JEE Mathematics Matrices MCQs Set 21

Answer: (d) \( \frac12I \)

Explanation:
1. \( A-B=\frac1\pi\begin{bmatrix}\sin^{-1}(x\pi)+\cos^{-1}(x\pi)&0\\0&\cot^{-1}(\pi x)+\tan^{-1}(\pi x)\end{bmatrix} \).
2. Both diagonal entries equal \( \frac\pi2 \), so \( A-B=\frac1\pi\cdot\frac\pi2I=\frac12I \).

Teacher's Note:
1. The identities \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \) and \( \tan^{-1}u+\cot^{-1}u=\frac\pi2 \) do all the work.

Answer: (a) \( A \)

Explanation:
1. Since \( A^2=I \), \( A^3=A \).
2. \( (A-I)^3=A^3-3A^2+3A-I=4A-4I \) and \( (A+I)^3=A^3+3A^2+3A+I=4A+4I \).
3. The sum is \( 8A \), so the expression equals \( 8A-7A=A \).

Teacher's Note:
1. Expand using \( A^2=I \) before adding; the I terms cancel.

Answer: (b) \( 7A+I \)

Explanation:
1. \( (A+I)^3=A^3+3A^2+3A+I \).
2. With \( A^2=A \) we get \( A^3=A \), so the sum is \( A+3A+3A+I=7A+I \).

Teacher's Note:
1. For an idempotent matrix every power of A equals A.

Answer: (c) \( I+A \)

Explanation:
1. Let \( t=\tan\frac\alpha2 \). Then \( \cos\alpha=\frac{1-t^2}{1+t^2} \) and \( \sin\alpha=\frac{2t}{1+t^2} \).
2. \( I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix} \). Multiplying, the first row is \( \left(\cos\alpha+t\sin\alpha,\ -\sin\alpha+t\cos\alpha\right)=\left(\frac{1-t^2+2t^2}{1+t^2},\ \frac{-2t+t-t^3}{1+t^2}\right)=(1,-t) \).
3. Similarly the second row is \( (t,1) \), so the product is \( \begin{bmatrix}1&-t\\t&1\end{bmatrix} \).
4. This is \( I+A \).

Teacher's Note:
1. Express cos and sin through \( \tan\frac\alpha2 \) and multiply carefully.

Answer: (a) 1

Explanation:
1. \( A^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix} \) and \( kA-2I=\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix} \).
2. Comparing, \( -2k=-2 \) gives \( k=1 \), and the other entries agree (\( 3k-2=1 \), \( 4k=4 \), \( -2k-2=-4 \)).

Teacher's Note:
1. Compare any one off-diagonal entry, then verify the rest.

Answer: (c) \( -2I \)

Explanation:
1. Characteristic polynomial: \( \lambda^3-6\lambda^2+7\lambda+2 \), since \( \operatorname{tr}A=6 \), the sum of principal minors is \( 7 \) and \( |A|=-2 \).
2. By the Cayley-Hamilton theorem \( A^3-6A^2+7A+2I=O \), so \( A^3-6A^2+7A=-2I \).

Teacher's Note:
1. Cayley-Hamilton: a matrix satisfies its own characteristic equation.

Answer: (d) \( \begin{bmatrix}\frac25&-\frac{12}5\\-\frac{11}5&3\end{bmatrix} \) and \( \begin{bmatrix}\frac25&\frac{13}5\\\frac{14}5&-2\end{bmatrix} \)

Explanation:
1. Eliminate Y: \( 3(3X+2Y)-2(2X+3Y)=5X \), so \( X=\frac15\left(3\begin{bmatrix}2&-2\\-1&5\end{bmatrix}-2\begin{bmatrix}2&3\\4&0\end{bmatrix}\right)=\frac15\begin{bmatrix}2&-12\\-11&15\end{bmatrix} \).
2. Eliminate X: \( Y=\frac15\left(3\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\right)=\frac15\begin{bmatrix}2&13\\14&-10\end{bmatrix} \).

