JEE Mathematics Matrices MCQs Set 18

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

Explanation:
1. \(A=\begin{bmatrix}2&1\\3&2\end{bmatrix}^{-1}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}^{-1}\).
2. The inverses are \(\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\) and \(\begin{bmatrix}3&2\\5&3\end{bmatrix}\); multiplying gives \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (d) \(I+P\)

Explanation:
1. Let \(t=\tan\dfrac\theta2\): \(\cos\theta=\dfrac{1-t^2}{1+t^2},\ \sin\theta=\dfrac{2t}{1+t^2}\).
2. \(I-P=\begin{bmatrix}1&t\\-t&1\end{bmatrix}\) and the rotation matrix is \(\dfrac1{1+t^2}\begin{bmatrix}1-t^2&-2t\\2t&1-t^2\end{bmatrix}\).
3. Multiplying: first row \(\dfrac1{1+t^2}(1+t^2,\ -t-t^3)=(1,-t)\), second row \(\dfrac1{1+t^2}(t+t^3,\ 1+t^2)=(t,1)\).
4. The result is \(\begin{bmatrix}1&-t\\t&1\end{bmatrix}=I+P\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (c) \(\begin{bmatrix}1&0\\25&1\end{bmatrix}\)

Explanation:
1. \(A^n=\begin{bmatrix}1&0\\n/2&1\end{bmatrix}\) by induction.
2. \(A^{50}=\begin{bmatrix}1&0\\25&1\end{bmatrix}\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) 0

Explanation:
1. \(a_{ji}=(j^2+i^2-ij)(i-j)=-a_{ij}\), so \(A\) is skew-symmetric.
2. The diagonal entries are \(a_{ii}=0\), hence \(\operatorname{tr}A=0\) (and, since \(n\) is odd, \(|A|=0\) too).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) \(\pm\dfrac1{\sqrt2},\pm\dfrac1{\sqrt6},\pm\dfrac1{\sqrt3}\)

Explanation:
1. Rows must be orthonormal: \(4b^2+c^2=1\), \(a^2+b^2+c^2=1\), \(a^2-b^2-c^2=0\), \(2b^2-c^2=0\).
2. \(c^2=2b^2\) gives \(6b^2=1\): \(b^2=\dfrac16,\ c^2=\dfrac13,\ a^2=\dfrac12\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) \(pI\)

Explanation:
1. The characteristic equation of \(A\) is \(\lambda^3-r\lambda^2-q\lambda-p=0\).
2. By the Cayley-Hamilton theorem \(A^3-rA^2-qA=pI\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

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

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

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) both A and B are idempotent

Explanation:
1. \(A^2=(AB)A=A(BA)=AB=A\).
2. \(B^2=(BA)B=B(AB)=BA=B\). Both are idempotent.

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) skew-symmetric

Explanation:
1. An odd power of a skew-symmetric matrix is skew-symmetric: \((A^k)^T=(A^T)^k=(-A)^k=-A^k\) for odd \(k\).
2. A sum of skew-symmetric matrices is skew-symmetric.

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (c) \(-YX\)

Explanation:
1. \(X^T=C^TB^TA^T+A^TB^TC^T=-CBA-ABC=-X\).
2. \(Y^T=-CBA+ABC=Y\).
3. \((XY)^T=Y^TX^T=Y(-X)=-YX\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

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

Explanation:
1. \(A^T=A=A^{-1}\) because \(A^2=I\).
2. \(C^n=AB^nA^T\), so \(A^TC^nA=A^TAB^nA^TA=B^n\).
3. \(B^n=\begin{bmatrix}1&0\\-n&1\end{bmatrix}\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) \(-1006\)

Explanation:
1. \(X^TAX=0\) for all \(X\) means \(A\) is skew-symmetric: \(a_{ji}=-a_{ij}\).
2. \(a_{21}=-2008,\ a_{31}=-1010,\ a_{32}=2012\); the sum is \(-1006\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (c) 1023

Explanation:
1. \(BA=A^2B\) gives \(B^jA=A^{2^j}B^j\).
2. Moving every \(A\) to the left: \((AB)^{10}=A^{1+2+4+\dots+2^9}B^{10}\).
3. \(k=2^{10}-1=1023\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (d) 27

