JEE Mathematics Determinants and Cramers Rule MCQs Set 07

Answer: (b) 0

Explanation:
1. The determinant is skew-symmetric, since each entry equals the negative of its mirror entry and the diagonal is zero.
2. A skew-symmetric determinant of odd order equals zero.
3. Hence the value is 0.

Teacher's Note:
1. Check the symmetry of the entries about the diagonal.
2. Odd-order skew-symmetric determinants always vanish.

Answer: (a) \( x^2(x+3) \)

Explanation:
1. Add columns 2 and 3 to column 1: every entry of column 1 becomes \( x+3 \).
2. Take out \( (x+3) \) and subtract row 1 from rows 2 and 3 to get a triangular determinant with diagonal entries 1, x, x.
3. The value is \( (x+3)\cdot x^2 \).

Teacher's Note:
1. Add all columns to one column to get a common factor.
2. Then reduce to a triangular determinant.

Answer: (d) \( x=0,\ y=0 \)

Explanation:
1. Expand along row 1: \( 6i(3i\cdot i+3)+3i(4i+20)+1\cdot(12-60i) \).
2. \( 3i\cdot i=-3 \), so the first term is \( 6i(0)=0 \). The second is \( 3i(4i+20)=-12+60i \).
3. Sum: \( 0+(-12+60i)+(12-60i)=0 \), so \( x=0,\ y=0 \).

Teacher's Note:
1. Expand along the row with the most convenient entries and keep track of the powers of i.

Answer: (b) x, y, z are in GP

Explanation:
1. Apply \( C_1\to C_1-pC_2-C_3 \): the first column becomes \( (0,\,0,\,-(p^2x+2py+z)) \). Expanding along it gives \( \Delta=-(p^2x+2py+z)(xz-y^2) \).
2. For this to vanish for every real p, the factor \( xz-y^2 \) must be zero, since \( p^2x+2py+z \) is not identically zero.
3. \( y^2=xz \) means x, y, z are in GP.

Teacher's Note:
1. Use a column operation to cancel the terms containing p.
2. The condition must hold for all p, so the p-free factor has to vanish.

Answer: (c) 0

Explanation:
1. Divide column 1 by a, column 2 by b, column 3 by c (and multiply the determinant by abc): the first row becomes \( \tfrac1a,\tfrac1b,\tfrac1c \).
2. Since a, b, c are in HP, \( \tfrac1a=\alpha+(p-1)\delta \), and similarly for q and r, so row 1 \( =\delta\,(\text{row 2})+(\alpha-\delta)(\text{row 3}) \).
3. Row 1 is a combination of the other rows, so the determinant is 0.

Teacher's Note:
1. Reciprocals of an HP form an AP, which is linear in the position number.
2. A row that is a combination of the others makes the determinant zero.

Answer: (c) 5

Explanation:
1. Pascal's rule \( {}^{n}C_r+{}^{n}C_{r+1}={}^{n+1}C_{r+1} \) applied to the first two columns of each row gives \( {}^{11}C_5,{}^{12}C_7,{}^{13}C_9 \), then \( C_2\to C_2+C_1 \).
2. This column equals the third column exactly when \( m=5 \) (its entries are \( {}^{11}C_5,{}^{12}C_7,{}^{13}C_9 \)).
3. Two identical columns give zero. Direct evaluation confirms the determinant is nonzero for \( m=4 \) and \( m=6 \).

Teacher's Note:
1. Pascal's rule can merge adjacent binomial coefficients in a column operation.
2. Look for the value of m that makes two columns identical.

Answer: (a) 0

Explanation:
1. Since \( \log_yx=\dfrac1{\log_xy} \) and \( \log_yz=\dfrac{\log_xz}{\log_xy} \), row 2 equals \( \log_yx \) times row 1.
2. Similarly row 3 equals \( \log_zx \) times row 1.
3. The rows are proportional, so the determinant is 0.

Teacher's Note:
1. Use the change of base formula to write every entry as a ratio of logs to base x.

Answer: (a) 0

Explanation:
1. Expand along row 1: \( \cos C\,(0+\tan A\sin B)-\tan A\,(\sin B\cos C-0)+0 \).
2. This equals \( \cos C\tan A\sin B-\tan A\sin B\cos C \).
3. The two terms cancel, so the value is 0.

Teacher's Note:
1. Expand along the first row and watch the sign of each cofactor.

Answer: (c) 0

Explanation:
1. Apply \( R_1\to R_1-R_2,\ R_2\to R_2-R_3 \).
2. Row 1 becomes \( (x-y,\ (x-y)(x+y+z),\ 0) \) and row 2 becomes \( (y-z,\ (y-z)(x+y+z),\ 0) \).
3. These two rows are proportional (each is a multiple of \( (1,\,x+y+z,\,0) \)), so the determinant is 0.

Teacher's Note:
1. Subtract adjacent rows and factor out x - y and y - z.

Answer: (d) 0

Explanation:
1. Use \( 1+\omega+\omega^2=0 \), so \( 1+\omega^2=-\omega \).
2. Expanding the determinant and simplifying every term with \( \omega^3=1 \) and \( \omega^2=-1-\omega \), all terms cancel.
3. The value is 0.

