JEE Mathematics Determinants and Cramers Rule MCQs Set 10

Answer: (c) \(-2\)

Explanation:
1. Adding the equations: \((\lambda+2)(x_1+x_2+x_3)=3\).
2. For \(\lambda=-2\) this reads \(0=3\), so the system is inconsistent; for the other values the coefficient determinant \((\lambda+2)(\lambda-1)^2\ne0\) or the system is consistent.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

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

Explanation:
1. Expanding along a row, the determinant is a sum of three \(2\times2\) minors, each of size at most \(2\); the largest value attainable is \(2\), for example \(\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=2\).
2. Swapping two rows gives \(-2\); no \( 0 - 1 \) determinant of order \(3\) exceeds \(2\) in magnitude.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) independent of \(\alpha\), \(\beta\) and \(\gamma\)

Explanation:
1. \((e^{i\theta}+e^{-i\theta})^2=4\cos^2\theta\) and \((e^{i\theta}-e^{-i\theta})^2=-4\sin^2\theta\).
2. So in every row \(C_1-C_2=4\cos^2\theta+4\sin^2\theta=4=C_3\); the columns are dependent and \(\Delta=0\), independent of all three angles.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(A\)

Explanation:
1. Multiply \(A^2=I\) by \(A^{-1}\): \(A=A^{-1}\).
2. Hence \(A^{-1}=A\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) none of these

Explanation:
1. The determinant is a circulant: \(\Delta=\left(s_0+s_1+s_2\right)\left(s_0^2+s_1^2+s_2^2-s_0s_1-s_1s_2-s_2s_0\right)\) with \(s_k=\sin(x+kh)\).
2. For small \(h\), \(s_0+s_1+s_2\approx3\sin x\) and the second factor is \(\approx3h^2\cos^2x\), so \(\Delta/h^2\to9\sin x\cos^2x\), which is not among the first three options.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) \(0\)

Explanation:
1. \(\sin77^\circ=\cos13^\circ\), so \(\sin^213^\circ+\sin^277^\circ=1\), and \(\tan135^\circ=-1\).
2. Every row sums to \(0\), so \(C_1+C_2+C_3\) is the zero column and the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(PAQ\)

Explanation:
1. \(\operatorname{adj}(XYZ)=\operatorname{adj}Z\cdot\operatorname{adj}Y\cdot\operatorname{adj}X\), so the adjoint equals \(\operatorname{adj}(P^{-1})\operatorname{adj}(B)\operatorname{adj}(Q^{-1})\).
2. With \(|P|=|Q|=1\): \(\operatorname{adj}(P^{-1})=P\) and \(\operatorname{adj}(Q^{-1})=Q\), giving \(PAQ\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(0\)

Explanation:
1. Differentiating the first row \(n\) times at \(x=0\): \(\dfrac{d^n}{dx^n}x^{n-1}=0\) (degree below \(n\)), \(\cos x\to\cos\dfrac{n\pi}2\) and \(\dfrac1{x+3}\to\dfrac{(-1)^nn!}{3^{n+1}}\).
2. The first row becomes \(\left(0,\ \cos\frac{n\pi}2,\ \frac{(-1)^nn!}{3^{n+1}}\right)\), identical to the second row, so the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) \(0\)

Explanation:
1. Add \(C_2\) to \(C_3\): the third column becomes \(ab+bc+ca\) in every row.
2. It is then a multiple of the first column, so the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) \(|A|^m\)

Explanation:
1. \(C\) is the transpose of \(\operatorname{adj}A\), so \(|C|=|\operatorname{adj}A|=|A|^{m-1}\).
2. \(|AC|=|A|\cdot|A|^{m-1}=|A|^m\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(a^6\)

Explanation:
1. \(|A|=a^3\).
2. \(|\operatorname{adj}A|=|A|^{2}=a^6\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) none of these

Explanation:
1. The matrix equals \(VV^T\) with \(V=\begin{pmatrix}1&1&1\\\alpha&\beta&\gamma\\\alpha^2&\beta^2&\gamma^2\end{pmatrix}\).
2. So \(\Delta=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2\), which is none of the given power sums in general.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) \(0\)

Explanation:
1. Only the first column depends on \(r\), so the sum is the determinant with first column \(\sum(r-1)\), \(\sum(r-1)^2\), \(\sum(r-1)^3\).
2. Those sums are \(\dfrac{n(n-1)}2\), \(\dfrac{n(n-1)(2n-1)}6\), \(\dfrac{n^2(n-1)^2}4\); with them the determinant vanishes identically (checked symbolically).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(6\)

Explanation:
1. \(|\operatorname{adj}A|=|A|^{n-1}\), so \(3^{n-1}=243=3^5\).
2. Hence \(n-1=5\) and \(n=6\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(n\)

Explanation:
1. Each of the \(n\) rows gets multiplied by \(\lambda\).
2. So \(\det(\lambda A)=\lambda^n\det A\) and \(s=n\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(g(x)+g(-x)=0\)

Explanation:
1. \(e^{kx\log_ea}=a^{kx}\), so replacing \(x\) by \(-x\) swaps the first two columns while \(x^2,x^4\) are unchanged.
2. Swapping two columns changes the sign, so \(g(-x)=-g(x)\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) \(0\)

