JEE Mathematics Determinants and Cramers Rule MCQs Set 08

Answer: (b) \( 8 \)

Explanation:
1. Solve the first two equations: \( 2x+3y=8 \) and \( 7x-5y=-3 \). The determinant of coefficients is \( -10-21=-31 \).
2. By Cramer's rule \( x=\dfrac{(8)(-5)-(3)(-3)}{-31}=1 \). Substituting \( x=1 \) in \( 2x+3y=8 \) gives \( y=2 \).
3. The third equation must also hold: \( 4(1)-6(2)+\lambda=0 \), so \( \lambda=8 \).

Teacher's Note:
1. Three equations in two unknowns are consistent only if the solution of two of them satisfies the third.
2. You can also use the condition that the 3 by 3 determinant of the augmented coefficients is zero.

Answer: (b) infinitely many solutions

Explanation:
1. Three lines \( ax+by+c=0 \), \( bx+cy+a=0 \), \( cx+ay+b=0 \) meeting in a common point requires \( \begin{vmatrix} a&b&c\\ b&c&a\\ c&a&b\end{vmatrix}=-(a^{3}+b^{3}+c^{3}-3abc)=0 \), i.e. \( (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0 \).
2. For real a, b, c either \( a+b+c=0 \) or \( a=b=c \).
3. If \( a+b+c=0 \), the entries of the second system become \( -a,-b,-c \) (in the same cyclic pattern), so its determinant is \( a^{3}+b^{3}+c^{3}-3abc=0 \).
4. If \( a=b=c \), all three equations of the second system reduce to \( 2a(x+y+z)=0 \), which also has infinitely many solutions.
5. A homogeneous system with zero determinant has infinitely many solutions.

Teacher's Note:
1. Turn the first condition into a determinant equal to zero and factorise it.
2. A homogeneous system always has the trivial solution, so the only question is whether the determinant vanishes.

Answer: (b) \( 9! \) and \( 0 \)

Explanation:
1. Each arrangement of nine numbers in nine places is a distinct determinant, so \( k=9! \).
2. Pair every arrangement with the one obtained by interchanging the first two rows. The two determinants are negatives of each other.
3. Hence the determinants cancel in pairs and \( \sum \Delta_i=0 \).

Teacher's Note:
1. Interchanging two rows changes the sign of a determinant.
2. Use this to pair the arrangements up.

Answer: (c) (i), (ii) and (iv) only

Explanation:
1. Expanding the determinant (checked symbolically) gives \( (x-y)(y-z)(z-x)(x+y+z)(x^{2}+y^{2}+z^{2}) \).
2. So it is divisible by \( x^{2}+y^{2}+z^{2} \), \( x-y \) and \( x+y+z \).
3. It is not divisible by \( x-y-z \), since that factor does not appear. Hence (i), (ii) and (iv).

Teacher's Note:
1. The determinant vanishes when two of x, y, z are equal, which shows that the differences are factors.
2. Check the remaining factors by comparing degrees: the determinant has degree 6.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. The determinant expands to \( (1-x^{3})^{2}=(x-1)^{2}(x^{2}+x+1)^{2} \).
2. The roots are \( 1,\omega,\omega^{2} \), each repeated twice: three distinct roots in three equal pairs. So (i) is false and (iv) is true.
3. The pair \( x=1 \) is a repeated real root, so (ii) is true.
4. All three roots are cube roots of unity, so each has modulus 1 and (iii) is true.

Teacher's Note:
1. Expand along the first row, or notice that the determinant is a square of \( 1-x^{3} \).
2. Cube roots of unity all have modulus 1.

Answer: (c) (i) and (iii) only

Explanation:
1. Use \( {}^{m}P_{m}=m! \) and \( {}^{m}C_{m}=1 \). The determinant is \( \begin{vmatrix} n&n+1&n+2\\ n!&(n+1)!&(n+2)!\\ 1&1&1\end{vmatrix} \).
2. Apply \( C_3\to C_3-C_2,\ C_2\to C_2-C_1 \): row 1 becomes \( n,1,1 \), row 2 becomes \( n!,\ n\cdot n!,\ (n+1)^{2}n! \), row 3 becomes \( 1,0,0 \).
3. Expanding along row 3 gives \( n!\,[(n+1)^{2}-n]=n!\,(n^{2}+n+1) \).
4. Hence it is divisible by \( n! \) and by \( n^{2}+n+1 \).
5. For \( n=1 \) the value is 3, which is not divisible by \( 2!=(n+1)! \), so (ii) fails.

Teacher's Note:
1. Replace the permutation and combination symbols by factorials and 1 first.
2. Test small values of n to rule out a divisibility claim.

Answer: (d) \( abcx \)

Explanation:
1. Expanding gives \( abc\,x^{2}(x+3) \). (Take a, b, c out of the columns and use \( C_1\to C_1+C_2+C_3 \).)
2. The factor \( abc\,x \) divides it, because \( x^{2}(x+3) \) contains x.
3. It has no factor \( (1+x) \), since \( x\neq -1 \) is not a root of the result, so the other three options fail.