Explanation:
1. \(\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix}=\begin{bmatrix}1&a+b\\0&1\end{bmatrix}\).
2. The product has top-right entry \(1+2+\dots+n=\dfrac{n(n+1)}2=378\), so \(n=27\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (c) 1

Explanation:
1. \(P=I+N\) with \(N=\begin{bmatrix}0&0&0\\4&0&0\\16&4&0\end{bmatrix}\), \(N^2=\begin{bmatrix}0&0&0\\0&0&0\\16&0&0\end{bmatrix}\), \(N^3=O\).
2. \(P^{50}=I+50N+1225N^2\): \(q_{21}=200,\ q_{32}=200,\ q_{31}=800+19600=20400\).
3. \(\dfrac{20400+200}{103\cdot200}=1\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) 2

Explanation:
1. \(A^2=I\), so \(A^x=I\) exactly for even \(x\).
2. \(\sum_{x=2,4,\dots}(\cos^x\theta+\sin^x\theta)=\dfrac{\cos^2\theta}{\sin^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta+\cot^2\theta\ge2\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) symmetric matrix

Explanation:
1. \((A^{-1}B)^T=B^T(A^{-1})^T=BA^{-1}\).
2. Since \(A,B\) commute, so do \(A^{-1},B\): \(BA^{-1}=A^{-1}B\). Hence symmetric.

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) \(A+B\)

Explanation:
1. \(A^2=A(BA)=(AB)A=BA=A\).
2. \(B^2=B(AB)=(BA)B=AB=B\), so \(A^2+B^2=A+B\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (c) \(5\text{ or }-1\)

Explanation:
1. \(A^2=\begin{bmatrix}\alpha^2&0\\\alpha+1&1\end{bmatrix}\), so \(\alpha^2=9,\ a=0,\ b=\alpha+1,\ c=1\).
2. \(\alpha=3\): \(a+b+c=5\); \(\alpha=-3\): \(a+b+c=-1\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) 1

Explanation:
1. The eigenvalues are \(1,-1,-1\) (block \(\begin{bmatrix}-5&-8\\3&5\end{bmatrix}\) has trace 0, determinant \(-1\); and \(-1\)).
2. \(\operatorname{tr}A^k=1+2(-1)^k\): \(-1\) for odd \(k\), \(3\) for even \(k\).
3. Sum \(=50(-1)+50(3)=100\); digit sum 1.

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (d) 0

Explanation:
1. The characteristic polynomial \(x^2-2x+2\) gives \(A^2=2A-2I\), \(A^3=2A-4I\) and \(A^4=-4I\).
2. Using \(A^4=-4I\): \(A^{10}=16A^2,\ A^9=16A,\ A^8=16I,\ A^7=-4A^3,\ A^6=-4A^2,\ A^5=-4A\), so \(B=5A^3-A^2-7A+17I\).
3. Substituting \(A^2\) and \(A^3\): \(B=A-I=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\), so \(a+b+c+d=0\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (a) 4

Explanation:
1. \(a=b=5110=2\cdot5\cdot7\cdot73\).
2. \(ab=5110^2\) has the distinct prime factors 2, 5, 7 and 73: four.

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) 8

Explanation:
1. \(X=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}^{-1}\begin{bmatrix}-16&-6\\7&2\end{bmatrix}=\dfrac1{10}\begin{bmatrix}-2&4\\-3&1\end{bmatrix}\begin{bmatrix}-16&-6\\7&2\end{bmatrix}=\begin{bmatrix}6&2\\\frac{11}2&2\end{bmatrix}\).
2. Trace \(=6+2=8\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.

Answer: (b) 2

Explanation:
1. \(\begin{bmatrix}5&0\\-a&5\end{bmatrix}^{-1}=\begin{bmatrix}1/5&0\\a/25&1/5\end{bmatrix}\).
2. Squaring: \(\begin{bmatrix}1/25&0\\2a/125&1/25\end{bmatrix}\), so \(x=\dfrac{2a}{125}\) and \(\dfrac{125x}a=2\).

Teacher's Note:
1. Use the defining properties (symmetry, orthogonality, idempotence) and reduce the matrix equation to simple scalar conditions.
2. For powers of a matrix, look for a pattern, a nilpotent part, or use the Cayley-Hamilton theorem.