JEE Mathematics Matrices MCQs Set 07

Answer: (b) \( \begin{bmatrix}8&6\\6&16\end{bmatrix} \) (a symmetric matrix)

Explanation:
1. \( A^T=\begin{bmatrix}4&5\\1&8\end{bmatrix} \).
2. \( A+A^T=\begin{bmatrix}8&6\\6&16\end{bmatrix} \).
3. Its transpose is the same matrix, so it is symmetric.

Teacher's Note:
1. For any square matrix A, the matrix A + A^T is always symmetric.
2. Check by comparing the (1,2) and (2,1) entries.

Answer: (d) \( \begin{bmatrix}0&-1\\1&0\end{bmatrix} \) (a skew-symmetric matrix)

Explanation:
1. \( A^T=\begin{bmatrix}3&5\\4&1\end{bmatrix} \).
2. \( A-A^T=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \).
3. Its transpose is the negative of itself, so it is skew-symmetric.

Teacher's Note:
1. For any square matrix A, the matrix A - A^T is always skew-symmetric.
2. The diagonal entries of a skew-symmetric matrix are always zero.

Answer: (a) \( \begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix} \)

Explanation:
1. \( A+B+C=O \Rightarrow C=-(A+B) \).
2. \( A+B=\begin{bmatrix}3&-4&1\\3&0&1\end{bmatrix} \).
3. \( C=-(A+B)=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix} \).

Teacher's Note:
1. Add the matrices entry by entry, then change every sign.
2. The additive inverse of a matrix is obtained by negating all its entries.

Answer: (c) \( \begin{bmatrix}\frac{1}{2}&\frac{9}{2}&\frac{25}{2}\\0&2&8\end{bmatrix} \)

Explanation:
1. \( a_{11}=\frac{(1-2)^2}{2}=\frac12,\ a_{12}=\frac{9}{2},\ a_{13}=\frac{25}{2} \).
2. \( a_{21}=0,\ a_{22}=\frac{(2-4)^2}{2}=2,\ a_{23}=\frac{(2-6)^2}{2}=8 \).
3. So the matrix is \( \begin{bmatrix}\frac{1}{2}&\frac{9}{2}&\frac{25}{2}\\0&2&8\end{bmatrix} \).

Teacher's Note:
1. Substitute i and j carefully: i is the row number and j the column number.
2. The factor 2 in the denominator applies to every entry.

Answer: (c) \( \begin{bmatrix}7&-3\\-2&1\end{bmatrix} \)

Explanation:
1. Write A = IA and apply \( R_2\to R_2-2R_1 \): \( \begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}A \).
2. Apply \( R_1\to R_1-3R_2 \): \( \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}A \).
3. Hence \( A^{-1}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix} \) (\( |A|=1 \)).

Teacher's Note:
1. Row operations applied to A = IA turn the left side into I and the right side into the inverse.
2. Always verify by multiplying A with the answer to get I.

Answer: (a) 1

Explanation:
1. \( A^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix} \).
2. Compare entry (1,1) of \( kA-2I \): \( 3k-2=1 \Rightarrow k=1 \).
3. Check (1,2): -2k = -2 and (2,1): 4k = 4, both give k = 1.

Teacher's Note:
1. Compute A^2 first, then compare one or two entries.
2. Verify with the remaining entries to be safe.

Answer: (d) \( \begin{bmatrix}\cos nA&\sin nA\\-\sin nA&\cos nA\end{bmatrix} \)

Explanation:
1. \( X^2=\begin{bmatrix}\cos 2A&\sin 2A\\-\sin 2A&\cos 2A\end{bmatrix} \) using \( \cos^2A-\sin^2A=\cos2A \) and \( 2\sin A\cos A=\sin2A \).
2. \( X^3=X\cdot X^2 \) gives the same pattern with 3A.
3. By induction, multiplying by X adds A to the angle, so \( X^n \) has angle nA.

Teacher's Note:
1. Compute X^2 and X^3 and spot the pattern, then confirm with induction.
2. This is the matrix form of De Moivre's theorem.

Answer: (b) \( \frac15\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix} \)

Explanation:
1. \( A^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix} \) and \( A^2-4A-5I=O \).
2. Multiply by \( A^{-1} \): \( A-4I-5A^{-1}=O \Rightarrow A^{-1}=\frac15(A-4I) \).
3. \( A-4I=\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix} \), so \( A^{-1}=\frac15\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix} \).

