JEE Mathematics Matrices MCQs Set 05

Answer: (a) -2 and 2

Explanation:
1. \( |\operatorname{adj}P|=|P|^{2} \) for a \( 3\times3 \) matrix.
2. \( |\operatorname{adj}P|=1(3-7)-4(6-7)+4(2-1)=-4+4+4=4 \).
3. \( |P|^2=4 \), so \( |P|=\pm2 \).

Teacher's Note:
1. For order n, \( |\operatorname{adj}P|=|P|^{n-1} \).
2. An even exponent means both signs are possible.

Answer: (c) 11

Explanation:
1. \( |\operatorname{adj}A|=|A|^2=16 \).
2. Expanding along the first row, \( |P|=1(12-12)-\alpha(4-6)+3(4-6)=2\alpha-6 \).
3. \( 2\alpha-6=16 \) gives \( \alpha=11 \).

Teacher's Note:
1. For a \( 3\times3 \) matrix, \( |\operatorname{adj}A|=|A|^2 \).
2. Take care with the cofactor signs when expanding.

Answer: (c) R and S only

Explanation:
1. (P) \( (N^TMN)^T=N^TM^TN \), which is \( \pm N^TMN \) according to M. So (P) is correct.
2. (Q) \( (MN-NM)^T=NM-MN \), so it is skew-symmetric. (Q) is correct.
3. (R) \( (MN)^T=NM \), which need not equal MN. (R) is not correct.
4. (S) \( \operatorname{adj}(MN)=(\operatorname{adj}N)(\operatorname{adj}M) \), the order is reversed. (S) is not correct.

Teacher's Note:
1. Apply the reversal law of transposes and adjoints to each statement.
2. Matrix multiplication is not commutative, so the product of two symmetric matrices need not be symmetric.

Answer: (c) I

Explanation:
1. \( B'=(A^{-1}A')'=A(A^{-1})'=A(A')^{-1} \).
2. \( BB'=A^{-1}A'A(A')^{-1} \).
3. Since \( A'A=AA' \): \( BB'=A^{-1}AA'(A')^{-1}=I\cdot I=I \).

Teacher's Note:
1. Use \( (XY)'=Y'X' \) and \( (A^{-1})'=(A')^{-1} \).
2. The condition \( AA'=A'A \) lets you swap the middle two factors.

Answer: (c) R and S only

Explanation:
1. Let \( M=\begin{bmatrix}a&b\\b&c\end{bmatrix} \), so \( |M|=ac-b^2 \).
2. (P) and (Q) only say that M is symmetric. For example \( \begin{bmatrix}1&1\\1&1\end{bmatrix} \) satisfies them but is singular, so they do not guarantee invertibility.
3. (R) If \( b=0 \) and a, c are non-zero, then \( |M|=ac\ne0 \). Correct.
4. (S) If ac is not the square of an integer, it cannot equal \( b^2 \) (b is an integer), so \( |M|\ne0 \). Correct.

Teacher's Note:
1. Write the general symmetric matrix and test each condition against \( ac-b^2\ne0 \).
2. A single counter-example is enough to reject a statement.

Answer: (a) P and Q only

Explanation:
1. Since M and N commute, \( (M-N^2)(M+N^2)=M^2-N^4=O \).
2. Because \( M-N^2\ne O \), the matrix \( M+N^2 \) must be singular, so \( M^2+MN^2=M(M+N^2) \) has determinant 0. (P) is correct and (R) is false.
3. A singular matrix has a non-zero U with \( (M^2+MN^2)U=O \). So (Q) is correct and (S) is false.

Teacher's Note:
1. Factorise using the fact that commuting matrices obey the usual algebraic identities.
2. If the product of two matrices is the zero matrix and one of them is non-zero, the other is singular.

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

Explanation:
1. Rows 1 and 3 are perpendicular: \( a+4+2b=0 \). Rows 2 and 3 are perpendicular: \( 2a+2-2b=0 \).
2. From the second equation \( b=a+1 \); substituting, \( a+4+2a+2=0 \), so \( a=-2,\ b=-1 \).
3. Check: \( a^2+4+b^2=4+4+1=9 \).

