JEE Mathematics Determinants MCQs Set 11

Answer: (c) \( \frac1{16} \)

Explanation:
1. \( \det(ABA^T)=(\det A)^2\det B=8 \) and \( \det(AB^{-1})=\frac{\det A}{\det B}=8 \).
2. So \( \det A=8\det B \) and \( 64(\det B)^3=8 \), giving \( \det B=\frac12 \), \( \det A=4 \).
3. \( \det(BA^{-1}B^T)=\frac{(\det B)^2}{\det A}=\frac{1/4}{4}=\frac1{16} \).

Teacher's Note:
1. Use det(XY) = det X det Y, det(Xᵀ) = det X and det(X⁻¹) = 1/det X.

Answer: (d) \( \left(\frac32,3\right] \)

Explanation:
1. Expanding: \( \det A=2+2\sin^2\theta \).
2. For \( \theta\in\left(\frac{3\pi}4,\frac{5\pi}4\right) \), \( \sin^2\theta\in\left[0,\frac12\right] \), so \( \det A\in[2,3] \).
3. This set lies inside \( \left(\frac32,3\right] \) and inside none of the other intervals.

Teacher's Note:
1. Expand along a row and find the range of sin²θ on the given interval.

Answer: (b) \( 2\sqrt3 \)

Explanation:
1. \( \det A=2(2b^2+2-b^2)-b(2b-b)+1(b^2-b^2-1)=b^2+3 \) (computed by expansion).
2. \( \frac{\det A}{b}=b+\frac3b\ge2\sqrt3 \) by AM-GM, with equality at \( b=\sqrt3 \).

Teacher's Note:
1. After expansion the ratio is b + 3/b, which AM-GM bounds below.

Answer: (b) \( 24 \)

Explanation:
1. \( R_2\to R_2-2R_1 \) gives \( (x-1,\ x-2,\ x+1) \) and \( R_3\to R_3-R_1 \) gives \( (x+3,\ x-2,\ x+1) \).
2. Then \( R_3\to R_3-R_2 \) gives \( (4,0,0) \). Expanding along this row: \( 4\begin{vmatrix}x+1&x-2\\x-2&x+1\end{vmatrix}=4(6x-3)=24x-12 \), so \( a=24 \).

Teacher's Note:
1. Row operations remove the x² terms before expanding.

Answer: (c) \( 4\lambda^2 \)

Explanation:
1. \( R_2-R_1 \) has entries \( 2\lambda a+\lambda^2,\ldots \) and \( R_3-R_1 \) has entries \( -2\lambda a+\lambda^2,\ldots \).
2. Their sum is \( 2\lambda^2(1,1,1) \) and their difference is \( 4\lambda(a,b,c) \). Replacing the two rows by their sum and difference multiplies the determinant by \( -2 \).
3. So the determinant is \( -\frac12\cdot2\lambda^2\cdot4\lambda\begin{vmatrix}a^2&b^2&c^2\\1&1&1\\a&b&c\end{vmatrix}=4\lambda^3\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{vmatrix} \), so \( k\lambda=4\lambda^3 \) and \( k=4\lambda^2 \).

Teacher's Note:
1. Subtract the first row from the others and combine rows to separate the factor λ³.

Answer: (c) \( -4-2\sqrt3 \)

Explanation:
1. Expanding, the determinant is \( \cos^3x-\sin^3x \), which is zero when \( \tan x=1 \).
2. In \( [0,2\pi] \): \( x=\frac\pi4 \) and \( \frac{5\pi}4 \).
3. \( \tan\frac{7\pi}{12}=-(2+\sqrt3) \) and \( \tan\left(\frac\pi3+\frac{5\pi}4\right)=\tan\frac{19\pi}{12}=-(2+\sqrt3) \), so the sum is \( -4-2\sqrt3 \).

Teacher's Note:
1. There are two solutions in [0, 2π]; add the tangent value for both.

Answer: (a) \( n\pi+(-1)^n\frac\pi6 \)

Explanation:
1. \( |A_k|=\sin^kx(5-4)-2^{-k}(5-4)=\sin^kx-2^{-k} \).
2. \( \sum_{k\ge0}\sin^kx=\frac1{1-\sin x} \) and \( \sum 2^{-k}=2 \). So \( \frac{1}{1-\sin x}=2 \) and \( \sin x=\frac12 \).
3. \( x=n\pi+(-1)^n\frac\pi6 \).

Teacher's Note:
1. Sum the two geometric series after expanding the determinant.

Answer: (a) \( a=0 \)

Explanation:
1. The constant term is the value at \( x=0 \).
2. At \( x=0 \) the determinant is \( \begin{vmatrix}1&0&1\\ 1&0&1\\ 0&0&1\end{vmatrix}=0 \) (a column of zeros), so \( a=0 \).

Teacher's Note:
1. Put x = 0 to read off the constant term.

Answer: (a) \( [50,250] \)

Explanation:
1. The matrix is \( 5I+\mathbf 1w^T \) with \( w=(\sin^2x,\cos^2x,4\sin2x) \), where \( \mathbf 1 \) is the column of ones. For such a matrix \( \det=5^2\,(5+w\cdot\mathbf 1) \).
2. \( w\cdot\mathbf1=1+4\sin2x \), so \( f(x)=25(6+4\sin2x)=150+100\sin2x \).
3. Since \( -1\le\sin2x\le1 \), \( f(x)\in[50,250] \).

