JEE Mathematics Matrices MCQs Set 22

Answer: (b) \( \frac13\begin{bmatrix}5\\7\end{bmatrix} \)

Explanation:
1. \( |A|=1-4=-3 \) and \( A^{-1}=\frac13\begin{bmatrix}1&2\\2&1\end{bmatrix} \).
2. \( X=A^{-1}B=\frac13\begin{bmatrix}3+2\\6+1\end{bmatrix}=\frac13\begin{bmatrix}5\\7\end{bmatrix} \).

Teacher's Note:
1. Multiply by the inverse on the left: \( X=A^{-1}B \).

Answer: (d) \( \begin{bmatrix}-1\\2\\3\end{bmatrix} \)

Explanation:
1. \( |A|=1(0-2)+1(2-3)+2(4-0)=5\ne0 \), so the solution is unique.
2. Testing \( x=(-1,2,3) \): \( -1-2+6=3 \), \( -2+0+3=1 \), \( -3+4+3=4 \). All three equations hold, so \( X=\begin{bmatrix}-1\\2\\3\end{bmatrix} \).

Teacher's Note:
1. Substituting the candidate solutions is a quick check for 3x3 systems.

Answer: (c) \( G^{-1}(y)\,F^{-1}(x) \)

Explanation:
1. For invertible matrices \( (PQ)^{-1}=Q^{-1}P^{-1} \).
2. So \( [F(x)G(y)]^{-1}=G^{-1}(y)F^{-1}(x) \); the order of the factors is reversed.

Teacher's Note:
1. The inverse of a product reverses the order of the factors.

Answer: (b) \( a=1,\ c=-1 \)

Explanation:
1. From \( AA^{-1}=I \), the (3,1) entry gives \( 3\cdot\frac12-4a+\frac52=4-4a=0 \), so \( a=1 \).
2. The (1,3) entry gives \( 0\cdot\frac12+c+2\cdot\frac12=c+1=0 \), so \( c=-1 \).

Teacher's Note:
1. Pick one entry of \( AA^{-1} \) that involves each unknown and set it equal to the identity entry.

Answer: (b) \( \cos^2\frac\theta2\cdot A^T \)

Explanation:
1. \( B=A^{-1}=\frac1{|A|}\operatorname{adj}A \) with \( |A|=1+\tan^2\frac\theta2=\sec^2\frac\theta2 \).
2. \( \operatorname{adj}A=\begin{bmatrix}1&-\tan\frac\theta2\\\tan\frac\theta2&1\end{bmatrix}=A^T \), so \( B=\cos^2\frac\theta2\cdot A^T \).

Teacher's Note:
1. For a 2x2 matrix the adjoint swaps the diagonal and changes signs of the off-diagonal entries.

Answer: (b) a, b, c are in GP

Explanation:
1. For a non-trivial solution the determinant must vanish. Expanding gives \( -(ac-b^2)(a\alpha^2+2b\alpha+c)=0 \).
2. So either \( b^2=ac \) (a, b, c in GP) or \( \alpha \) is a root of \( ax^2+2bx+c=0 \).
3. Among the options, only the GP condition is a correct sufficient condition; (d) states the opposite of the second possibility.

Teacher's Note:
1. Expand the 3x3 determinant and factor it; it splits into two simple factors.

Answer: (c) \( -2 \)

Explanation:
1. The determinant is \( (\alpha+2)(\alpha-1)^2 \).
2. For \( \alpha=1 \) all three equations become \( x+y+z=0 \), which has infinitely many solutions.
3. For \( \alpha=-2 \), adding the equations gives \( 0=3(-3)=-9 \), which is impossible. So there is no solution only for \( \alpha=-2 \).

Teacher's Note:
1. Check both values that make the determinant zero; they behave differently.

Answer: (a) zero

Explanation:
1. Adding the first two equations gives \( x_2=-1 \) and then \( x_1=0 \).
2. The third equation would need \( 0-2(-1)=2=3 \), which is false. So there is no solution.

Teacher's Note:
1. Solve two equations and test the third.

Answer: (d) \( 1,2,-1 \)

Explanation:
1. Adding the second and third equations gives \( 3x-y=1 \), so \( y=3x-1 \).
2. The first equation then gives \( 6x-1+2z=3 \), so \( z=2-3x \).
3. Substituting in \( x+2y+z=4 \): \( x+6x-2+2-3x=4 \), so \( x=1 \), \( y=2 \), \( z=-1 \).

