JEE Mathematics Matrices MCQs Set 08

Answer: (b) 27,897

Explanation:
1. m(k) is a k x k matrix filled with consecutive integers, so the first entry of m(17) is 1 + (1² + 2² + ... + 16²) = 1 + 1496 = 1497.
2. Consecutive diagonal entries differ by 17 + 1 = 18, so the diagonal is an A.P. with first term 1497 and common difference 18.
3. \( \text{Trace}=\frac{17}{2}[2(1497)+16(18)]=\frac{17}{2}(3282)=27897 \)

Teacher's Note:
1. Check the pattern with m(2) and m(3): the first entries are 2 and 6, which are 1 + 1 and 1 + 1 + 4.
2. Along a diagonal the entries differ by (order + 1).

Answer: (c) \( -YX \)

Explanation:
1. \( X^T=B^TA^T+A^TB^T=BA+AB=X \), so X is symmetric.
2. \( Y^T=BA-AB=-Y \), so Y is skew-symmetric.
3. \( (XY)^T=Y^TX^T=(-Y)(X)=-YX \).

Teacher's Note:
1. The transpose reverses the order of a product.
2. The sum AB + BA of two symmetric matrices is symmetric and the difference AB - BA is skew-symmetric.

Answer: (b) 1

Explanation:
1. P is orthogonal, so \( P^TP=PP^T=I \).
2. \( Q^{2012}=PA^{2012}P^T \), hence \( P^TQ^{2012}P=A^{2012} \).
3. \( A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix} \), so \( A^{2012}=\begin{bmatrix}1&2012\\0&1\end{bmatrix} \) giving \( \alpha=1,\ \beta=0 \) and \( \alpha+\beta=1 \).

Teacher's Note:
1. An orthogonal matrix cancels in the middle of a product P(...)P^T.
2. Powers of an upper unit triangular matrix are found by induction.

Answer: (a) \( \frac{7}{24}I-\frac{1}{24}A \)

Explanation:
1. Write \( A^2-4A+3I=0 \) as \( (A+3I)(A-7I)=-24I \) (expand: \( A^2-4A-21I \) differs from the given equation by \( -24I \)).
2. So \( (A+3I)\left(\frac{7I-A}{24}\right)=I \).
3. Hence \( (A+3I)^{-1}=\frac{7}{24}I-\frac{1}{24}A \).

Teacher's Note:
1. Factor the polynomial in A so that it produces a multiple of I.
2. Then read the inverse off directly.

Answer: (b) 0

Explanation:
1. \( A^2=A^{-1} \) gives \( A^3=I \).
2. Powers of 2 modulo 3 alternate: \( 2^m\equiv(-1)^m\pmod3 \). The exponents \( 2^n \) and \( 2^{n-2} \) have the same parity of m (n and n-2), so they leave the same remainder mod 3.
3. Hence \( A^{2^n}=A^{2^{n-2}} \), i.e. \( B=C \) and \( \det(B-C)=0 \).

Teacher's Note:
1. A^3 = I means only the exponent modulo 3 matters.
2. 2^n and 2^(n-2) differ by 3 times 2^(n-2), a multiple of 3.

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

Explanation:
1. \( |\operatorname{adj}A|=|A|^2=16 \).
2. \( |(\operatorname{adj}A)^{-1}|=\frac{1}{16} \).

Teacher's Note:
1. For an n x n matrix |adj A| = |A|^(n-1).
2. The determinant of an inverse is the reciprocal.

Answer: (b) \( M^2 \)

Explanation:
1. Since M and N are symmetric, \( M^T=M \) and \( (MN^{-1})^T=(N^{-1})^TM^T=N^{-1}M \).
2. The expression becomes \( M^2N^2(MN)^{-1}N^{-1}M=M^2N^2N^{-1}M^{-1}N^{-1}M \).
3. Because M and N commute this simplifies to \( M^2N^2N^{-2}M^{-1}M=M^2 \).

Teacher's Note:
1. Use symmetric + commuting to rearrange freely.
2. (XY)^-1 = Y^-1 X^-1.

Answer: (b) \( (|A|+k)^n \)

Explanation:
1. \( AB=A(\operatorname{adj}A)=|A|I_n \).
2. \( AB+kI_n=(|A|+k)I_n \).
3. Its determinant is \( (|A|+k)^n \).

Teacher's Note:
1. A(adj A) = |A| I is the key identity.
2. det(cI_n) = c^n.

Answer: (a) 0

Explanation:
1. For a 2 x 2 matrix, \( A^3=0 \) implies \( \det A=0 \) and all eigenvalues are 0.
2. By Cayley-Hamilton, \( A^2=(\operatorname{tr}A)A-(\det A)I=0 \) since the trace and determinant are both 0.
3. So \( A^2 \) is the zero matrix and the sum of its elements is 0.

Teacher's Note:
1. A nilpotent 2 x 2 matrix always satisfies A^2 = 0.
2. Cayley-Hamilton: A^2 - (trace)A + (det)I = 0.

Answer: (b) 1

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&-2&1\\2&-1&2\\1&1&\lambda\end{vmatrix}=3(\lambda-1) \), so a unique solution exists unless \( \lambda=1 \).
2. For \( \lambda=1 \): subtract the first equation from the third to get \( 3y=8 \), and subtract twice the first from the second to get \( 3y=10 \). These contradict each other.
3. So the system has no solution only for \( \lambda=1 \).

Teacher's Note:
1. First find where the determinant vanishes, then test consistency at that value.
2. Different values of the same variable from two eliminations signal inconsistency.