Teacher's Note:
1. Treat the matrix equations like simultaneous linear equations.

Answer: (c) \( B^2=-I \)

Explanation:
1. \( A^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=-I \), so (a) and (b) are false.
2. \( B^2=\begin{bmatrix}i^2&0\\0&i^2\end{bmatrix}=-I \).

Teacher's Note:
1. Square each matrix directly; \( i^2=-1 \).

Answer: (a) 14

Explanation:
1. \( X=\begin{bmatrix}5&0\\1&4\end{bmatrix} \) and \( Y=\begin{bmatrix}2&0\\1&1\end{bmatrix} \).
2. \( 3X-4Y=\begin{bmatrix}7&0\\-1&8\end{bmatrix} \), whose elements add up to 14.

Teacher's Note:
1. Solve for X and Y first, then compute the combination.

Answer: (b) \( A^3-3A^2-I=O \)

Explanation:
1. The characteristic equation is \( \lambda^3-\operatorname{tr}(A)\lambda^2+(\text{sum of principal minors})\lambda-|A|=0 \) with \( \operatorname{tr}A=3 \), \( |A|=1 \) and the sum of principal minors \( =0 \).
2. So \( \lambda^3-3\lambda^2-1=0 \), and by Cayley-Hamilton \( A^3-3A^2-I=O \) (direct multiplication gives \( A^2=\begin{bmatrix}2&3&1\\5&6&2\\3&4&1\end{bmatrix} \), \( A^3=\begin{bmatrix}7&9&3\\15&19&6\\9&12&4\end{bmatrix} \), and \( A^3-3A^2=I \)).

Teacher's Note:
1. Use Cayley-Hamilton to get a relation among powers of A.

Answer: (b) \( \begin{bmatrix}2&-2\\-1&1\\-1&1\end{bmatrix} \)

Explanation:
1. Row 1 of P times the columns of Q: \( i(-i)+(-i)(i)=1+1=2 \) and \( i\cdot i+(-i)(-i)=-1-1=-2 \).
2. Row 2: \( i\cdot i=-1 \) and \( i(-i)=1 \). Row 3: \( (-i)(-i)=-1 \) and \( (-i)(i)=1 \).
3. So \( PQ=\begin{bmatrix}2&-2\\-1&1\\-1&1\end{bmatrix} \).

Teacher's Note:
1. Use \( i^2=-1 \) and \( -i\cdot i=1 \) while multiplying.

Answer: (c) 6

Explanation:
1. Write \( P^n=a_nI+b_nP \). Since \( P^2=I-P \), \( P^{n+1}=b_nI+(a_n-b_n)P \).
2. Starting with \( P=(0,1) \): \( P^2=(1,-1) \), \( P^3=(-1,2) \), \( P^4=(2,-3) \), \( P^5=(-3,5) \), \( P^6=(5,-8) \).
3. So \( P^6=5I-8P \) and \( n=6 \).

Teacher's Note:
1. Track the coefficients of I and P step by step (they follow a Fibonacci-type pattern).

Answer: (c) \( \theta=\phi+(2n+1)\frac\pi2 \), \( n=0,1,2,\ldots \)

Explanation:
1. \( AB=\cos(\theta-\phi)\begin{bmatrix}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{bmatrix} \) (each entry carries the factor \( \cos(\theta-\phi) \)).
2. So \( AB=O \) iff \( \cos(\theta-\phi)=0 \), i.e. \( \theta-\phi=(2n+1)\frac\pi2 \).

Teacher's Note:
1. Every entry of the product has the common factor \( \cos(\theta-\phi) \).

Answer: (c) \( A^n=nA-(n-1)I \)

Explanation:
1. \( A^2=\begin{bmatrix}1&0\\2&1\end{bmatrix} \) and in general \( A^n=\begin{bmatrix}1&0\\n&1\end{bmatrix} \).
2. \( nA-(n-1)I=\begin{bmatrix}n-(n-1)&0\\n&n-(n-1)\end{bmatrix}=\begin{bmatrix}1&0\\n&1\end{bmatrix} \), which matches.