Teacher's Note:
1. When options are numeric triples, substitute them in all equations.

Answer: (c) \( 1,2 \)

Explanation:
1. Subtracting: \( y+3z=\alpha-1 \) and \( 2y+6z=\alpha^2-\alpha \).
2. Consistency needs \( 2(\alpha-1)=\alpha^2-\alpha \), i.e. \( \alpha^2-3\alpha+2=0 \), so \( \alpha=1 \) or \( 2 \).

Teacher's Note:
1. The coefficient rows are dependent, so consistency is a condition on the right-hand sides.

Answer: (a) \( A^3=9A \)

Explanation:
1. \( A^2=3A \), so \( A^3=3A^2=9A \).
2. \( A+A=2A\ne3A=A^2 \), and \( |A|=0 \) so \( A^{-1} \) does not exist.

Teacher's Note:
1. Compute \( A^2 \) first and notice it is a multiple of A.

Answer: (c) \( AB=O\Rightarrow|A|=0 \) or \( |B|=0 \)

Explanation:
1. \( AB=O \) gives \( |A||B|=0 \), so at least one of the determinants is zero.
2. The other statements are false in general: \( \operatorname{adj}(AB)=\operatorname{adj}B\operatorname{adj}A \), and the inverse of a sum is not the sum of inverses.

Teacher's Note:
1. Take determinants of both sides of \( AB=O \).

Answer: (a) \( a=\cos2\theta,\ b=\sin2\theta \)

Explanation:
1. \( \begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix}^{-1}=\cos^2\theta\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix} \).
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. Rotation-type matrices multiply by adding angles.

Answer: (c) \( |A|\,|I_n-A| \)

Explanation:
1. \( A-I=A-AA^T=A(I-A^T) \) because \( AA^T=I \).
2. \( |A-I|=|A|\,|I-A^T|=|A|\,|I-A| \).

Teacher's Note:
1. Write \( I \) as \( AA^T \) and factor A out.

Answer: (c) no solution (inconsistent)

Explanation:
1. Adding the first two equations gives \( 3x+2y+5z=4 \).
2. The third equation says \( 3x+2y+5z=3 \), which contradicts it, so the system is inconsistent.

Teacher's Note:
1. Look for a row that is the sum of two others on the left but not on the right.

Answer: (a) \( (14)^2 \)

Explanation:
1. \( |A|=14 \).
2. \( |\operatorname{adj}A|=|A|^{n-1}=14^2 \) for \( n=3 \).

Teacher's Note:
1. The determinant of the adjoint is \( |A|^{n-1} \).

Answer: (a) 2

Explanation:
1. \( A^2 \): the third row is \( (3\cdot1+4\cdot0+(-1)\cdot3,\ 0+4-4,\ 1)=(0,0,1) \).
2. So \( A^2=I \) and the least positive n is 2.

Teacher's Note:
1. Square the matrix and see if it already returns to I.

Answer: (a) \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \)

Explanation:
1. P is orthogonal, so \( P^TP=PP^T=I \).
2. \( Q^{2005}=PA^{2005}P^T \), hence \( P^TQ^{2005}P=A^{2005} \).
3. \( A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix} \), so the result is \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \).

Teacher's Note:
1. Conjugation by an orthogonal matrix commutes with taking powers.

Answer: (c) \( \pm3 \)

Explanation:
1. \( \det A^3=(\det A)^3=(\alpha^2-4)^3=125 \).
2. So \( \alpha^2-4=5 \), giving \( \alpha=\pm3 \).

Teacher's Note:
1. \( \det A^3=(\det A)^3 \).

Answer: (d) \( (A(\alpha,\beta))'=A(\alpha,-\beta) \)

Explanation:
1. The transpose swaps the signs of the sine terms and keeps \( e^\beta \), so \( A'=A(-\alpha,\beta) \); option (a) is true and option (d) is false.
2. The inverse is \( A(-\alpha,-\beta) \) (rotation by \( -\alpha \), and \( e^{-\beta} \)), and \( |A|=e^\beta \), so \( \operatorname{adj}A=|A|A^{-1}=e^\beta A(-\alpha,-\beta) \).

Teacher's Note:
1. Compare the transpose and the inverse with the matrix at the negated parameters.