Teacher's Note:
1. Take a, b, c common from the three columns first.
2. Add all the columns to the first one and factorise.

Answer: (d) \( p \)

Explanation:
1. Expand along the first row. The minor of 1 is \( \cos pa\sin(p+d)a-\cos(p+d)a\sin pa=\sin da \).
2. The minor of \( \alpha \) (taken with a minus sign) is \( \cos(p-d)a\sin(p+d)a-\cos(p+d)a\sin(p-d)a=\sin 2da \).
3. The minor of \( \alpha^{2} \) is \( \cos(p-d)a\sin pa-\cos pa\sin(p-d)a=\sin da \).
4. So the determinant equals \( \sin da-\alpha\sin 2da+\alpha^{2}\sin da \), which has no p. It does depend on \( \alpha \), d and a.

Teacher's Note:
1. Use the identity \( \sin(A-B)=\sin A\cos B-\cos A\sin B \) on each 2 by 2 minor.
2. Notice that p cancels in every minor.

Answer: (d) \( 2 \) or \( -2 \)

Explanation:
1. Let the AP be \( a, a+r, a+2r, a+3r \). Substituting and expanding, \( \Delta(x)=-2r^{2} \), a constant independent of x (checked symbolically).
2. Then \( \int_{0}^{2}\Delta\,dx=-4r^{2}=-16 \), so \( r^{2}=4 \).
3. Therefore \( r=2 \) or \( r=-2 \); both are possible.

Teacher's Note:
1. Write all entries in terms of a and the common difference r.
2. Subtract rows to see that the x terms disappear.

Answer: (d) \( \operatorname{Re}(z)=-4 \) and \( \operatorname{Im}(z)=0 \)

Explanation:
1. Put \( p=e^{iA},q=e^{iB},r=e^{iC} \) with \( pqr=e^{i\pi}=-1 \).
2. Since \( pqr=-1 \), we have \( e^{-iC}=1/r=-pq \), \( e^{-iB}=-pr \), \( e^{-iA}=-qr \). The determinant becomes \( \begin{vmatrix} p^{2}&-pq&-pr\\ -pq&q^{2}&-qr\\ -pr&-qr&r^{2} \end{vmatrix} \).
3. Expanding: \( p^{2}(q^{2}r^{2}-q^{2}r^{2})+pq(-pqr^{2}-pqr^{2})-pr(pq^{2}r+pq^{2}r)=-4p^{2}q^{2}r^{2}=-4(pqr)^{2}=-4 \) (confirmed numerically for sample angles).
4. So \( \operatorname{Re}(z)=-4,\ \operatorname{Im}(z)=0 \).

Teacher's Note:
1. Use \( e^{iA}e^{iB}e^{iC}=-1 \) to simplify.
2. Check your answer with a numerical example such as A = 0.7, B = 1.3.

Answer: (c) \( 0 \) and \( 3a \)

Explanation:
1. Add all columns to the first: each entry of the first column becomes \( 3a-x \).
2. Take \( 3a-x \) out and subtract rows. The determinant becomes \( (3a-x)(2x)(2x)=4x^{2}(3a-x) \).
3. Setting it to zero gives \( x=0 \) (double) or \( x=3a \).

Teacher's Note:
1. Add all three columns to the first column; the first column becomes constant.
2. Then subtract the first row from the others.

Answer: (b) \( c=2 \) or \( c=4 \) only

Explanation:
1. For consistency the augmented determinant \( \begin{vmatrix} 1&1&1\\ c+2&c+4&6\\ (c+2)^{2}&(c+4)^{2}&36\end{vmatrix} \) must vanish.
2. It is a Vandermonde determinant in \( c+2,\,c+4,\,6 \) and equals \( 2(c-2)(c-4) \).
3. So \( c=2 \) or \( c=4 \). Check: for \( c=2 \) the solution is \( x=0,\ y=1 \); for \( c=4 \) it is \( x=1,\ y=0 \), and both satisfy all three equations.

Teacher's Note:
1. A system of three equations in two unknowns is consistent if the augmented determinant is zero and a common solution exists.
2. Verify each candidate value of c by direct substitution.

Answer: (c) (iii) and (iv) only

Explanation:
1. The determinant of coefficients is \( 7\lambda+35 \), which is zero only for \( \lambda=-5 \).
2. For \( \lambda\neq -5 \) the homogeneous system has only the trivial solution, so (iv) is true; (i) fails at \( \lambda=-5 \).
3. For \( \lambda=-5 \) the system has infinitely many nontrivial solutions, so (iii) is true.
4. A homogeneous system never has exactly one nontrivial solution, so (ii) is false.

Teacher's Note:
1. Homogeneous systems always have the trivial solution.
2. A nontrivial solution exists exactly when the determinant is zero.