Teacher's Note:
1. Reduce the determinant to a linear function of sin 2x.

Answer: (c) \( 0 \)

Explanation:
1. \( \sin\alpha+i\cos\alpha=ie^{-i\alpha} \) and \( \sin\beta+i\cos\beta=ie^{-i\beta} \).
2. The determinant is \( e^{i\alpha}\cdot ie^{-i\alpha}-e^{i\beta}\cdot ie^{-i\beta}=i-i=0 \), so the real part is \( 0 \).

Teacher's Note:
1. Write each entry in exponential form so the products simplify.

Answer: (b) \( \frac12 \)

Explanation:
1. The columns of the right-hand determinant are \( b+c \), \( c+a \), \( a+b \) (in each row), i.e. the left-hand columns multiplied by the matrix \( \begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix} \), whose determinant is \( 2 \).
2. So the right-hand determinant is \( 2\times \) the left-hand one, and \( k=\frac12 \).

Teacher's Note:
1. Reduce the right-hand determinant with column operations to twice the left-hand one.

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

Explanation:
1. Expanding fully gives \( (a+b+c)(a^2+b^2+c^2) \).
2. For a check, \( a=b=c=1 \) gives determinant \( \begin{vmatrix}1&0&2\\2&1&0\\0&2&1\end{vmatrix}=9=3\cdot3 \).

Teacher's Note:
1. Substitute simple numbers to rule out the other options as a quick check.

Answer: (d) none of these

Explanation:
1. \( (a^x+a^{-x})^2-(a^x-a^{-x})^2=4 \), so \( C_1-C_2=4C_3 \).
2. Then column 1 is a combination of columns 2 and 3, so the determinant is \( 0 \), which is none of the listed expressions.

Teacher's Note:
1. Look at C₁ − C₂ − 4C₃ to see that the columns are dependent.

Answer: (c) isosceles

Explanation:
1. The determinant equals \( 4(a-b)(b-c)(a-c) \).
2. It is zero when at least two of \( a,b,c \) are equal, so the triangle is isosceles (not necessarily equilateral).

Teacher's Note:
1. Reduce by R₂ − R₁ and R₃ − R₁ to a Vandermonde-type determinant.

Answer: (d) \( \pm1 \)

Explanation:
1. Replace \( C_1,C_2 \) by \( C_1+C_2=(1+x)(a_i+b_i) \) and \( C_1-C_2=(1-x)(a_i-b_i) \); this multiplies the determinant by \( -2 \).
2. So \( -2\Delta_2=(1-x^2)\cdot\det[a+b,\ a-b,\ c]=(1-x^2)(-2)\det[a,b,c] \), giving \( \Delta_2=(1-x^2)\det[a,b,c] \).
3. Since \( \det[a,b,c]\neq0 \), \( 1-x^2=0 \) and \( x=\pm1 \).

Teacher's Note:
1. Apply C₁ → C₁ + C₂-type operations to take out the factor (1 − x²).

Answer: (b) purely imaginary

Explanation:
1. \( a=\omega^2 \), so \( a^2=\omega \). Expanding: \( (a^2-a^4)-(a-a^2)+(a^2-a)=3(a^2-a)=3(\omega-\omega^2) \).
2. \( \omega-\omega^2=i\sqrt3 \), so the value is \( 3\sqrt3\,i \), purely imaginary.

Teacher's Note:
1. Identify a as ω² and simplify using 1 + ω + ω² = 0.

Answer: (a) \( 0 \)

Explanation:
1. \( \log2x=\log2+\log x \), so \( R_2-R_1=\log2\,(1,1,1) \); similarly \( R_3-R_1=\log3\,(1,1,1) \).
2. These two rows are proportional, so \( \Delta=0 \).

Teacher's Note:
1. Subtract the first row from the others.

Answer: (a) \( 0 \)

Explanation:
1. The first determinant is \( 2abc \), so its square is \( 4a^2b^2c^2 \).
2. The second determinant is \( 4a^2b^2c^2 \) (it equals the product of the first with its transpose).
3. The difference is \( 0 \).

Teacher's Note:
1. The second matrix is the product of the first with its transpose.

Answer: (b) \( 2s^3 \)

Explanation:
1. Add all rows to the first row: each entry becomes \( 2s \), so \( \Delta=2s\begin{vmatrix}1&1&1\\c&s+a&b\\c&a&s+b\end{vmatrix} \).
2. Subtracting the first column from the others gives \( 2s\cdot s\cdot s=2s^3 \).

Teacher's Note:
1. Add the rows to produce a common factor, then reduce to a triangular form.

Answer: (a) \( 0 \)

Explanation:
1. The matrix equals \( uu^T+vv^T \) where \( u=(\cos\alpha,\cos\beta,\cos\gamma)^T \) and \( v=(\sin\alpha,\sin\beta,\sin\gamma)^T \), because \( \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta \).
2. A \( 3\times3 \) matrix of rank at most 2 has determinant \( 0 \).

Teacher's Note:
1. Write the entries as sums of two products to see the matrix has rank at most 2.