JEE Mathematics Determinants and Cramers Rule MCQs Set 05

Answer: (c) 4

Explanation:
1. Let \( c=\operatorname{cosec}\alpha \). Expanding: \( c(4c^2-1)-1\cdot2c=4c^3-3c \).
2. With \( t=\tan\frac\alpha2 \), \( \operatorname{cosec}\alpha=\frac{1+t^2}{2t}=\frac s2 \) where \( s=t+\frac1t \).
3. \( 4c^3-3c=\frac{s^3-3s}2=\frac12\left(t^3+\frac1{t^3}\right)=\frac12\left(\tan^3\frac\alpha2+\cot^3\frac\alpha2\right) \). So \( \lambda=\mu=2 \) and \( \lambda+\mu=4 \).

Teacher's Note:
1. Reduce cosec alpha to tan(alpha/2) + cot(alpha/2).
2. The identity s^3 - 3s = t^3 + 1/t^3 finishes it.

Answer: (a) 0

Explanation:
1. In each row of \( \Delta_1 \) the three entries add to \( \log\frac ab+\log\frac bc+\log\frac ca=\log1=0 \).
2. Adding all columns to the first gives a zero column, so \( \Delta_1=0 \).
3. \( \Delta_2=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ne0 \) for distinct positive numbers, so the ratio is 0.

Teacher's Note:
1. Sum the columns to expose a zero column.
2. The denominator is non-zero because a, b, c are positive and distinct.

Answer: (c) isosceles

Explanation:
1. Let \( x=\tan\frac A2,\ y=\tan\frac B2,\ z=\tan\frac C2 \) with \( xy+yz+zx=1 \).
2. The third row can be written \( (S-x,\ S-y,\ S-z) \) with \( S=x+y+z \); subtracting S times row 1 turns it into \( (-x,-y,-z) \).
3. The determinant becomes \( -\begin{vmatrix}1&1&1\\1/x&1/y&1/z\\x&y&z\end{vmatrix} \) and multiplying columns by x, y, z gives a multiple of \( \begin{vmatrix}x&y&z\\1&1&1\\x^2&y^2&z^2\end{vmatrix} \), i.e. \( (x-y)(y-z)(z-x) \) up to sign.
4. It vanishes only if two of x, y, z are equal, so two angles are equal: the triangle is isosceles.

Teacher's Note:
1. Express everything in tan of half angles.
2. A Vandermonde-type determinant vanishes when two variables are equal.

Answer: (d) The graph of g(x) is symmetrical about the origin

Explanation:
1. \( e^{kx\log_ea}=a^{kx} \), so the first two columns are \( (a^{-x},a^{-3x},a^{-5x}) \) and \( (a^x,a^{3x},a^{5x}) \).
2. Replacing x by -x interchanges the first two columns and leaves the even column \( (x^2,x^4,1) \) unchanged, so \( g(-x)=-g(x) \): g is odd.
3. An odd function is symmetrical about the origin; its fourth derivative at 0 is 0; and \( \log\frac{a-x}{a+x} \) is also odd, so the product is even, not odd.

Teacher's Note:
1. Swapping two columns changes the sign of a determinant.
2. Odd times odd is even.

Answer: (a) Statement I is true, Statement II is true, and Statement II is a correct explanation for Statement I

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&-2&3\\1&-3&4\\-1&1&-2\end{vmatrix}=0 \).
2. Adding equations 1 and 3 gives \( -y+z=k-1 \), while equation 2 minus equation 1 gives \( -y+z=2 \). These agree only if \( k=3 \), so there is no solution for \( k\ne3 \) (Statement I true).
3. The determinant in Statement II equals \( 3-k \), which is non-zero for \( k\ne3 \). It is the determinant obtained by replacing a column of the coefficient matrix with the constants, and a non-zero such determinant with a zero coefficient determinant means inconsistency. Hence Statement II explains Statement I.

Teacher's Note:
1. With zero main determinant, a non-zero auxiliary determinant signals no solution.
2. Check by eliminating variables directly.

Answer: (d) 512

Explanation:
1. Each of the 9 entries has 2 choices.
2. Total = \( 2^9=512 \).

Teacher's Note:
1. Count independent choices and multiply.
2. The question counts matrices, not distinct values.

Answer: (a) 3

Explanation:
1. The trace is the sum of three diagonal entries, each at most 1.
2. The maximum is \( 1+1+1=3 \), for example the matrix with all entries 1.