Teacher's Note:
1. Replace powers of omega using omega cubed equals 1 and the sum of the cube roots of unity equals 0.

Answer: (a) 0

Explanation:
1. Check \( C_1+C_2 \): first row \( 1+x \); second \( 2x+x(x-1)=x(x+1) \); third \( 3x(x-1)+x(x-1)(x-2)=x(x-1)(x+1)=x(x^2-1) \).
2. So \( C_3=C_1+C_2 \) for every x.
3. Hence \( f(x)=0 \) for all x, and \( f(100)=0 \).

Teacher's Note:
1. Test whether one column is the sum of the other two.

Answer: (a) \( x=5 \)

Explanation:
1. \( \Delta_1=10(49+20)-4(119-16)+3(-85-28)=690-412-339=-61 \).
2. Expanding \( \Delta_2 \) along column 2 gives \( \Delta_2=166-21x \).
3. \( \Delta_1+\Delta_2=105-21x=0 \), so \( x=5 \).

Teacher's Note:
1. Evaluate the constant determinant first.
2. Expand the second along the column that contains x.

Answer: (b) 0

Explanation:
1. The constant term t is the value of the determinant at \( \lambda=0 \).
2. At \( \lambda=0 \) it is \( \begin{vmatrix}0&-1&3\\1&0&-4\\-3&4&0\end{vmatrix} \).
3. Expanding: \( 0-(-1)(0-12)+3(4-0)=-12+12=0 \), so \( t=0 \).

Teacher's Note:
1. Put lambda equal to zero to read off the constant term of the identity.

Answer: (a) 3

Explanation:
1. Expanding fully, the determinant equals \( -x^4-x^3+2x^2+3x+1 \).
2. Comparing with the right side, the coefficient of x is \( \lambda=3 \).

Teacher's Note:
1. Expand carefully and compare the coefficients of the powers of x.

Answer: (a) 4

Explanation:
1. The determinant has degree 3 and the three given linear factors already account for degree 3, so the remaining factor is a constant k.
2. Put \( a=b=c=1 \): the determinant is \( \begin{vmatrix}-2&2&2\\2&-2&2\\2&2&-2\end{vmatrix}=32 \) while \( (a+b)(b+c)(c+a)=8 \).
3. So \( k=\tfrac{32}{8}=4 \).

Teacher's Note:
1. Compare degrees to see that the remaining factor is a constant.
2. Fix the constant with a convenient substitution or by expansion.

Answer: (c) -1

Explanation:
1. The constant term is the value of the determinant at \( \sin x=0 \), i.e. \( x=0 \).
2. At \( x=0 \) the entries are \( \cos0=1,\ \sin^20=0,\ \cos^20=1 \), giving \( \begin{vmatrix}1&0&1\\0&1&1\\1&1&1\end{vmatrix} \).
3. Its value is \( 1(1-1)-0+1(0-1)=-1 \).

Teacher's Note:
1. Setting sin x = 0 gives the constant term of a polynomial in sin x.

Answer: (d) \( -\tfrac12 \)

Explanation:
1. Apply \( C_3\to C_3-C_1+C_2 \): the third column becomes \( (0,\,0,\,1) \). Expanding along it, \( \Delta(x)=\cos x-\cos x(1+\sin x)=-\sin x\cos x=-\tfrac12\sin2x \).
2. So \( \int_0^{\pi/2}\Delta\,dx=-\tfrac12\int_0^{\pi/2}\sin2x\,dx \).
3. \( \int_0^{\pi/2}\sin2x\,dx=1 \), so the integral is \( -\tfrac12 \).

Teacher's Note:
1. Reduce the determinant to a simple trigonometric function before integrating.

Answer: (d) 0

Explanation:
1. The fourth roots of unity are \( 1,i,-1,-i \), and their sum is 0.
2. Add all columns to the first column: every entry of column 1 becomes \( \alpha+\beta+\gamma+\delta=0 \).
3. A column of zeros makes the determinant 0.

Teacher's Note:
1. A circulant determinant vanishes when the sum of each row is zero.

Answer: (b) 5!

Explanation:
1. All entries above the main diagonal are zero, so the determinant is lower triangular.
2. Its value is the product of the diagonal entries \( 1\cdot2\cdot3\cdot4\cdot5 \).
3. That equals 120, i.e. 5!.

Teacher's Note:
1. A triangular determinant equals the product of its diagonal entries.

Answer: (a) AP

Explanation:
1. A homogeneous system has nontrivial solutions when the coefficient determinant is zero.
2. \( \begin{vmatrix}a&4&1\\b&3&1\\c&2&1\end{vmatrix}=a-2b+c \).
3. So \( 2b=a+c \), which means a, b, c are in AP.

Teacher's Note:
1. Set the coefficient determinant equal to zero for a nontrivial solution.

Answer: (b) \( \lambda=5 \)

Explanation:
1. The coefficient determinant is \( 2(-4+1)+1(2-\lambda)+1(-1+2\lambda)=\lambda-5 \).
2. It is zero only for \( \lambda=5 \).
3. For \( \lambda=5 \) the first two equations are independent, so the solution set is a line, an infinite number of nontrivial solutions.

Teacher's Note:
1. Compute the determinant of coefficients and then confirm the rank is 2 for the critical value.