Teacher's Note:
1. \( AA^T=9I \) means the rows are mutually perpendicular and each has length 3.
2. Two perpendicularity conditions are enough to determine a and b; the length condition is a check.

Answer: (c) R and S only

Explanation:
1. (P) \( (Y^3Z^4-Z^4Y^3)^T=Z^4(-Y)^3-(-Y)^3Z^4=-Z^4Y^3+Y^3Z^4 \), which equals the original, so it is symmetric.
2. (Q) Even powers of a skew-symmetric matrix are symmetric, so it is symmetric.
3. (R) \( (X^4Z^3-Z^3X^4)^T=Z^3X^4-X^4Z^3 \), the negative of the original, so it is skew-symmetric.
4. (S) Odd powers of a skew-symmetric matrix are skew-symmetric, and so is their sum.

Teacher's Note:
1. Take the transpose of each expression and compare with the original.
2. Remember: \( (A^n)^T=(A^T)^n \).

Answer: (b) 5

Explanation:
1. \( A\operatorname{adj}A=|A|I \), so \( AA^T \) must be the diagonal matrix \( |A|I \) with \( |A|=10a+3b \).
2. The off-diagonal entry of \( AA^T \) is \( 15a-2b=0 \), so \( b=\frac{15a}{2} \).
3. The (2,2) entry is \( 9+4=13=|A|=10a+3b=10a+\frac{45a}{2} \), so \( \frac{65a}{2}=13 \), giving \( a=\frac25,\ b=3 \).
4. \( 5a+b=2+3=5 \).

Teacher's Note:
1. \( A\operatorname{adj}A \) is always a scalar matrix, which restricts \( AA^T \) as well.
2. Use the off-diagonal entry and one diagonal entry.

Answer: (b) Q and R only

Explanation:
1. \( Q=kP^{-1} \) and \( |P|=12\alpha+20 \). The (2,3) entry of \( kP^{-1} \) works out to \( q_{23}=\frac{k(-3\alpha-4)}{12\alpha+20} \).
2. Setting \( q_{23}=-\frac k8 \): \( 8(3\alpha+4)=12\alpha+20 \), so \( \alpha=-1 \) and \( |P|=8 \).
3. \( |Q|=\frac{k^3}{|P|}=\frac{k^3}{8}=\frac{k^2}{2} \) gives \( k=4 \). So (P) is false, and \( 4\alpha-k+8=-4-4+8=0 \) makes (Q) true.
4. \( \det(P\operatorname{adj}Q)=|P||Q|^2=8\cdot64=2^9 \), so (R) is true. \( \det(Q\operatorname{adj}P)=|Q||P|^2=8\cdot64=2^9\ne2^{13} \), so (S) is false.

Teacher's Note:
1. Use \( Q=kP^{-1} \) and the identities \( |\operatorname{adj}A|=|A|^{2} \) and \( |kA|=k^3|A| \) for \( 3\times3 \) matrices.
2. Solve for the two unknowns from the two given conditions before testing the statements.

Answer: (b) 103

Explanation:
1. Write \( P=I+N \) with \( N=\begin{bmatrix}0&0&0\\4&0&0\\16&4&0\end{bmatrix} \). Then \( N^2=\begin{bmatrix}0&0&0\\0&0&0\\16&0&0\end{bmatrix} \) and \( N^3=O \).
2. \( P^{50}=I+50N+\binom{50}{2}N^2 \), so \( Q=P^{50}-I=50N+1225N^2 \).
3. \( q_{21}=200,\ q_{32}=200,\ q_{31}=800+1225\cdot16=20400 \).
4. \( \frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=103 \).

Teacher's Note:
1. Split P as I plus a nilpotent matrix and use the binomial theorem, since I commutes with N.
2. The series stops because \( N^3=O \).