Teacher's Note:
1. First find the quadratic relation satisfied by A (Cayley-Hamilton), then use it to get the inverse.
2. Check: A times the answer must give I.

Answer: (b) the zero matrix \( O \)

Explanation:
1. \( A^2=\begin{bmatrix}5&4\\4&5\end{bmatrix} \).
2. \( f(A)=A^2-2A-3I=\begin{bmatrix}5&4\\4&5\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix} \).
3. Every entry cancels, so f(A) = O.

Teacher's Note:
1. To evaluate a polynomial at a matrix, the constant term becomes a multiple of I.
2. Here the characteristic polynomial of A is x^2-2x-3, so Cayley-Hamilton predicts f(A)=O.

Answer: (d) \( -\frac53 \)

Explanation:
1. Multiply the last two matrices first: \( \begin{bmatrix}1&2&3\\4&5&6\\3&2&5\end{bmatrix}\begin{bmatrix}1\\-2\\3\end{bmatrix}=\begin{bmatrix}6\\12\\14\end{bmatrix} \).
2. Then \( \begin{bmatrix}1&x&1\end{bmatrix}\begin{bmatrix}6\\12\\14\end{bmatrix}=[6+12x+14]=[20+12x] \).
3. \( 20+12x=0 \Rightarrow x=-\frac53 \).

Teacher's Note:
1. Matrix multiplication is associative, so you may group the product in the convenient way.
2. A 1 x 1 matrix equal to zero means its single entry is zero.

Answer: (a) \( \begin{bmatrix}1&- \frac{3}{2}&\frac{1}{2}\\- \frac{3}{2}&8&\frac{9}{2}\\\frac{1}{2}&\frac{9}{2}&5\end{bmatrix} \)

Explanation:
1. \( A+A^T=\begin{bmatrix}2&-3&1\\-3&16&9\\1&9&10\end{bmatrix} \).
2. Half of it: \( P=\begin{bmatrix}1&- \frac{3}{2}&\frac{1}{2}\\- \frac{3}{2}&8&\frac{9}{2}\\\frac{1}{2}&\frac{9}{2}&5\end{bmatrix} \), which equals its transpose, so P is symmetric.
3. The skew-symmetric part is \( \frac{A-A^T}{2}=\begin{bmatrix}0&\frac{9}{2}&\frac{9}{2}\\- \frac{9}{2}&0&- \frac{3}{2}\\- \frac{9}{2}&\frac{3}{2}&0\end{bmatrix} \) and \( A=P+\frac{A-A^T}{2} \).

Teacher's Note:
1. Every square matrix is the sum of a symmetric and a skew-symmetric matrix.
2. Do not forget the factor 1/2.

Answer: (c) (-4, 1)

Explanation:
1. \( A^2=\begin{bmatrix}11&8\\4&3\end{bmatrix} \).
2. \( A^2+aA+bI=\begin{bmatrix}3 a + b + 11&2 a + 8\\a + 4&a + b + 3\end{bmatrix}=O \).
3. From the (2,1) entry \( 4+a=0\Rightarrow a=-4 \); then \( 3+a+b=0\Rightarrow b=1 \).

Teacher's Note:
1. Equate every entry to zero and solve the resulting linear equations.
2. The relation found is the characteristic equation x^2 - 4x + 1 = 0 of A.

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

Explanation:
1. From \( A^2-4A+I=O \), multiply by \( A^{-1} \): \( A-4I+A^{-1}=O \).
2. So \( A^{-1}=4I-A=\begin{bmatrix}1&-2\\-1&3\end{bmatrix} \).
3. Check: \( |A|=3-2=1 \) and \( \operatorname{adj}A=\begin{bmatrix}1&-2\\-1&3\end{bmatrix} \).

Teacher's Note:
1. Using the matrix equation avoids computing the adjoint.
2. In a 2 x 2 matrix, swap the diagonal entries and change the signs of the other two.

Answer: (a) \( \begin{bmatrix}-1&1\\-2&\frac{3}{2}\end{bmatrix} \)

Explanation:
1. Multiply \( A^2=A-2I \) by \( A^{-1} \): \( A=I-2A^{-1} \).
2. Hence \( 2A^{-1}=I-A=\begin{bmatrix}-2&2\\-4&3\end{bmatrix} \).
3. \( A^{-1}=\begin{bmatrix}-1&1\\-2&\frac{3}{2}\end{bmatrix} \). Check: \( |A|=-6+8=2 \).

Teacher's Note:
1. The relation A^2 = A - 2I holds for this matrix with lambda = 1.
2. Remember to divide by 2 at the end.