Teacher's Note:
1. Each diagonal entry can be chosen independently.
2. Off-diagonal entries do not affect the trace.

Answer: (b) 6

Explanation:
1. Apply \( R_2\to R_2-10R_3 \) to get the second row \( (-1,0,0) \) and expand along it.
2. Then \( R_2\to R_2-2R_1 \) in the remaining 2 x 2 determinant gives \( D=2x^2+3x+1 \).
3. \( a+b+c=2+3+1=6 \) (also \( D(1)=6 \)).

Teacher's Note:
1. a + b + c is the value of the polynomial at x = 1.
2. Reduce the determinant first, then substitute.

Answer: (b) \( (p-1)^2 \)

Explanation:
1. Trace \( 2a \) is not divisible by the odd prime p exactly when \( a\ne0 \): \( p-1 \) choices.
2. \( \det A=a^2-bc\equiv0 \) means \( bc\equiv a^2 \).
3. For each non-zero a and each of the \( p-1 \) non-zero values of b, c is determined uniquely.
4. Total \( (p-1)(p-1)=(p-1)^2 \).

Teacher's Note:
1. Trace 2a is non-zero mod p whenever a is non-zero, since p is odd.
2. Count solutions of bc = a^2 for b non-zero.

Answer: (d) 1

Explanation:
1. Since \( 1+S_n=1^n+\alpha^n+\beta^n \), the matrix is \( VV^T \) where \( V=\begin{bmatrix}1&1&1\\1&\alpha&\beta\\1&\alpha^2&\beta^2\end{bmatrix} \) (the (i, j) entry of \( VV^T \) is \( 1+\alpha^{i+j-2}+\beta^{i+j-2} \) up to the indexing of the rows as powers 0, 1, 2).
2. So \( \Delta=(\det V)^2=\Delta_1^2 \) and the ratio is 1.

Teacher's Note:
1. Power-sum matrices are Gram matrices of Vandermonde matrices.
2. det(V V^T) = (det V)^2.

Answer: (c) \( \frac{(a+b+c)^2(b^2-4ac)}{a^4} \)

Explanation:
1. \( \Delta=[(\alpha-1)(\beta-1)(\alpha-\beta)]^2 \).
2. \( (\alpha-1)(\beta-1)=\alpha\beta-(\alpha+\beta)+1=\frac ca+\frac ba+1=\frac{a+b+c}{a} \).
3. \( (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{b^2-4ac}{a^2} \).
4. Hence \( \Delta=\frac{(a+b+c)^2}{a^2}\cdot\frac{b^2-4ac}{a^2}=\frac{(a+b+c)^2(b^2-4ac)}{a^4} \).

Teacher's Note:
1. Use sum and product of roots.
2. (alpha - beta)^2 is the discriminant over a^2.

Answer: (b) less than 3

Explanation:
1. A non-zero solution needs \( \begin{vmatrix}1&-c&-b\\-c&1&-a\\-b&-a&1\end{vmatrix}=0 \), i.e. \( a^2+b^2+c^2+2abc=1 \).
2. The relation forces each of a, b, c to lie in [-1, 1] (as in the cosines of the angles of a triangle), and not all of them have absolute value 1 when one is a proper fraction.
3. Hence \( a^2+b^2+c^2<3 \). Values such as \( a=b=-0.95,\ c\approx-0.805 \) give a sum about 2.45, so it is neither always less than 2 nor greater than 2.

Teacher's Note:
1. The determinant condition is a^2 + b^2 + c^2 + 2abc = 1.
2. Test one admissible triple to discard the other options.

Answer: (b) 7

Explanation:
1. Expanding the determinant gives \( D_i=(n+1)^3-2i\,n(n+2) \).
2. Sum over i = 1 to n: \( n(n+1)^3-2n(n+2)\cdot\frac{n(n+1)}{2}=n(n+1)\left[(n+1)^2-n(n+2)\right]=n(n+1) \).
3. \( n(n+1)=56 \) gives n = 7.

Teacher's Note:
1. Only the factor containing i changes with the summation index.
2. (n + 1)^2 - n(n + 2) = 1.

Answer: (b) 0

Explanation:
1. For each row, \( \sum_ja_{ij}c_{ij}=\Delta \), so \( f(x)=3\Delta(x) \).
2. Expanding along the third row: \( \Delta=(x+4)\cdot(-(2x-2)(x-1))=-2(x+4)(x-1)^2 \).
3. \( f(x)=-6(x+4)(x-1)^2 \) with \( f'(x)=-6(x-1)(3x+7) \).
4. On [-3, 18]: f decreases on [-3, -7/3], increases to 0 at x = 1, then decreases.
5. Comparing \( f(-3)=-96 \), \( f(1)=0 \) and \( f(18)<0 \), the greatest value is 0.