Teacher's Note:
1. Find \( A^2 \) and \( A^3 \), guess the pattern, then test it against the options.

Answer: (b) not possible to find

Explanation:
1. \( 3x+7=0 \) gives \( x=-\frac73 \); \( 5=y-2 \) gives \( y=7 \); \( y+1=8 \) agrees.
2. But \( 2-3x=2+7=9\ne4 \), so the last entries cannot be equal.
3. No values of x and y make the matrices equal.

Teacher's Note:
1. Equal matrices need all four corresponding entries equal; test the last one too.

Answer: (b) \( x=2,\ y=3 \)

Explanation:
1. \( 4x=x+6 \) gives \( x=2 \); \( 2x+y=7 \) gives \( y=3 \).
2. Check: \( 5x-7=3=y \) and \( 4x=8=7y-13 \). So \( x=2 \), \( y=3 \) and \( x+y=5 \).

Teacher's Note:
1. Pick two entries that give x and y directly and check the others.

Answer: (d) none of these

Explanation:
1. \( a+4=2a+2 \) gives \( a=2 \).
2. \( 3b=b^2+2 \) gives \( b=1 \) or \( 2 \); \( -6=b^2-5b \) gives \( b=2 \) or \( 3 \).
3. Both equations hold only for \( b=2 \), so \( (a,b)=(2,2) \), which is not one of the options (the first option wrongly allows \( b=1 \)).

Teacher's Note:
1. Each unknown must satisfy every equation in which it appears.

Answer: (d) no real values

Explanation:
1. \( A^2=\begin{bmatrix}\alpha^2&0\\\alpha+1&1\end{bmatrix} \).
2. \( A^2=B \) needs \( \alpha^2=1 \) and \( \alpha+1=5 \), i.e. \( \alpha=\pm1 \) and \( \alpha=4 \) together, which is impossible.

Teacher's Note:
1. Two different entries give contradictory values of the unknown.

Answer: (c) \( \{-3,2\} \)

Explanation:
1. Trace \( =(x-1)+(x^2-2)+(x-3)+(x^2-6)=2x^2+2x-12 \).
2. \( 2x^2+2x-12=0 \) gives \( x^2+x-6=0 \), so \( x=2 \) or \( x=-3 \).

Teacher's Note:
1. The trace is the sum of the diagonal entries only.

Answer: (a) \( \begin{bmatrix}0&4\\3&-8\end{bmatrix} \)

Explanation:
1. Transposing the second relation: \( A+B=\begin{bmatrix}1&0\\2&-1\end{bmatrix} \).
2. Subtracting from the first: \( B=\begin{bmatrix}1&-4\\-1&7\end{bmatrix} \).
3. Then \( A=(A+B)-B=\begin{bmatrix}0&4\\3&-8\end{bmatrix} \).

Teacher's Note:
1. Take the transpose of the second equation to bring it to the same form as the first.

Answer: (d) \( x=0,3 \)

Explanation:
1. Expanding the determinant gives \( -x(x-3)^2 \).
2. It is zero for \( x=0 \) or \( x=3 \).

Teacher's Note:
1. A matrix is singular exactly when its determinant is zero.

Answer: (d) all of the above

Explanation:
1. \( A \) equals its transpose, so it is symmetric.
2. \( |A|=abc+ab+bc+ca=abc\left(1+\frac1a+\frac1b+\frac1c\right)=abc\ne0 \), so it is non-singular.
3. Every square matrix is the sum of a symmetric and a skew-symmetric matrix. So all statements hold.

Teacher's Note:
1. The determinant of this matrix simplifies to \( abc\left(1+\sum\frac1a\right) \).