Answer: (d) \( \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix} \)

Explanation:
1. Start with A = IA. Use \( R_2\to R_2+R_1 \), \( R_2\to\frac15R_2 \), \( R_3\to R_3+2R_2 \), \( R_3\to5R_3 \).
2. Then \( R_1\to R_1+2R_3 \), \( R_2\to R_2+\frac25R_3 \) and \( R_1\to R_1-2R_2 \) reduce the left side to I.
3. The right-hand matrix is \( A^{-1}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix} \), and \( |A|=1 \).

Teacher's Note:
1. Do the row operations in a fixed order: make zeros below the diagonal, then above it.
2. Check the answer by confirming that A times it equals I.

Answer: (b) \( \begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix} \)

Explanation:
1. \( AB=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix} \) (a 3 x 3 matrix).
2. Transposing: \( (AB)^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix} \).
3. This equals \( B^TA^T \), verifying the reversal law \( (AB)^T=B^TA^T \).

Teacher's Note:
1. The transpose of a product reverses the order: (AB)^T = B^T A^T.
2. A column times a row gives a square matrix.

Answer: (b) \( I \)

Explanation:
1. \( (I+A)^3=I+3A+3A^2+A^3 \).
2. Since \( A^2=A \), also \( A^3=A\cdot A^2=A^2=A \).
3. So \( (I+A)^3-7A=I+3A+3A+A-7A=I \).

Teacher's Note:
1. Idempotent matrices satisfy A^n = A for every n >= 1.
2. I and A commute, so the binomial expansion is valid.

Answer: (d) \( \begin{bmatrix}4&-7\\-3&10\end{bmatrix} \)

Explanation:
1. Pre-multiply by \( A^{-1} \): \( X=A^{-1}B \). Here \( |A|=1 \) and \( A^{-1}=\begin{bmatrix}1&-3\\-1&4\end{bmatrix} \).
2. \( X=\begin{bmatrix}1&-3\\-1&4\end{bmatrix}\begin{bmatrix}7&2\\1&3\end{bmatrix}=\begin{bmatrix}4&-7\\-3&10\end{bmatrix} \).

Teacher's Note:
1. In AX = B the unknown is on the right of A, so multiply on the left by A^{-1}.
2. Multiplying on the wrong side gives a different (wrong) matrix.

Answer: (a) \( A_{\alpha+\beta} \)

Explanation:
1. Multiply: the (1,1) entry is \( \cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta) \).
2. The (1,2) entry is \( \cos\alpha\sin\beta+\sin\alpha\cos\beta=\sin(\alpha+\beta) \).
3. The second row follows the same way, giving the matrix with angle alpha + beta.

Teacher's Note:
1. Rotation matrices compose by adding their angles.
2. This is why the n-th power of such a matrix has angle n times alpha.

Answer: (c) a symmetric matrix

Explanation:
1. A is skew-symmetric: \( A^T=-A \). Let \( B=A^{2n} \).
2. \( B^T=(A^T)^{2n}=(-A)^{2n}=A^{2n}=B \).
3. So B equals its own transpose, i.e. B is symmetric.

Teacher's Note:
1. Use (A^n)^T = (A^T)^n.
2. An odd power of a skew-symmetric matrix is skew-symmetric.

Answer: (c) 1

Explanation:
1. \( \begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix}^{-1}=\frac{1}{1+\tan^2\theta}\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix} \).
2. The product is \( \frac{1}{1+\tan^2\theta}\begin{bmatrix}1-\tan^2\theta&-2\tan\theta\\2\tan\theta&1-\tan^2\theta\end{bmatrix}=\begin{bmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{bmatrix} \).
3. So \( a=\cos2\theta,\ b=\sin2\theta \) and \( a^2+b^2=1 \).

Teacher's Note:
1. Find the inverse of the second matrix first using the 2 x 2 formula.
2. Use (1 - tan^2)/(1 + tan^2) = cos 2 theta and 2 tan/(1 + tan^2) = sin 2 theta.

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

Explanation:
1. \( EF=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix} \) (row i of E picks rows of F).
2. For comparison \( FE=\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix} \), so EF and FE are different.
3. Hence E and F do not commute.

Teacher's Note:
1. Multiply row by column carefully; EF and FE are generally not equal.
2. Here EF and FE are complementary diagonal matrices whose sum is I.

Answer: (d) \( E \)

Explanation:
1. \( E^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} \).
2. \( E^2F=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix} \) and \( FE^2=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix} \).
3. Their sum is \( \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}=E \).