Teacher's Note:
1. Each row-wise sum of element times cofactor equals the determinant.
2. Check the end points as well as the stationary points.

Answer: (a) 0

Explanation:
1. For a triangle \( \tan A+\tan B+\tan C=\tan A\tan B\tan C=p \) and AM-GM gives \( \sum\tan A\tan B\ge9 \), with equality only when \( A=B=C=60^\circ \).
2. So the triangle is equilateral and \( x=y=z=\sqrt3 \).
3. Rows of the determinant are then identical, so the value is 0. (Also \( \det=-(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0 \).)

Teacher's Note:
1. For an acute triangle, the sum of the products of tangents is at least 9.
2. Equality pins down the equilateral triangle.

Answer: (a) 3

Explanation:
1. Taking out x, y, z from the rows and then from columns: \( \Delta=1+x^3+y^3+z^3 \) (the matrix is \( I+uv^T \) with \( u=(x^2,y^2,z^2)^T \) and \( v=(x,y,z)^T \), so \( \det=1+v^Tu \)).
2. \( 1+x^3+y^3+z^3=11 \) gives \( x^3+y^3+z^3=10 \).
3. The only positive integer cubes summing to 10 are 1, 1, 8, so (x, y, z) is a permutation of (1, 1, 2): 3 solutions.

Teacher's Note:
1. Recognise identity + rank one, whose determinant is 1 + v.u.
2. Test cubes 1, 8 for the sum 10.

Answer: (b) 0

Explanation:
1. \( (1+x+x^2)^8 \) is a palindromic polynomial of degree 16, so \( a_k=a_{16-k} \).
2. Hence \( a_{13}=a_3,\ a_{12}=a_4,\ a_{11}=a_5 \).
3. The third column equals the first column, so the determinant is 0.

Teacher's Note:
1. Palindromic polynomial means coefficients read the same forwards and backwards.
2. Equal columns make a determinant zero.

Answer: (b) 2

Explanation:
1. \( a_4=5,\ a_5=4 \): the reciprocals are in A.P. with first values \( \frac15,\frac14 \), common difference \( \frac1{20} \), so \( \frac1{a_n}=\frac{n}{20} \) and \( a_n=\frac{20}n \).
2. \( D=\begin{vmatrix}20&10&\frac{20}3\\5&4&\frac{10}3\\\frac{20}7&\frac52&\frac{20}9\end{vmatrix}=\frac{50}{21}\approx2.38 \).
3. \( [D]=2 \).

Teacher's Note:
1. Write the HP as the reciprocal of an AP.
2. Evaluate the determinant exactly and then take the integer part.

Answer: (c) 1 : 1

Explanation:
1. Put \( a_n=A+(n-1)d \). In each determinant subtract the second column from the third column and then the (shifted) differences become constant \( d \).
2. Expanding with these reductions (and checking symbolically) gives \( \Delta_1=-2d^4 \) and \( \Delta_2=-2d^4 \).
3. So the ratio is 1 : 1.

Teacher's Note:
1. Substitute the A.P. in terms of A and d.
2. Both determinants reduce to the same multiple of d to the fourth power.

Answer: (a) 2

Explanation:
1. Row i of Q is row i of P multiplied by \( 2^i \) and column j gets a further factor \( 2^j \).
2. Total factor \( 2^{(1+2+3)}\cdot2^{(1+2+3)}=2^{12} \).
3. \( \det Q=2^{12}\cdot2=2^{13} \), so \( k=2 \).

Teacher's Note:
1. Each scalar multiple of a row or column multiplies the determinant.
2. Add exponents.

Answer: (b) 1

Explanation:
1. Expanding the determinant gives \( z^3 \) (using \( 1+\omega+\omega^2=0 \) and \( \omega^3=1 \)).
2. \( z^3=0 \) has the single distinct root \( z=0 \).

Teacher's Note:
1. Add all rows to expose the common factors.
2. Powers of omega reduce modulo 3.

Answer: (a) 4

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

Teacher's Note:
1. A skew-symmetric matrix of odd order is singular.
2. det(adj A) = (det A)^2 for a 3 x 3 matrix.