JEE Mathematics Matrices MCQs Set 14

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

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

Teacher's Note:
1. Choose entries of the product AA inverse that contain only one unknown.
2. Check another entry: row 2 of A times column 2 of the inverse gives \( -\frac{1}{2}+6-\frac{9}{2}=1 \).

Answer: (d) PAQ

Explanation:
1. \( \operatorname{adj}(XYZ)=\operatorname{adj}Z\,\operatorname{adj}Y\,\operatorname{adj}X \).
2. \( \operatorname{adj}(P^{-1})=|P^{-1}|P=P \) and \( \operatorname{adj}(Q^{-1})=Q \) since the determinants are 1.
3. So \( \operatorname{adj}(Q^{-1}BP^{-1})=P\,A\,Q \).

Teacher's Note:
1. The adjoint of a product reverses the order.
2. For a matrix with determinant 1, the adjoint equals the inverse.

Answer: (d) \( I+2A+2A^2 \)

Explanation:
1. \( f(A)=(I+A)(I-A)^{-1} \).
2. Since \( A^3=O \), \( (I-A)^{-1}=I+A+A^2 \).
3. \( f(A)=(I+A)(I+A+A^2)=I+2A+2A^2+A^3=I+2A+2A^2 \).

Teacher's Note:
1. Use the geometric series for a nilpotent matrix.
2. Multiply out and discard A cubed.

Answer: (c) \( P^n \)

Explanation:
1. Multiply the given relation by \( P^{-1} \): \( P^{-1}+I+P+\dots+P^{n-1}=O \).
2. Since \( I+P+\dots+P^{n-1}=-P^n \), we get \( P^{-1}-P^n=O \).
3. So \( P^{-1}=P^n \).

Teacher's Note:
1. Multiplying by the inverse shifts every power down by one.
2. The remaining terms are the original sum without its last term.

Answer: (d) \( G(-y)F(-x) \)

Explanation:
1. \( \operatorname{adj}(FG)=\operatorname{adj}G\,\operatorname{adj}F \), and both determinants are 1, so this equals \( G^{-1}F^{-1} \).
2. \( F^{-1}(x)=F(-x) \) and \( G^{-1}(y)=G(-y) \).
3. So \( \operatorname{adj}(FG)=G(-y)F(-x) \).

Teacher's Note:
1. The inverse of a rotation by an angle is the rotation by the negative angle.
2. Remember to reverse the order of the two factors.

Answer: (d) non-singular

Explanation:
1. \( B=A^{11}(I+A+A^2+A^3) \) and \( A^{11} \) is invertible.
2. The eigenvalues of \( I+A+A^2+A^3 \) are \( 1+z+z^2+z^3 \) with \( z=e^{\pm i\pi/5} \).
3. This is zero only when \( z^4=1 \), z not 1; here \( z^4\ne1 \), so B is non-singular.

Teacher's Note:
1. A is a rotation matrix; its eigenvalues are complex numbers on the unit circle.
2. Singularity would need a root of z cubed + z squared + z + 1 \( = \) 0, which are 4th roots of unity (not 1).

Answer: (c) an orthogonal matrix

Explanation:
1. Let the matrix be \( M \). It is symmetric, and \( M^2=I \) after expanding: the (1,1) entry is \( \frac{(b^2-a^2)^2+4a^2b^2}{(a^2+b^2)^2}=1 \).
2. So \( MM^T=M^2=I \), which means M is orthogonal.
3. It is not the unit matrix (off-diagonal entries are non-zero) and not idempotent or nilpotent.

Teacher's Note:
1. Check the rows: each has squared length 1 and they are perpendicular.
2. M is a reflection matrix, so it is orthogonal and involutory.

Answer: (a) \( \begin{bmatrix} a & i \\ i & b \end{bmatrix} \)

Explanation:
1. \( a \) and \( b \) satisfy \( t^2=t+1 \), so \( a^2=a+1 \) and \( b^2=b+1 \), \( a+b=1 \).
2. For \( A=\begin{bmatrix}a&i\\i&b\end{bmatrix} \): \( A^2=\begin{bmatrix}a^2-1&i(a+b)\\i(a+b)&b^2-1\end{bmatrix}=\begin{bmatrix}a&i\\i&b\end{bmatrix}=A \).
3. The other options fail: for the first, \( (2,2) \) entry of \( A^2 \) is \( b^2-1=b\ne-b \).

Teacher's Note:
1. Use the relation t squared \( = \) t + 1 to simplify the diagonal entries of A squared.
2. The off-diagonal entries become i times (a + b) \( = \) i.

