JEE Mathematics Determinants and Cramers Rule MCQs Set 09

Answer: (c) \(\triangle ABC\text{ is isosceles}\)

Explanation:
1. \(R_2-R_3\) gives \(4(a,b,c)\), and \(R_2+R_3-2R_1\) gives \(2(1,1,1)\).
2. So the determinant is \(4(a-b)(b-c)(a-c)\) up to sign; it vanishes when two sides are equal: isosceles.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \(0\)

Explanation:
1. Expanding: \(\Delta=a^2\sin^2A-b^2\sin^2A+2bc\sin^2A\cos A-c^2\sin^2A\).
2. \(=\sin^2A\,(a^2-b^2-c^2+2bc\cos A)=0\) by the cosine rule.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \(\text{can not be determined}\)

Explanation:
1. \(C_2-C_1\) and \(C_3-C_2\) reduce it to the Vandermonde determinant \((\sin Q-\sin P)(\sin R-\sin P)(\sin R-\sin Q)\).
2. The sines are distinct, so it is non-zero, but its sign depends on the order of the angles; it can not be determined.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \(\text{is independent of both }a\text{ and }n\)

Explanation:
1. Only the first row depends on \(r\), and it is linear in \(r\): \(\Delta_r=\dfrac{(2r-n)(n-1)(2a-3n+4)}4\).
2. \(\sum_{r=1}^{n-1}(2r-n)=n(n-1)-n(n-1)=0\), so the sum is 0.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) 65

Explanation:
1. \(a^2+b^2+c^2+ab+bc+ca=\dfrac12[(a+b)^2+(b+c)^2+(c+a)^2]\le0\) forces \(a=b=c=0\).
2. The determinant becomes \(\begin{vmatrix}4&0&1\\1&4&0\\0&1&4\end{vmatrix}=4(16)+1=65\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) 25

Explanation:
1. \(|B|=|A|^2=25\) and \(AB=|A|I=5I\), so \(|AB|=125\).
2. \(|A^{-1}|\,|(AB)^T|=\dfrac15\cdot125=25\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) one

Explanation:
1. The determinant of coefficients is \(7\lambda+35\), zero only for \(\lambda=-5\).
2. Since \(t^2\pm3t+4>0\) for all real \(t\), \(\lambda>0\); the determinant is non-zero and the solution is unique.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (c) \(\text{one value of }t\)

Explanation:
1. The coefficient determinant equals \(-4(2t+1)\).
2. It vanishes only for \(t=-\dfrac12\): one value.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \(\alpha^{-1}=-(a^{-1}+b^{-1}+c^{-1})\)

Explanation:
1. The coefficient determinant is \(abc+\alpha(ab+bc+ca)\).
2. Setting it to zero and dividing by \(\alpha abc\): \(\dfrac1\alpha=-\left(\dfrac1a+\dfrac1b+\dfrac1c\right)\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \(b\in\left(-\dfrac34,3\right)\)

Explanation:
1. The first two give \(x=\dfrac{1+c}3,\ y=\dfrac{2-c}3\). Substituting in the third: \(b(5-4c)=3c\), so \(b=\dfrac{3c}{5-4c}\).
2. \(\dfrac{db}{dc}=\dfrac{15}{(5-4c)^2}>0\); as \(c\to-\infty\), \(b\to-\dfrac34\), and at \(c=1\), \(b=3\). So \(b\in\left(-\dfrac34,3\right)\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \(\text{only }x=1,\ y=1,\ z=1\)

Explanation:
1. The coefficient determinant is \(\begin{vmatrix}1&1&1\\2&1&2\\1&-1&3\end{vmatrix}=1(3+2)-1(6-2)+1(-2-1)=-2\ne0\).
2. So there is a unique solution; \((1,1,1)\) satisfies all three equations.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (c) \(D'=D(1+pqr)\)

Explanation:
1. Write \(D'=D\cdot\begin{vmatrix}1&0&r\\p&1&0\\0&q&1\end{vmatrix}\) (product of determinants).
2. The second determinant is \(1(1-0)-0+r(pq)=1+pqr\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \(\begin{bmatrix}68&0&0\\0&68&0\\0&0&68\end{bmatrix}\)

Explanation:
1. \(A(\operatorname{adj}A)=|A|I\).
2. \(|A|=xyz-8x-4y-3z+28=60-20+28=68\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) 20

Explanation:
1. \(S\) has \(3\cdot3\cdot3=27\) matrices. Singular ones satisfy \(a_{11}^2=a_{12}a_{21}\).
2. \(a_{11}=0\): products zero in 5 ways; \(a_{11}=1\): product 1 in 1 way; \(a_{11}=2\): product 4 in 1 way. Singular: 7.
3. Non-singular: \(27-7=20\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) \(0.20\)

Explanation:
1. \(A=3I\), \(|A|=27\), and \(|\operatorname{adj}(\operatorname{adj}A)|=|A|^4=3^{12}=531441\).
2. \(531441=5\cdot106288+1\), so the fractional part of \(\dfrac{531441}5\) is \(0.2\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) 4

Explanation:
1. Expanding along the first row: \(f(x)=x(-5x\tan x+x^3\sin2x)-\sin x\,(5x^3+2x^4)+\cos x\,(x^2\sin2x+2x\tan x)\).
2. The lowest-order term is \(\cos x\cdot2x\tan x\approx2x^2\), so \(f(x)=2x^2+O(x^3)\) and \(f'(x)=4x+O(x^2)\).
3. Hence \(\lim_{x\to0}\dfrac{f'(x)}x=4\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) 0