Answer: (a) 1

Explanation:
1. \( P^2=\begin{bmatrix}z^{2r}+z^{4s}&z^{2s}\left((-z)^r+z^r\right)\\z^{2s}\left((-z)^r+z^r\right)&z^{4s}+z^{2r}\end{bmatrix} \).
2. For the off-diagonal entries to vanish, \( (-z)^r+z^r=0 \), so r must be odd: r = 1 or 3.
3. The diagonal needs \( z^{2r}+z^{4s}=-1 \). Since \( z^3=1 \), this needs \( z^{2r} \) and \( z^{4s} \) to be \( z \) and \( z^2 \) in some order. For r = 3, \( z^{2r}=1 \), which fails. For r = 1: \( z^2+z^{4s}=-1 \) requires \( z^{4s}=z \), that is \( s=1 \).
4. Only (1, 1) works, so the number of pairs is 1.

Teacher's Note:
1. First kill the off-diagonal entries (this fixes the parity of r), then match the diagonal using \( 1+z+z^2=0 \).
2. Reduce exponents modulo 3 because \( z^3=1 \).

Answer: (b) \( \begin{bmatrix}51&63\\84&72\end{bmatrix} \)

Explanation:
1. \( A^2=\begin{bmatrix}16&-9\\-12&13\end{bmatrix} \), so \( 3A^2=\begin{bmatrix}48&-27\\-36&39\end{bmatrix} \) and \( 12A=\begin{bmatrix}24&-36\\-48&12\end{bmatrix} \).
2. \( 3A^2+12A=\begin{bmatrix}72&-63\\-84&51\end{bmatrix} \).
3. For a \( 2\times2 \) matrix \( \begin{bmatrix}p&q\\r&s\end{bmatrix} \), the adjoint is \( \begin{bmatrix}s&-q\\-r&p\end{bmatrix} \), giving \( \begin{bmatrix}51&63\\84&72\end{bmatrix} \).

Teacher's Note:
1. For a \( 2\times2 \) matrix, swap the diagonal entries and change the signs of the off-diagonal entries to get the adjoint.
2. Compute the matrix first, then take its adjoint.

Answer: (b) 198

Explanation:
1. The sum of the diagonal entries of \( M^TM \) equals the sum of the squares of all nine entries of M.
2. We need squares summing to 5 using entries 0, 1, 2 (squares 0, 1, 4).
3. Case 1: one entry is 2 and one entry is 1, the other seven are 0: \( 9\times8=72 \) matrices.
4. Case 2: five entries are 1 and four are 0: \( \binom95=126 \) matrices.
5. Total \( 72+126=198 \).

Teacher's Note:
1. \( \operatorname{tr}(M^TM)=\sum m_{ij}^2 \), which turns the problem into a counting problem.
2. List the ways of writing 5 as a sum of squares 1 and 4.

Answer: (b) P and R only

Explanation:
1. If \( B^2 \) equals the given matrix, then \( |B^2|=|B|^2\ge0 \) for real B.
2. The determinants are: (P) -1, (Q) 1, (R) -1, (S) 1. So (P) and (R) cannot be squares.
3. (Q) is the square of \( \begin{bmatrix}1&0&0\\0&0&1\\0&-1&0\end{bmatrix} \), and (S) is the square of I.

Teacher's Note:
1. A square of a real matrix always has a non-negative determinant.
2. For the matrices with positive determinant, find an explicit square root to confirm.

Answer: (d) 4

Explanation:
1. The columns of A must be mutually perpendicular, each with squared length 3.
2. Column 1: \( 8x^2=3 \), so \( x^2=\frac38 \). Column 2: \( 4y^2+y^2+y^2=6y^2=3 \), so \( y^2=\frac12 \). Column 3: \( 1+1+1=3 \) (automatically satisfied).
3. Perpendicularity: col 1 and col 2: \( 0+2xy-2xy=0 \); col 1 and col 3: \( 0-2x+2x=0 \); col 2 and col 3: \( 2y\cdot1+y(-1)+(-y)(1)=0 \). All satisfied.
4. So x = ± the root and y = ± the root: 2 x 2 = 4 matrices (x is never equal to y).

Teacher's Note:
1. \( A^TA=3I \) is equivalent to the columns being mutually perpendicular with the same length.
2. Each independent sign choice for x and y gives a different matrix.