Answer: (b) A

Explanation:
1. \( Q^n=PA^nP^T \) since \( P^TP=I \).
2. \( X=P^TPA^{2021}P^TP=A^{2021} \).
3. \( A^5=A \) gives \( A^{2021}=A^{1+4\cdot505}=A \).

Teacher's Note:
1. The similarity transformation passes through powers.
2. Reduce the exponent modulo 4 because \( A^{k+4} = A^k \) for k at least 1.

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

Explanation:
1. Put \( Y=\omega y \) and \( Z=\omega^2z \); the system becomes linear in \( x,Y,Z \) with right sides \( 1-\omega^2 \), \( \omega^2-\omega \), \( \omega-1 \).
2. Solving gives \( x=-1 \), \( Y=-\omega \), \( Z=-\omega^2 \).
3. So \( y=-1 \) and \( z=-1 \): the point is \( (-1,-1,-1) \).

Teacher's Note:
1. Check the first plane at \( (-1,-1,-1) \): \( -(1+2\omega+3\omega^2)=-(\omega+2\omega^2)=1-\omega^2=(1+\omega)(1-\omega) \).
2. The other two planes are verified in the same way.

Answer: (a) there is only one solution

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&2&3\\2&1&3\\5&5&9\end{vmatrix}=1(9-15)-2(18-15)+3(10-5)=3\ne0 \).
2. So the system has a unique solution.

Teacher's Note:
1. A non-zero determinant of coefficients gives a unique solution by Cramer's rule.
2. Options (c) and (d) mean the same thing.

Answer: (a) an infinite number of non-trivial solutions for one value of \( \lambda \)

Explanation:
1. The coefficient determinant is \( 6(3-8)-5(-9-4)+\lambda(6+1)=35+7\lambda \).
2. It is zero only for \( \lambda=-5 \).
3. For that value the homogeneous system has infinitely many non-trivial solutions (all multiples of one solution).

Teacher's Note:
1. A homogeneous system has either only the trivial solution or infinitely many solutions.
2. Exactly one non-trivial solution is never possible.

Answer: (a) 2

Explanation:
1. Three planes through the origin meet in a line if the coefficient determinant is zero.
2. \( \begin{vmatrix}k&4&1\\4&k&2\\2&2&1\end{vmatrix}=(k-2)(k-4) \).
3. Among the options only \( k=2 \) appears.

Teacher's Note:
1. Planes through the origin always share the origin; a line needs a zero determinant.
2. The other root k \( = \) 4 is not in the options.

Answer: (a) 2

Explanation:
1. \( A^2=I \), so \( A^x=I \) exactly for even \( x=2,4,6,\dots \).
2. The sum is \( \sum_{k\ge1}(\cos^{2k}\theta+\sin^{2k}\theta)=\frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta+\cot^2\theta \).
3. By AM-GM this is at least 2, with equality at \( \tan^2\theta=1 \).

Teacher's Note:
1. Check that A squared is the identity before anything else.
2. The sum of a geometric series with ratio cos squared or sin squared converges when both are less than 1.

Answer: (a) \( a=2 \)

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}2a&-2&3\\1&a&2\\2&0&a\end{vmatrix}=2a^3-4a-8 \).
2. \( 2a^3-4a-8=2(a-2)(a^2+2a+2) \), and the quadratic factor has no real root.
3. So the real value is \( a=2 \).

Teacher's Note:
1. Set the determinant of coefficients to zero.
2. Check a \( = \) 2: 16 - 8 - 8 \( = \) 0.

Answer: (d) \( a-b=2 \)

Explanation:
1. Subtract the second equation from the first: \( 2y+2=0 \), so \( y=-1 \) and then \( x=1 \).
2. The third equation must hold at \( (1,-1) \): \( a-2-b=0 \).
3. So \( a-b=2 \).

Teacher's Note:
1. Solve two of the equations and substitute into the third.
2. Consistent lines are concurrent at a single point.

Answer: (a) A and B only

Explanation:
1. \( A^2+kI=O \) with A not a scalar matrix means the characteristic polynomial is \( x^2+k \), so trace \( A=0 \).
2. Then \( \beta_2=-\alpha_1 \) and \( \alpha_1\beta_2=-\alpha_1^2<0 \).
3. \( \det A=k \), which is not necessarily 0 or -1. So only A and B must hold.

Teacher's Note:
1. A 2x2 matrix that is not a scalar multiple of I has minimal polynomial equal to its characteristic polynomial.
2. The constant term of the characteristic polynomial is the determinant, which is k.