Explanation:
1. The matrix of power sums is \(VV^T\) with \(V=\begin{bmatrix}\alpha&\beta\\\alpha^2&\beta^2\\\alpha^3&\beta^3\end{bmatrix}\) of size \(3\times2\).
2. Its rank is at most 2, so the determinant is 0.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (c) 0

Explanation:
1. With \(u=1+x\) and \(w_j=u^j\), the entry in row \(i\), column \(j\) is \(w_j^{\,i+1}\): the rows are \(w_j^2,w_j^3,w_j^4\).
2. The determinant is \(w_1^2w_2^2w_3^2\,(w_2-w_1)(w_3-w_1)(w_3-w_2)\), which contains \((u^2-u)(u^3-u)(u^3-u^2)\), i.e. the factor \((u-1)^3=x^3\). So the coefficient of \(x\) is 0.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) 2

Explanation:
1. The determinant is a Vandermonde determinant in \(x,x^2,x^3\): \((x^2-x)(x^3-x)(x^3-x^2)=x^4(x-1)^3(x+1)\).
2. \(x^4\left[(x-1)^3(x+1)-3\right]=0\): \(x=0\) or \((x-1)^3(x+1)=3\), whose only integer root is \(x=2\).
3. Two integers: 0 and 2.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) 0

Explanation:
1. The quadratic in \(x\) is \(\le0\) only if its discriminant is non-negative; with Cauchy-Schwarz this forces \(\dfrac ab=\dfrac bc=\dfrac cd\): \(a,b,c\) in \( G.P \).
2. So \(\log a,\log b,\log c\) are in \( A.P \).; all three columns are in \( A.P \)., giving \(C_1+C_3=2C_2\) and the determinant 0.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) 10

Explanation:
1. Expanding, \(U_n=-(2k^2+4k)n+(k^3+3k^2+3k+1)\), which is linear in \(n\).
2. \(\sum_{n=1}^kU_n=-(2k^2+4k)\dfrac{k(k+1)}2+k(k+1)^3=k(k+1)\left[(k+1)^2-(k^2+2k)\right]=k(k+1)\).
3. \(k(k+1)=110\) gives \(k=10\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \((x-1)^{n-1}(x+n-1)\)

Explanation:
1. Add all rows to the first row: every entry becomes \(x+n-1\).
2. Take it out and subtract the first row from the others: the determinant is \((x+n-1)(x-1)^{n-1}\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \(48a^4\)

Explanation:
1. \(f(x)=(x-a)^3(x+3a)\).
2. \(f(3a)=(2a)^3\cdot6a=48a^4\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) 1

Explanation:
1. The determinant equals \((x-1)^2\left[1+(a^2+b^2+c^2+2)x\right]\).
2. With \(a^2+b^2+c^2=-2\): \((x-1)^2=0\), one distinct root.

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \((a-b)(b-c)(c-a)(a+b+c)\)

Explanation:
1. Multiply \(C_1,C_2,C_3\) by \(a,b,c\) to get \(\dfrac1{abc}\begin{vmatrix}a&b&c\\a^3&b^3&c^3\\abc&abc&abc\end{vmatrix}\), then use \(\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}\).
2. The result is \((a-b)(b-c)(c-a)(a+b+c)\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (b) \((a-b)(b-c)(c-a)(ab+bc+ca)\)

Explanation:
1. Multiply the columns by \(a,b,c\) (dividing by \(abc\)): the determinant becomes \(\dfrac1{abc}\begin{vmatrix}a^2&b^2&c^2\\a^3&b^3&c^3\\abc&abc&abc\end{vmatrix}=\begin{vmatrix}a^2&b^2&c^2\\a^3&b^3&c^3\\1&1&1\end{vmatrix}\).
2. This equals \((a-b)(b-c)(c-a)(ab+bc+ca)\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (c) \(only the solution x = 1, y = 1, z = 1\)

Explanation:
1. Its coefficient determinant is \(-2\ne0\).
2. The unique solution is \((1,1,1)\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (c) \(\cos2\theta\)

Explanation:
1. \(\begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix}^{-1}=\dfrac1{\sec^2\theta}\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix}\).
2. The product is \(\cos^2\theta\begin{bmatrix}1-\tan^2\theta&-2\tan\theta\\2\tan\theta&1-\tan^2\theta\end{bmatrix}=\begin{bmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{bmatrix}\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (d) \(\sin2\theta\)

Explanation:
1. The product equals \(\begin{bmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{bmatrix}\) (computed from the inverse via the adjoint).
2. Comparing entries, \(b=\sin2\theta\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.

Answer: (a) \(2+\sin2\theta\)

Explanation:
1. Apply \(R_1\to R_1-R_2\): first row \((1,-1,0)\).
2. Then \(C_2\to C_2+C_1\): the determinant becomes \(\begin{vmatrix}1&0&0\\\sin^2\theta&1+1&\sin2\theta\\\sin^2\theta&1&1+\sin2\theta\end{vmatrix}=2(1+\sin2\theta)-\sin2\theta=2+\sin2\theta\).

Teacher's Note:
1. Use row and column operations to create zeros and factorise before expanding.
2. For a system of linear equations, link consistency to the value of the determinant of the coefficient matrix.