEE Mathematics Determinants MCQs Set 10

Answer: (c) \( -2(a+b+c) \)

Explanation:
1. The determinant equals \( (a+b+c)^3 \).
2. So \( (x+a+b+c)^2=(a+b+c)^2 \), giving \( x=0 \) or \( x=-2(a+b+c) \).
3. Since \( x\neq0 \), \( x=-2(a+b+c) \).

Teacher's Note:
1. Add all rows to column 1 to pull out the factor (a + b + c).

Answer: (a) exactly three values of \( \lambda \)

Explanation:
1. A non-trivial solution of a homogeneous system needs the coefficient determinant to vanish.
2. \( \begin{vmatrix}1&\lambda&-1\\ \lambda&-1&-1\\ 1&1&-\lambda\end{vmatrix}=\lambda(\lambda-1)(\lambda+1) \), which is zero for \( \lambda=0,1,-1 \).

Teacher's Note:
1. Set the determinant of coefficients equal to zero.

Answer: (a) \( 9 \)

Explanation:
1. The second determinant is the determinant of the cofactor (adjoint) matrix of the first, because each entry is a cofactor of the first determinant.
2. For a \( 3\times3 \) matrix, \( \det(\operatorname{adj}M)=(\det M)^2=3^2=9 \).

Teacher's Note:
1. The determinant of the matrix of cofactors of a 3×3 matrix is the square of the determinant.

Answer: (c) \( 4 \)

Explanation:
1. \( B \) is skew-symmetric of odd order, so \( \det B=0 \). Also \( \det A=(2k+1)^3 \).
2. \( \det(\operatorname{adj}M)=(\det M)^2 \), so \( (2k+1)^6=10^6 \), giving \( 2k+1=10 \) and \( k=4.5 \).
3. Hence \( [k]=4 \).

Teacher's Note:
1. A skew-symmetric matrix of odd order has zero determinant.

Answer: (a) the graph of \( g(x) \) is symmetric about the origin

Explanation:
1. \( e^{x\ln a}=a^x \), so replacing \( x \) by \( -x \) swaps the first two columns and keeps the third (even powers): \( g(-x)=-g(x) \).
2. So \( g \) is odd, symmetric about the origin. Then \( f(x)=g(x)\log\frac{a-x}{a+x} \) is a product of two odd functions, which is even, so (d) is false.

Teacher's Note:
1. Swapping two columns of a determinant reverses its sign, which shows g is odd.

Answer: (b) \( 2 \)

Explanation:
1. Expanding the determinant gives \( \Delta(x)=2x^3-6x^2-2x+10 \).
2. The coefficient of \( x \) is \( -2 \), with absolute value \( 2 \).

Teacher's Note:
1. Expand fully, or differentiate at x = 0.

Answer: (d) \( 0 \)

Explanation:
1. The determinant equals \( (a^3+b^3+c^3-3abc)^2 \).
2. Here \( a+b+c=3 \), \( ab+bc+ca=3 \), so \( a^2+b^2+c^2=3 \) and \( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=3\cdot0=0 \).

Teacher's Note:
1. Use the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Answer: (d) \( \begin{bmatrix}36&-32\\0&4\end{bmatrix} \)

Explanation:
1. \( A\begin{bmatrix}1&2\\0&3\end{bmatrix}=\lambda I \) means \( A=\lambda\begin{bmatrix}1&2\\0&3\end{bmatrix}^{-1}=\lambda\begin{bmatrix}1&-\frac23\\0&\frac13\end{bmatrix} \).
2. \( |3A|=9|A|=108 \) gives \( |A|=12=\frac{\lambda^2}{3} \), so \( \lambda^2=36 \).
3. \( A^2=\lambda^2\begin{bmatrix}1&-\frac89\\0&\frac19\end{bmatrix}=\begin{bmatrix}36&-32\\0&4\end{bmatrix} \).

Teacher's Note:
1. Express A through the inverse of the given matrix and use |kA| = k²|A| for 2×2.

Answer: (d) exists and equals \( -2 \)

Explanation:
1. Expanding: \( f(x)=-x^2\cos x+x^2\tan x \).
2. \( f'(x)=-2x\cos x+x^2\sin x+2x\tan x+x^2\sec^2x \), so \( \frac{f'(x)}{x}\to-2+0+0+0=-2 \).

Teacher's Note:
1. Expand first; then divide the derivative by x term by term.

Answer: (d) \( 0 \)

Explanation:
1. Let \( p=\omega^n \), so \( p^3=1 \). Row 2 is \( (p,p^2,1)=p(1,p,p^2) \), which is \( p \) times row 1.
2. Two proportional rows give \( \Delta=0 \).

Teacher's Note:
1. Look for proportional rows after using ωⁿ cubed = 1.

Answer: (c) \( 9 \)

Explanation:
1. The determinant equals \( -8\alpha^3 \) (it factorises as a product of Vandermonde-type determinants).
2. \( -8\alpha^3=-648\alpha \) gives \( \alpha=0 \) or \( \alpha^2=81 \), i.e. \( \alpha=\pm9 \).
3. Among the options only \( 9 \) qualifies.