Teacher's Note:
1. Find E^2 first; it has a single non-zero entry.
2. Add the two products entry by entry.

Answer: (b) 8

Explanation:
1. Condition (ii) holds automatically when all entries are 1 or -1, so there are 16 matrices to test.
2. Condition (iii) says \( a_{11}a_{21}+a_{12}a_{22}=0 \), i.e. the two products have opposite signs.
3. Choose the first row in 4 ways; the second row must be orthogonal to it, and exactly 2 of the 4 possible rows are. So 4 x 2 = 8.

Teacher's Note:
1. Count the first row freely, then count how many second rows are orthogonal to it.
2. These 8 matrices are exactly the 2 x 2 orthogonal matrices with entries 1 or -1.

Answer: (b) \( x=\frac32,\ y=2 \)

Explanation:
1. The left side is \( \begin{bmatrix}3y-2x&3y-2x\\3y&3y\\2y+4x&2y+4x\end{bmatrix} \).
2. Comparing the first row: 3y - 2x = 3; the second row: 3y = 3y (no information); the third row: 2y + 4x = 10.
3. Solving \( 3y-2x=3 \) and \( 2y+4x=10 \) gives \( x=\frac32,\ y=2 \).

Teacher's Note:
1. Multiply the matrices first and then equate corresponding entries.
2. Two of the three row equations are enough to find x and y.

Answer: (d) 5049

Explanation:
1. AB = B gives \( (A-I)B=O \) with \( B\ne O \), so \( A-I \) is singular: \( |A-I|=0 \).
2. \( (a-1)(d-1)-bc=0\Rightarrow ad-bc-(a+d)+1=0 \).
3. Hence \( ad-bc=(a+d)-1=5050-1=5049 \).

Teacher's Note:
1. A homogeneous system with a non-zero solution has a zero determinant.
2. Expand |A - I| as a 2 x 2 determinant.

Answer: (a) \( \begin{bmatrix}0\\\frac1{11}\end{bmatrix} \)

Explanation:
1. \( A^2=\begin{bmatrix}3&0\\0&3\end{bmatrix}=3I \).
2. So \( A^4=9I,\ A^6=27I,\ A^8=81I \) and the bracket is \( (81+27+9+3+1)I=121I \).
3. \( 121V=\begin{bmatrix}0\\11\end{bmatrix}\Rightarrow V=\begin{bmatrix}0\\\frac1{11}\end{bmatrix} \).

Teacher's Note:
1. Spot that A^2 is a scalar matrix; then every even power is a scalar matrix.
2. Add the scalars before dividing.

Answer: (c) \( O \)

Explanation:
1. The characteristic polynomial of A is \( |xI-A|=x^3-6x^2+7x+2 \) (trace 6, sum of principal 2 x 2 minors 7, determinant -2).
2. By the Cayley-Hamilton theorem A satisfies its own characteristic equation.
3. Direct calculation: \( A^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix} \), \( A^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix} \), and \( A^3-6A^2+7A+2I=O \).

Teacher's Note:
1. Cayley-Hamilton: every square matrix satisfies its characteristic equation.
2. For a 3 x 3 matrix it reads x^3 - (trace)x^2 + (sum of minors)x - det = 0.

Answer: (c) \( \frac{n(n-1)}{2}\lambda^{n-2} \)

Explanation:
1. Write \( J=\lambda I+N \) where \( N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix} \), \( N^2 \) has only a 1 in position (1,3) and \( N^3=O \).
2. Since \( \lambda I \) commutes with N, \( J^n=\lambda^nI+n\lambda^{n-1}N+\binom n2\lambda^{n-2}N^2 \).
3. The (1,3) entry comes only from \( N^2 \): \( \binom n2\lambda^{n-2}=\frac{n(n-1)}{2}\lambda^{n-2} \).

Teacher's Note:
1. Split the matrix into a scalar part and a nilpotent part and use the binomial theorem.
2. The binomial theorem works because the two parts commute.

Answer: (a) \( a^nI+na^{n-1}bA \)

Explanation:
1. \( A^2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=O \), so higher powers of A vanish.
2. I commutes with A, so the binomial theorem gives \( (aI+bA)^n=a^nI+\binom n1a^{n-1}bA+0 \).
3. \( =a^nI+na^{n-1}bA \).

Teacher's Note:
1. When N^2 = O only the first two terms of the binomial expansion survive.
2. Check n = 1: aI + bA, as expected.