Answer: (a) A, B and C only

Explanation:
1. \( A^n=\begin{bmatrix}1&0\\2n&1\end{bmatrix} \), so \( A^{2012}=\begin{bmatrix}1&0\\4024&1\end{bmatrix} \).
2. So \( a=d=1 \), \( b=0 \), \( c=4024 \).
3. (A) true; (B) \( 1+0+4024+1=4026 \) true; (C) \( 1+0+1=2 \) true; (D) false.

Teacher's Note:
1. Find the pattern \( A^n = [[1 \), \( 0] \), \( [2n \), \( 1]] \) from the first two powers.
2. Substitute the values into each statement.

Answer: (a) A, B and C only

Explanation:
1. \( A=\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix} \) is symmetric (A true) with trace \( 6 \) (B true).
2. \( \det A=2(4-1)+1(-2)=4 \), a perfect square (C true).
3. \( A^{-1} \) of a symmetric matrix is symmetric, not skew-symmetric (D false).

Teacher's Note:
1. The inverse of a symmetric matrix is symmetric.
2. Compute the determinant by expanding along the first row.

Answer: (a) A and B only

Explanation:
1. The characteristic equation is \( A^3-(\operatorname{tr}A)A^2+(\text{sum of principal minors})A-|A|I=O \).
2. \( \operatorname{tr}A=11 \), the sum of principal 2x2 minors is \( 18-4+1=15 \) and \( |A|=1 \).
3. So \( A^3-11A^2+15A=I \), giving \( k=11 \) and \( \ell=15 \).

Teacher's Note:
1. Cayley-Hamilton: the matrix satisfies its characteristic equation.
2. The coefficient of A is the sum of the principal minors of order 2.

Answer: (a) All of 1, 2, 3 and 4 are true

Explanation:
1. The coefficient of the second highest power in the characteristic polynomial gives (1); the constant term gives (2).
2. A zero eigenvalue forces determinant zero, which gives (3).
3. If \( Ax=\lambda x \) then \( A^{-1}x=\frac1\lambda x \), which gives (4).

Teacher's Note:
1. These are the four standard facts connecting eigenvalues with trace and determinant.
2. All four hold for every square matrix.

Answer: (d) A, B, C and D

Explanation:
1. B is a reflection, so \( B^2=I \) and \( B=B^{-1} \); AB is also a reflection, hence \( (AB)^2=I \) and \( (AB)^{-1}=AB \) (A and B true).
2. \( BAB=BAB^{-1}=A^{-1} \) because conjugating a rotation by a reflection reverses it (C true).
3. \( BA^4B=A^{-4} \), and \( A^{-4}=A^{-1} \) means \( A^3=I \), i.e. \( 3\alpha=2\pi \), so the least positive value is \( \frac{2\pi}3 \) (D true).

Teacher's Note:
1. Reflections square to the identity; the product of a rotation and a reflection is a reflection.
2. All four statements are true.

Answer: (a) A, B and C only

Explanation:
1. \( A^2=I \), so A is involutory (C true).
2. Trace of A is \( -1 \) (A true) and \( A^{2012}=I \) has trace \( 3 \) (B true).
3. \( A^{2013}=A \) has trace \( -1 \), not 1 (D false).

Teacher's Note:
1. Find the order of the matrix: A squared is the identity.
2. Even powers give I and odd powers give A.

Answer: (a) A and B only

Explanation:
1. For an orthogonal matrix each row has \( l_i^2+m_i^2+n_i^2=1 \), so each row is a set of direction cosines (A true).
2. Different rows have dot product 0, so the lines are mutually perpendicular (B true).
3. Mutually perpendicular lines are neither parallel nor coplanar through a point in space (C and D false).

Teacher's Note:
1. Orthogonality means unit rows and perpendicular rows.
2. Three mutually perpendicular lines through a point span 3-dimensional space.

Answer: (a) B, C and D only

Explanation:
1. \( |A|=-1 \) and \( A^2=I \), so \( A^{-1}=A \) (C and D true).
2. \( \operatorname{adj}A=|A|A^{-1}=-A \), which is the matrix in (B) (true); it is not zero (A false).

Teacher's Note:
1. A is a permutation matrix swapping the first and third coordinates.
2. The adjoint equals the determinant times the inverse.

Answer: (a) A, B and C only

Explanation:
1. \( |B|=|P^{-1}||A||P|=|A| \), so A and B are invertible together (A true).
2. \( B^n=P^{-1}AP\cdot P^{-1}AP\cdots=P^{-1}A^nP \) (B true), and \( P^{-1}(A-\lambda I)P=B-\lambda I \) (C true).
3. A and B need not be singular (D false).

Teacher's Note:
1. Similar matrices share determinant, trace and eigenvalues.
2. The middle P and P inverse cancel in every power.