Answer: (b) non-singular

Explanation:
1. A is a rotation through \( -\alpha \), so \( A^k \) is a rotation through \( -k\alpha \) and \( A^5=-I \). Then \( (A-I)(I+A+A^2+A^3+A^4)=A^5-I=-2I \), so \( I+B=-2(A-I)^{-1} \).
2. A direct computation gives \( B=\begin{bmatrix}0&\cot\frac\alpha2\\-\cot\frac\alpha2&0\end{bmatrix} \) with \( \cot\frac\pi{10}\approx3.08 \).
3. Its determinant is \( \cot^2\frac\pi{10}\approx9.47\ne0 \) and B is skew-symmetric, so B is non-singular, not symmetric and \( |B|\ne1 \).

Teacher's Note:
1. Sum the rotations using the geometric-series identity and compute the resulting matrix.

Answer: (b) \( PAQ \)

Explanation:
1. \( \operatorname{adj}(XYZ)=\operatorname{adj}Z\,\operatorname{adj}Y\,\operatorname{adj}X \).
2. \( \operatorname{adj}(P^{-1})=P \) (as \( |P|=1 \)), \( \operatorname{adj}B=A \), and \( \operatorname{adj}(Q^{-1})=Q \).
3. So the adjoint is \( P\,A\,Q \).

Teacher's Note:
1. Reverse the order when taking the adjoint of a product.

Answer: (d) \( K=44 \)

Explanation:
1. \( \operatorname{tr}A=4 \) and \( |A|=5 \), so \( A^2=4A-5I \).
2. Repeated reduction gives \( A^3=11A-20I \), \( A^4=24A-55I \), \( A^5=41A-120I \) and \( A^6=44A-205I \).
3. So \( K=44 \).

Teacher's Note:
1. Use the Cayley-Hamilton relation to reduce each power to the form \( pA+qI \).

Answer: (a) 2009

Explanation:
1. \( X'AX=0 \) for all X means \( A \) is skew-symmetric, so \( a_{ij}=-a_{ji} \).
2. \( a_{32}=-a_{23}=2009 \).

Teacher's Note:
1. A quadratic form that vanishes identically belongs to a skew-symmetric matrix.

Answer: (b) \( -1 \)

Explanation:
1. The coefficient matrix is triangular with determinant \( k(k-1)(k+2) \).
2. A unique solution needs \( k\ne0,1,-2 \). Among the options only \( k=-1 \) satisfies this.

Teacher's Note:
1. For a triangular system the determinant is the product of the diagonal entries.

Answer: (d) \( 2007^2 \)

Explanation:
1. \( \det M_r=r^2-(r-1)^2=2r-1 \).
2. The sum of the first 2007 odd numbers is \( 2007^2 \).

Teacher's Note:
1. Each determinant is an odd number.

Answer: (c) \( -3 \)

Explanation:
1. Idempotent means \( A^2=A \). The (3,3) entry of \( A^2 \) is \( 1(-4)+(-2)(4)+x\cdot x=x^2-12 \), which must equal \( x \); so \( x=-3 \) or \( x=4 \).
2. The (1,3) entry of \( A^2 \) is \( 2(-4)+(-2)(4)+(-4)x=-16-4x \), which must equal \( -4 \); so \( x=-3 \).
3. Hence \( x=-3 \) (it satisfies both conditions).

Teacher's Note:
1. Use two different entries of \( A^2=A \) to remove the extra root.

Answer: (a) \( f(-x) \)

Explanation:
1. A is a rotation matrix, so \( A^{-1}=A^T=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix} \).
2. This is \( f(-x) \).

Teacher's Note:
1. The inverse of a rotation through x is the rotation through \( -x \).

Answer: (a) \( nA-(n-1)I \)

Explanation:
1. \( (A-I)^2=A^2-2A+I=O \). Write \( A=I+N \) with \( N^2=O \).
2. \( A^n=(I+N)^n=I+nN=I+n(A-I)=nA-(n-1)I \).

Teacher's Note:
1. The matrix \( A-I \) squares to zero, so the binomial expansion stops after two terms.

Answer: (c) \( \begin{bmatrix}5&2\\-8&-3\end{bmatrix} \)

Explanation:
1. \( A^2=\begin{bmatrix}4-4&2-2\\-8+8&-4+4\end{bmatrix}=O \), so all higher powers vanish.
2. The sum is \( I+2A=\begin{bmatrix}5&2\\-8&-3\end{bmatrix} \).

Teacher's Note:
1. Check whether A is nilpotent before summing a series.