JEE Mathematics Matrices MCQs Set 04

Answer: (c) 4

Explanation:
1. A is symmetric, so \( A^TA=A^2=I \). The (1,1) entry gives \( a^2+b^2+c^2=1 \) and the (1,2) entry gives \( ab+bc+ca=0 \).
2. Then \( (a+b+c)^2=1 \) and, as a, b, c are positive, \( a+b+c=1 \).
3. \( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=1 \).
4. So \( a^3+b^3+c^3=1+3abc=4 \).

Teacher's Note:
1. Multiply the matrix by its transpose and compare entries with the identity.
2. The identity for \( a^3+b^3+c^3-3abc \) is the key step.

Answer: (a) \( \pm 3 \)

Explanation:
1. \( |A^3|=|A|^3=125 \), so \( |A|=5 \).
2. \( |A|=\alpha^2-4=5 \), so \( \alpha^2=9 \) and \( \alpha=\pm3 \).

Teacher's Note:
1. Use \( |A^n|=|A|^n \) to reduce the cube to the determinant itself.
2. Keep both signs when taking the square root.

Answer: (c) \( -6,\ 11 \)

Explanation:
1. \( \det(A-\lambda I)=(1-\lambda)\left[(1-\lambda)(4-\lambda)+2\right]=(1-\lambda)(\lambda^2-5\lambda+6) \), which expands to \( -(\lambda^3-6\lambda^2+11\lambda-6) \).
2. By Cayley-Hamilton, \( A^3-6A^2+11A-6I=O \).
3. Multiplying by \( A^{-1} \): \( A^2-6A+11I=6A^{-1} \), so \( A^{-1}=\frac16(A^2-6A+11I) \).
4. Hence \( c=-6,\ d=11 \).

Teacher's Note:
1. Cayley-Hamilton expresses the inverse as a polynomial in A without computing adjoints.
2. The constant term of the characteristic polynomial equals the determinant, here 6.

Answer: (a) \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \)

Explanation:
1. P is orthogonal, so \( P^TP=PP^T=I \).
2. \( Q^n=PA^nP^T \) because the inner factors \( P^TP \) cancel.
3. \( X=P^TPA^{2005}P^TP=A^{2005} \).
4. \( A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix} \), so \( X=\begin{bmatrix}1&2005\\0&1\end{bmatrix} \).

Teacher's Note:
1. Spot orthogonal matrices first; they make products of the form \( P^T(\cdot)P \) collapse.
2. The powers of \( \begin{bmatrix}1&1\\0&1\end{bmatrix} \) follow the pattern \( \begin{bmatrix}1&n\\0&1\end{bmatrix} \).

Answer: (a) 3

Explanation:
1. Combining the three equations, \( AU=B=\begin{bmatrix}1&2&2\\0&3&3\\0&0&1\end{bmatrix} \).
2. \( |A||U|=|B| \) with \( |A|=1 \) and \( |B|=3 \) (triangular matrices).
3. Therefore \( |U|=3 \).

Teacher's Note:
1. Combine the column equations into the single matrix equation \( AU=B \).
2. The determinant of a triangular matrix is the product of its diagonal entries.

Answer: (b) 0

Explanation:
1. \( AU=B=\begin{bmatrix}1&2&2\\0&3&3\\0&0&1\end{bmatrix} \), so \( U=A^{-1}B \) and \( U^{-1}=B^{-1}A \).
2. Computing, \( U^{-1}=\begin{bmatrix}-\frac13&-\frac23&0\\-\frac73&-\frac53&-1\\3&2&1\end{bmatrix} \).
3. Sum of entries \( =-1+\left(-\frac{7}{3}-\frac53-1\right)+6=-1-5+6=0 \).

Teacher's Note:
1. Invert the product in reverse order: \( (A^{-1}B)^{-1}=B^{-1}A \).
2. Add the three row sums carefully: -1, -5 and 6.

Answer: (a) 5

Explanation:
1. \( U=A^{-1}B=\begin{bmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{bmatrix} \).
2. \( U\begin{bmatrix}3\\2\\0\end{bmatrix}=\begin{bmatrix}7\\-8\\-5\end{bmatrix} \).
3. \( \begin{bmatrix}3&2&0\end{bmatrix}\begin{bmatrix}7\\-8\\-5\end{bmatrix}=21-16=5 \).

Teacher's Note:
1. Evaluate the product from the right to keep the arithmetic short.
2. Row-by-column products give a single number here.

Answer: (a) 12

Explanation:
1. Symmetric: the three off-diagonal pairs each count 2 ones, each diagonal entry counts 1.
2. Five ones: either all three diagonal entries are 1 and one off-diagonal pair is 1 (total 3+2=5), giving 3 matrices.
3. Or exactly one diagonal 1 and two off-diagonal pairs are 1 (1+4=5), giving 3 x 3 = 9 matrices.
4. Total \( 3+9=12 \).