Teacher's Note:
1. Expand using the pattern of squares and compare with the right side.

Answer: (a) \( 0 \)

Explanation:
1. Column 1 is \( (b-a)(b,a,c) \) and column 3 is \( (b-a)(c,b,a) \).
2. Column 2 is \( (b-c,\ a-b,\ c-a) \), and \( (b,a,c)-(c,b,a)=(b-c,a-b,c-a) \). So \( (b-a)C_2=C_1-C_3 \): the columns are dependent and the determinant is \( 0 \).

Teacher's Note:
1. Factor (b − a) from the columns and check for dependence.

Answer: (d) \( x^3 \)

Explanation:
1. Adding \( R_2 \) and \( R_3 \) to \( R_1 \) and using \( 1+\omega+\omega^2=0 \) gives \( R_1=(x,x,x) \). Take out \( x \).
2. Then \( C_2\to C_2-C_1 \), \( C_3\to C_3-C_1 \) and expand: the remaining \( 2\times2 \) determinant is \( (x^2+3)-(1-\omega)(1-\omega^2)=(x^2+3)-3=x^2 \).
3. So \( \Delta=x\cdot x^2=x^3 \).

Teacher's Note:
1. Use 1 + ω + ω² = 0 and ω³ = 1 while expanding.

Answer: (a) \( x=\frac13(a+b+c) \)

Explanation:
1. Expanding the second determinant gives \( \Delta=(a-b)(a-c)(b-c)\,(3x-a-b-c) \).
2. This vanishes for distinct \( a,b,c \) when \( x=\frac13(a+b+c) \).

Teacher's Note:
1. Expand with a Vandermonde-type factorisation and set the remaining linear factor to zero.

Answer: (d) \( \frac ab \) is a cube root of \( -1 \)

Explanation:
1. Expanding along the first row: \( a(a^2-0)-b(0-b^2)=a^3+b^3 \).
2. \( a^3+b^3=0 \) gives \( \left(\frac ab\right)^3=-1 \).

Teacher's Note:
1. The determinant of this circulant-like matrix is a³ + b³.

Answer: (d) \( -\frac13(a+b+c) \)

Explanation:
1. Adding all rows to the first gives the factor \( (a+b+c+3x) \).
2. The remaining factor is \( a^2+b^2+c^2-ab-bc-ca \), which is non-zero because \( (b-c)^2\neq(a-b)(c-a) \).
3. So \( x=-\frac13(a+b+c) \).

Teacher's Note:
1. The given inequality says the second factor is not zero.

Answer: (d) \( 0,\pm\sqrt{\frac32(a^2+b^2+c^2)} \)

Explanation:
1. Adding all columns to the first gives \( (a+b+c-x) \) in every entry of column 1, i.e. \( -x \) since \( a+b+c=0 \): so \( x=0 \) is a root.
2. The remaining quadratic is \( x^2=3(a^2+ab+b^2)=\frac32(a^2+b^2+c^2) \) (using \( c=-a-b \)).
3. So \( x=0 \) or \( x=\pm\sqrt{\frac32(a^2+b^2+c^2)} \).

Teacher's Note:
1. Use C₁ → C₁ + C₂ + C₃ and the condition a + b + c = 0.

Answer: (b) \( 25 \)

Explanation:
1. The entries of \( \Delta' \) are the cofactors of \( \Delta \) arranged as the adjoint.
2. \( \det(\operatorname{adj}M)=(\det M)^2=25 \).

Teacher's Note:
1. The determinant of the adjoint of a 3×3 matrix is the square of its determinant.

Answer: (b) \( \pm1 \)

Explanation:
1. Let \( M \) be the matrix. The conditions give \( MM^T=I \).
2. So \( (\det M)^2=1 \) and \( \det M=\pm1 \).

Teacher's Note:
1. Orthonormal rows mean M Mᵀ = I.

Answer: (c) \( 2s^3(s-a)(s-b)(s-c) \)

Explanation:
1. Direct expansion in terms of \( a,b,c \) gives \( \frac{(a+b+c)^3(b+c-a)(c+a-b)(a+b-c)}{32} \).
2. With \( a+b+c=2s \) and \( b+c-a=2(s-a) \) etc.: \( \frac{8s^3\cdot8(s-a)(s-b)(s-c)}{32}=2s^3(s-a)(s-b)(s-c) \).

Teacher's Note:
1. Use the semi-perimeter identities and the known factorisation of this determinant.

Answer: (b) \( \frac{7\pi}{24},\frac{11\pi}{24} \)

Explanation:
1. The matrix is \( I+\mathbf 1w^T \) with \( w=(\sin^2\theta,\cos^2\theta,4\sin4\theta) \), so its determinant is \( 1+w\cdot\mathbf 1=2+4\sin4\theta \).
2. \( \sin4\theta=-\frac12 \) with \( 4\theta\in(0,2\pi) \) gives \( 4\theta=\frac{7\pi}6,\frac{11\pi}6 \), so \( \theta=\frac{7\pi}{24},\frac{11\pi}{24} \).

Teacher's Note:
1. Column additions turn the determinant into a simple expression in sin 4θ.