Answer: (a) \( \begin{bmatrix}1&-13\\0&1\end{bmatrix} \)

Explanation:
1. Matrices of the form \( \begin{bmatrix}1&a\\0&1\end{bmatrix} \) multiply by adding the upper-right entries.
2. So \( 1+2+\cdots+(n-1)=\frac{n(n-1)}2=78 \), which gives \( n(n-1)=156 \) and \( n=13 \).
3. \( \begin{bmatrix}1&13\\0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-13\\0&1\end{bmatrix} \).

Teacher's Note:
1. \( \begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix}=\begin{bmatrix}1&a+b\\0&1\end{bmatrix} \) turns the product into a sum.
2. The inverse of this type of matrix just changes the sign of the upper-right entry.

Answer: (d) \( \frac\pi{64} \)

Explanation:
1. A is the rotation matrix, so \( A^n=\begin{bmatrix}\cos n\alpha&-\sin n\alpha\\\sin n\alpha&\cos n\alpha\end{bmatrix} \).
2. \( A^{32} \) equals the matrix for rotation by \( \frac\pi2 \), so \( 32\alpha=\frac\pi2 \) is a solution.
3. \( \alpha=\frac\pi{64} \).

Teacher's Note:
1. Powers of rotation matrices add the angles.
2. Compare the entries cos and sin of \( 32\alpha \) with 0 and 1 to find the angle.

Answer: (a) 10

Explanation:
1. \( P=I+N \) with \( N=\begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix} \), \( N^2=\begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix} \), \( N^3=O \).
2. \( P^5=I+5N+10N^2 \), so \( Q=I+P^5 \) has \( q_{21}=15,\ q_{32}=15,\ q_{31}=45+90=135 \).
3. \( \frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=10 \).

Teacher's Note:
1. Write P as I plus a nilpotent matrix and apply the binomial theorem.
2. Only the off-diagonal entries of \( P^5 \) matter here, since Q differs from \( P^5 \) by I.

Answer: (c) \( \frac1{16} \)

Explanation:
1. \( |A|^2|B|=8 \) and \( \frac{|A|}{|B|}=8 \).
2. From the second, \( |A|=8|B| \). Substituting: \( 64|B|^3=8 \), so \( |B|=\frac12 \) and \( |A|=4 \).
3. \( \det(BA^{-1}B^T)=\frac{|B|^2}{|A|}=\frac{1/4}{4}=\frac1{16} \).

Teacher's Note:
1. Use \( |XY|=|X||Y| \), \( |X^T|=|X| \) and \( |X^{-1}|=\frac1{|X|} \).
2. Two equations in \( |A| \) and \( |B| \) are enough.

Answer: (c) \( \frac1{\sqrt2} \)

Explanation:
1. The columns of A form an orthonormal set (since \( A^TA=I \) also holds). Column 1: \( 2p^2=1 \).
2. Hence \( p^2=\frac12 \) and \( |p|=\frac1{\sqrt2} \).
3. Check with the other columns: \( 6q^2=1 \) and \( 3r^2=1 \), and the columns are mutually perpendicular.

Teacher's Note:
1. For a square matrix \( AA^T=I \) implies \( A^TA=I \), so columns can be used as well as rows.
2. The first column contains only p, which makes the equation immediate.

Answer: (c) \( \begin{bmatrix}\frac{\sqrt3}2&\frac12\\-\frac12&\frac{\sqrt3}2\end{bmatrix} \)

Explanation:
1. \( A^{-50} \) is the rotation by \( -50\theta=-\frac{50\pi}{12}=-\frac{25\pi}6 \).
2. \( -\frac{25\pi}6=-4\pi-\frac\pi6 \), so it equals the rotation by \( -\frac\pi6 \).
3. \( A^{-50}=\begin{bmatrix}\cos\frac\pi6&\sin\frac\pi6\\-\sin\frac\pi6&\cos\frac\pi6\end{bmatrix}=\begin{bmatrix}\frac{\sqrt3}2&\frac12\\-\frac12&\frac{\sqrt3}2\end{bmatrix} \).

Teacher's Note:
1. Rotation matrices add angles, and a full turn of \( 2\pi \) can be dropped.
2. The negative power rotates in the opposite direction.