Teacher's Note:
1. Count by the number of ones on the diagonal, since off-diagonal ones come in pairs.
2. Check that diagonal ones plus twice the number of ones pairs equals 5.

Answer: (b) at least 4 but less than 7

Explanation:
1. A unique solution exists exactly when \( |A|\ne0 \).
2. Of the 12 matrices (three with diagonal 1,1,1 and one off-diagonal pair, nine with one diagonal 1 and two off-diagonal pairs), checking each determinant shows that 6 are non-singular.
3. 6 lies in the range 4 to 6, so the answer is (b).

Teacher's Note:
1. A unique solution of \( AX=B \) needs a non-zero determinant, so test each of the 12 matrices.
2. Several of these matrices have two identical rows, which makes the determinant zero.

Answer: (b) more than 2

Explanation:
1. Inconsistency means the rank of A is less than the rank of the augmented matrix.
2. Testing all 12 matrices, exactly 4 give an inconsistent system.
3. 4 is more than 2.

Teacher's Note:
1. Compare the rank of the coefficient matrix with that of the augmented matrix.
2. Only singular matrices can give an inconsistent system.

Answer: (a) 0

Explanation:
1. A linear system has no solution, exactly one solution or infinitely many solutions.
2. If it had two distinct solutions \( X_1,X_2 \), then every point \( X_1+t(X_2-X_1) \) would also be a solution.
3. So exactly two solutions is impossible and the number of such matrices is 0.

Teacher's Note:
1. Think of the possible solution counts of a linear system before counting matrices.
2. Two solutions force infinitely many.

Answer: (d) \( 2p-1 \)

Explanation:
1. Skew-symmetric needs \( a=-a \) and \( b=-c \), which with entries in \( 0,\dots,p-1 \) means \( a=b=c=0 \). So the only skew-symmetric matrix is O, which is also symmetric.
2. Symmetric means \( b=c \), and \( \det A=a^2-b^2=(a-b)(a+b) \) is divisible by p when \( a=b \) or \( a+b=p \) or \( a=b=0 \).
3. With \( a=b=c \): p matrices. With \( a+b=p \) and \( a,b\ge1 \): p-1 matrices. Total \( p+(p-1)=2p-1 \) (these two sets do not overlap since p is odd).

Teacher's Note:
1. Remember that the entries are the integers 0 to p-1, so skew-symmetry forces a zero matrix.
2. Check with p=3: the formula gives 5, which matches a direct count.

Answer: (c) \( (p-1)^2 \)

Explanation:
1. \( \operatorname{tr}A=2a \), not divisible by p means \( a\ne0 \): so a has \( p-1 \) choices.
2. \( \det A=a^2-bc\equiv0 \pmod p \) means \( bc\equiv a^2 \). For each non-zero residue a, the pairs (b, c) with \( bc\equiv a^2 \) number \( p-1 \) (choose b non-zero, c is then determined).
3. Total \( (p-1)(p-1)=(p-1)^2 \).

Teacher's Note:
1. Because p is odd, \( 2a\not\equiv0 \) exactly when \( a\ne0 \).
2. For a fixed non-zero product, each non-zero b gives exactly one c.

Answer: (d) \( p^3-p^2 \)

Explanation:
1. Total matrices: \( p^3 \).
2. Matrices with det divisible by p: for a=0 we need \( bc\equiv0 \), giving \( 2p-1 \) pairs; for each of the \( p-1 \) non-zero a we get \( p-1 \) pairs. Total \( (2p-1)+(p-1)^2=p^2 \).
3. Matrices with det not divisible by p: \( p^3-p^2 \).

Teacher's Note:
1. Count the complement: the matrices whose determinant is divisible by p.
2. Verify with p=3: the formula gives 18 out of 27.

Answer: (d) 6

Explanation:
1. The equation gives \( a+8b+7c=0 \), \( 9a+2b+3c=0 \), \( 7a+7b+7c=0 \).
2. From the third, \( a+b+c=0 \). Subtracting from the first, \( 7b+6c=0 \), so \( b=-\frac67c \) and \( a=-\frac{c}{7} \). The second equation is then satisfied: \( -\frac97c-\frac{12}{7}c+3c=0 \).
3. On the plane: \( -\frac27c-\frac67c+c=-\frac17c=1 \), so \( c=-7,\ b=6,\ a=1 \).
4. \( 7a+b+c=7+6-7=6 \).

Teacher's Note:
1. Write the row-matrix equation as three linear equations in a, b, c.
2. Check the solution in all three equations, not just two.