Explanation:
1. The left side equals \((ax-b)^2+(bx-c)^2+(cx-d)^2\le0\), so \(\dfrac ba=\dfrac cb=\dfrac dc=x\): \(a,b,c\) are in \( G.P \).
2. Then \(\log a,\log b,\log c\) are in \( A.P \). and the first two columns are in \( A.P \). too (differences \(32,13\)), so \(R_3-R_2=R_2-R_1\) and the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(5\)

Explanation:
1. By Pascal's rule \(^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1}\), so \(C_1+C_2=\left({}^{11}C_5,\ {}^{12}C_7,\ {}^{13}C_9\right)\).
2. For \(m=5\) the third column is exactly \(\left({}^{11}C_5,\ {}^{12}C_7,\ {}^{13}C_9\right)\), equal to \(C_1+C_2\), so the determinant is \(0\); the other options give non-zero values.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(0\)

Explanation:
1. If the \( G.P \). has first term \(A\) and ratio \(R\), then \(\log a=\log A+(p-1)\log R\), and similarly for \(b\), \(c\).
2. So the first column is a combination of the second and third columns; the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) purely imaginary

Explanation:
1. Expanding along the first row: \(\Delta=xy\bar z-\bar x\bar yz\).
2. With \(w=xy\bar z\) this is \(w-\bar w=2i\,\operatorname{Im}w\), which is purely imaginary (it is \(0\) only when \(\operatorname{Im}w=0\)).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(\begin{vmatrix}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{vmatrix}=0\)

Explanation:
1. A unique solution means the coefficient determinant \(\Delta\ne0\).
2. By Cramer's rule \(x_0=\Delta_x/\Delta\), so \(x_0=0\) gives \(\Delta_x=0\), the determinant with the first column replaced by \(d\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(64\)

Explanation:
1. \(|\operatorname{adj}A|=|A|^{2}\) for order \(3\).
2. \(=8^2=64\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) \(0\)

Explanation:
1. The logarithms are in \( A.P \). with common difference \(d=\log r\).
2. \(C_2-C_1=C_3-C_2\) (each entry differs by \(d\)), so the columns are dependent and the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) consistent with a unique solution

Explanation:
1. From the first two equations: \(x=-2\), \(y=0\).
2. The third equation gives \(-8+0+8=0\), so it is satisfied: a unique common solution.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(0\)

Explanation:
1. \(\bar\omega=\omega^2\) and \(\omega^2(a+b\omega)=a\omega^2+b\).
2. So \(C_2=\omega^2C_1\) and the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) \(0\)

Explanation:
1. \(a_r=e^{2\pi ir/9}=\zeta^r\) with \(\zeta=e^{2\pi i/9}\).
2. Each row is \(\zeta^3\) times the previous row, so the determinant is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(0\)

Explanation:
1. \(\cot\dfrac A2=\dfrac{s-a}r\), so the middle column equals \(\dfrac{s\cdot1-a}r\), i.e. a combination of the first column and a column of ones.
2. Hence the given determinant equals \(\dfrac sr\begin{vmatrix}a&1&\lambda\\b&1&\mu\\c&1&\gamma\end{vmatrix}=0\), and expanding that determinant gives exactly \(-\left[a(\mu-\gamma)+b(\gamma-\lambda)+c(\lambda-\mu)\right]\), so the expression is \(0\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) \(0\)

Explanation:
1. Replace \(C_3\) by \(C_3-C_1-C_2\): it becomes \((0,0,-(a+2b+c))\).
2. The determinant is \(-(a+2b+c)(ac-b^2)=0\) since \(b^2=ac\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (c) divisible by \(k\)

Explanation:
1. Apply \(C_3\to100C_1+10C_2+C_3\); the new third column holds the three numbers, each divisible by \(k\).
2. Expanding along it shows \(\Delta\) is divisible by \(k\) (not necessarily by \(k^2\)).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) isosceles

Explanation:
1. The determinant is a Vandermonde: \((\sin B-\sin A)(\sin C-\sin B)(\sin C-\sin A)=0\).
2. So two sines are equal; since the angles are less than \(\pi\) and sum to \(\pi\), two angles are equal and the triangle is isosceles.

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(\alpha I\)

Explanation:
1. \(B=\operatorname{adj}A\) and \(A\operatorname{adj}A=|A|I\).
2. So \(AB=\alpha I\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (d) \(\begin{pmatrix}10&0\\0&10\end{pmatrix}\)

Explanation:
1. \(|A|=12-2=10\).
2. \(A\operatorname{adj}A=|A|I=10I\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (a) \(-12\)

Explanation:
1. Take \(2\) out of \(C_1\) and \(3\) out of \(C_2\): factor \(6\). Then take \(2\) out of \(R_2\) and \(-1\) out of \(R_3\): factor \(-2\).
2. Total factor \(6\times(-2)=-12\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).

Answer: (b) \(-1\)

Explanation:
1. \(ab+bc+ca=\dfrac{xy(x-y)+yz(y-z)+zx(z-x)}{(y-z)(z-x)(x-y)}\).
2. The numerator factorises as \((x-y)(x-z)(y-z)=-(x-y)(y-z)(z-x)\), so the ratio is \(-1\).

Teacher's Note:
1. Use row and column operations to create equal, proportional or zero rows or columns before expanding.
2. Keep the standard results handy: \( |adj A| = |A|^{n-1} \) and \( A adj A = |A| I \).