Answer: (b) \( |A|\ne0 \)

Explanation:
1. \( \omega^{100}=\omega \) and \( \omega^{200}=\omega^2 \), and \( 1+\omega+\omega^2=0 \).
2. The diagonal entries become \( 1+2\omega+\omega^2=\omega \), \( 1+\omega+2\omega^2=\omega^2 \) and \( 2+\omega+2\omega^2=-\omega \).
3. \( |A|=\begin{vmatrix}\omega&\omega^2&1\\1&\omega^2&\omega\\\omega&\omega^2&-\omega\end{vmatrix}=\omega^3=1\ne0 \).

Teacher's Note:
1. Reduce the powers of \( \omega \) using \( \omega^3=1 \) before expanding.

Answer: (a) \( \begin{bmatrix}6&6&7\\6&2&5\\7&5&1\end{bmatrix} \)

Explanation:
1. The symmetric part is \( B=\frac12(A+A') \).
2. Averaging \( a_{ij} \) with \( a_{ji} \): \( \frac{8+4}2=6 \), \( \frac{5+9}2=7 \), \( \frac{3+7}2=5 \), with the diagonal unchanged. So \( B=\begin{bmatrix}6&6&7\\6&2&5\\7&5&1\end{bmatrix} \).

Teacher's Note:
1. Symmetric part is \( \frac12(A+A') \) and skew part is \( \frac12(A-A') \).

Answer: (d) for every real value of \( k \)

Explanation:
1. \( |A|=1\cdot(1\cdot1-0)+(-k)(0-k)=1+k^2 \).
2. Since \( 1+k^2\ge1 \), the determinant is never zero, so A is invertible for every real k.

Teacher's Note:
1. Expand along the second column, which has two zeros.

Answer: (a) 0

Explanation:
1. For a skew-symmetric matrix of odd order, \( |A|=0 \).
2. \( |\operatorname{adj}A|=|A|^{n-1}=0 \).

Teacher's Note:
1. Odd-order skew-symmetric matrices are always singular.

Answer: (d) 1

Explanation:
1. \( |A|=1(0+3)+1(0+6)+1(0-4)=5 \).
2. \( |\operatorname{adj}B|=|B|^2=|\operatorname{adj}A|^2=(|A|^2)^2=|A|^4=625 \).
3. \( |C|=5^3|A|=625 \), so the ratio is 1.

Teacher's Note:
1. For a 3x3 matrix \( |\operatorname{adj}M|=|M|^2 \) and \( |kM|=k^3|M| \).

Answer: (c) \( 14^4 \)

Explanation:
1. \( |A|=1(1+2)-2(-1-4)+(-1)(1-2)=3+10+1=14 \).
2. \( |\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}=14^4 \) for \( n=3 \).

Teacher's Note:
1. Use \( |\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2} \).

Answer: (c) \( DD'=D^n \)

Explanation:
1. \( |\operatorname{adj}A|=|A|^{n-1} \), so \( D'=D^{n-1} \).
2. Hence \( DD'=D\cdot D^{n-1}=D^n \).

Teacher's Note:
1. The determinant of the adjoint is the (n-1)th power of the determinant.

Answer: (d) \( 3A' \)

Explanation:
1. Computing the cofactors gives \( \operatorname{adj}A=\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix} \).
2. \( A'=\begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix} \), so \( \operatorname{adj}A=3A' \).

Teacher's Note:
1. Here \( AA'=9I \), which is why the adjoint is a multiple of the transpose.

Answer: (b) \( I-A \)

Explanation:
1. \( A^2-A=-I \), i.e. \( A(A-I)=-I \).
2. So \( A\cdot(I-A)=I \) and \( A^{-1}=I-A \).

Teacher's Note:
1. Factor the equation to display A times something equal to I.

Answer: (d) 7

Explanation:
1. The determinant is 1. The cofactor \( C_{31}=\begin{vmatrix}2&-3\\1&2\end{vmatrix}=7 \).
2. \( (A^{-1})_{13}=\frac{C_{31}}{|A|}=7 \) (the inverse is \( \begin{bmatrix}1&-2&7\\0&1&-2\\0&0&1\end{bmatrix} \)).

Teacher's Note:
1. The (1,3) entry of the inverse uses the cofactor of the (3,1) entry.