Answer: (a) -2

Explanation:
1. The solutions are proportional to \( (1,6,-7) \), so a = 2 gives \( b=12,\ c=-14 \).
2. \( \frac{3}{\omega^2}=3\omega \) (since \( \omega^3=1 \)), \( \frac1{\omega^{12}}=1 \) and \( \frac{3}{\omega^{-14}}=3\omega^{14}=3\omega^2 \).
3. The sum is \( 3\omega+1+3\omega^2=1+3(\omega+\omega^2)=1-3=-2 \).

Teacher's Note:
1. Reduce exponents modulo 3 because \( \omega^3=1 \).
2. Use \( 1+\omega+\omega^2=0 \) to finish.

Answer: (b) 7

Explanation:
1. The solutions are proportional to \( (1,6,-7) \), so b = 6 gives \( a=1,\ c=-7 \).
2. The quadratic is \( x^2+6x-7=0 \): \( \alpha+\beta=-6,\ \alpha\beta=-7 \).
3. \( \frac1\alpha+\frac1\beta=\frac{-6}{-7}=\frac67 \).
4. The geometric series sums to \( \frac1{1-\frac67}=7 \).

Teacher's Note:
1. First solve for a and c, then use sum and product of roots.
2. The series converges because the common ratio is between -1 and 1.

Answer: (c) \( -M^2 \)

Explanation:
1. \( M^T=-M,\ N^T=-N,\ (N^{-1})^T=(N^T)^{-1}=-N^{-1} \).
2. \( (M^TN)^{-1}=(-MN)^{-1}=-(MN)^{-1}=-N^{-1}M^{-1} \).
3. \( (MN^{-1})^T=(N^{-1})^TM^T=(-N^{-1})(-M)=N^{-1}M \).
4. The product is \( M^2N^2(-N^{-1}M^{-1})(N^{-1}M)=-M^2N^2N^{-2}M^{-1}M \) (all commute) \( =-M^2 \).

Teacher's Note:
1. Convert every transpose using skew-symmetry, then use the reversal law for inverses.
2. Because MN = NM, the factors can be rearranged freely.

Answer: (a) 2

Explanation:
1. Expanding along the first row, \( |A|=1(1-c\omega)-a(\omega-c\omega^2)+b(\omega^2-\omega^2)=1-\omega(a+c)+ac\omega^2 \).
2. If \( a\ne c \), then \( a+c=\omega+\omega^2=-1 \) and \( ac=1 \), giving \( 1+\omega+\omega^2=0 \): singular.
3. If \( a=c=\omega^2 \): \( 1-2\omega^3+\omega^3=1-2+1=0 \), singular.
4. If \( a=c=\omega \): \( 1-2\omega^2+\omega^4=1+\omega-2\omega^2=-3\omega^2\ne0 \). Here b can be either value, so S has 2 matrices.

Teacher's Note:
1. List all eight possibilities and test the determinant for each; many give two proportional rows.
2. Use the relations of cube roots of unity to simplify powers.

Answer: (b) 9

Explanation:
1. The first equation gives the second column of M: \( (-1,2,3)^T \).
2. The second gives (first column) - (second column) = \( (1,1,-1)^T \), so the first column is \( (0,3,2)^T \).
3. The third gives the sum of columns \( =(0,0,12)^T \), so the third column is \( (0,0,12)-(0,3,2)-(-1,2,3)=(1,-5,7)^T \).
4. \( M=\begin{bmatrix}0&-1&1\\3&2&-5\\2&3&7\end{bmatrix} \) and the trace is \( 0+2+7=9 \).

Teacher's Note:
1. Multiplying M by a column of 0s and 1s picks out or combines its columns.
2. Only the diagonal entries are needed in the end.

Answer: (d) \( 2^{13} \)

Explanation:
1. \( b_{ij}=2^i\cdot a_{ij}\cdot2^j \), so \( Q=DPD \) with \( D=\operatorname{diag}(2,4,8) \).
2. \( |D|=64=2^6 \).
3. \( |Q|=|D|^2|P|=2^{12}\cdot2=2^{13} \).

Teacher's Note:
1. Writing the entrywise scaling as \( DPD \) with a diagonal matrix D is the standard trick.
2. The determinant of a diagonal matrix is the product of its diagonal entries.

Answer: (d) \( PX=-X \)

Explanation:
1. Take the transpose of \( P^T=2P+I \): \( P=2P^T+I \).
2. Substitute: \( P=2(2P+I)+I=4P+3I \), so \( P=-I \).
3. Then \( PX=-X \) for every X, in particular for a non-zero X.

Teacher's Note:
1. Transposing the given relation and substituting back gives P directly.
2. When P = -I, every non-zero column is mapped to its negative.