Answer: (a) -1

Explanation:
1. Rewrite as the homogeneous system \( a-xb+xc=0,\ ya+b-yc=0,\ -za+zb+c=0 \) in the unknowns a, b, c.
2. Since a, b, c are not all zero, the determinant of the coefficients is zero: \( \begin{vmatrix}1&-x&x\\y&1&-y\\-z&z&1\end{vmatrix}=0 \).
3. Expanding, \( (1+yz)+x(y-yz)+x(yz+z)=1+xy+yz+zx=0 \).

Teacher's Note:
1. A homogeneous system has a non-trivial solution only when its determinant is zero.
2. Check: a = 1, b = 2, c = 4 gives x = -1/2, y = 2/3, z = -4 and xy + yz + zx = -1.

Answer: (a) One

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}3&-1&4\\1&2&-3\\6&5&\lambda\end{vmatrix}=7\lambda+35=7(\lambda+5) \).
2. Since \( t^2\pm3t+4>0 \), \( \lambda \) is positive and (checking its range) lies between 1/7 and 7. So \( \lambda\ne-5 \) and the determinant is non-zero.
3. Hence the system has a unique solution.

Teacher's Note:
1. A non-zero determinant guarantees exactly one solution.
2. Only the sign and size of lambda matter, not its exact value.

Answer: (d) I and III only

Explanation:
1. (I): \( (I+A)^3=I+3A+3A^2+A^3=I+7A \) using \( A^2=A \), so \( (I+A)^3-7A=I \). True, hence (II) is false.
2. (III): as B and C commute, \( A^3=B^3+3B^2C+3BC^2+C^3=B^3+3B^2C \) since \( C^2=O \). So \( A^3-B^3-3B^2C=O \). True, hence (IV) is false.

Teacher's Note:
1. Idempotent matrices make every power of A equal to A.
2. The binomial theorem works only for commuting matrices.

Answer: (a) II and III only

Explanation:
1. Let \( C=(I-AB)^{-1} \). Compute \( (I-BA)(I+BCA)=I+BCA-BA-BABCA \).
2. \( BCA-BABCA=B(I-AB)CA=BA \), so the product is \( I+BA-BA=I \).
3. Hence \( I-BA \) is invertible with inverse \( I+B(I-AB)^{-1}A \) (II and III true). (IV) uses the inverse of the very matrix being inverted, so it is not a valid formula.

Teacher's Note:
1. The push-through identity A(I - BA) = (I - AB)A is behind this result.
2. Check any proposed inverse by multiplying it out.

Answer: (b) 2

Explanation:
1. Check whether row 3 is a combination of rows 1 and 2: \( R_2+2R_1=(1+2,\ 2+2,\ 3+2,\ 4-2)=(3,4,5,2)=R_3 \).
2. So \( R_3-R_2-2R_1 \) is the zero row.
3. Rows 1 and 2 are not proportional, hence the rank is 2.

Teacher's Note:
1. Look for a linear dependence among the rows.
2. Rank equals the number of independent rows.

Answer: (c) B

Explanation:
1. Computing: \( A^2=A \) (idempotent), \( B^2=O \) (nilpotent), \( C^2=I \) (involutory).
2. \( AC^2=AI=A \), which is idempotent and not nilpotent (it has non-zero powers equal to itself).
3. So only B is nilpotent.

Teacher's Note:
1. Multiply each matrix by itself and see which of A^2 = A, B^2 = 0, C^2 = I holds.
2. An idempotent non-zero matrix is never nilpotent.

Answer: (a) Only the product with P

Explanation:
1. \( P\begin{bmatrix}3\\4\\5\end{bmatrix}=\begin{bmatrix}21\\20\\29\end{bmatrix} \) and \( 21^2+20^2=441+400=841=29^2 \): a Pythagorean triplet.
2. \( Q\begin{bmatrix}3\\4\\5\end{bmatrix}=\begin{bmatrix}21\\-20\\29\end{bmatrix} \) and \( R\begin{bmatrix}3\\4\\5\end{bmatrix}=\begin{bmatrix}-21\\20\\29\end{bmatrix} \) contain a negative number, so they are not triplets of positive integers.

Teacher's Note:
1. Multiply each matrix with the column and check positivity and a^2 + b^2 = c^2.
2. These matrices generate the tree of primitive Pythagorean triplets.

Answer: (b) 6

Explanation:
1. The equation E means \( a+8b+7c=0,\ 9a+2b+3c=0,\ 7a+7b+7c=0 \). The third equation gives \( a+b+c=0 \); subtracting from the first gives \( 7b+6c=0 \), so \( (a,b,c)=t(1,6,-7) \).
2. Plane condition: \( 2t+6t-7t=t=1 \), so \( (a,b,c)=(1,6,-7) \).
3. \( 7a+b+c=7+6-7=6 \).

Teacher's Note:
1. Solve the homogeneous system for the ratio a : b : c first.
2. Then use the extra condition to fix the scale.

Answer: (d) 7

Explanation:
1. The equation E means \( a+8b+7c=0,\ 9a+2b+3c=0,\ 7a+7b+7c=0 \). The third equation gives \( a+b+c=0 \); subtracting from the first gives \( 7b+6c=0 \), so \( (a,b,c)=t(1,6,-7) \).
2. With b = 6: \( t=1 \), so \( (a,b,c)=(1,6,-7) \).
3. \( \frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{-b/a}{c/a}=-\frac bc=\frac67 \).
4. A G.P. with ratio \( \frac67 \) sums to \( \frac1{1-6/7}=7 \).

Teacher's Note:
1. The sum of the roots over the product equals -b/c.
2. An infinite G.P. with |r| < 1 sums to 1/(1 - r).

Answer: (d) -4

Explanation:
1. Enumerating all 16 admissible matrices, the determinant takes only the values 0, 4 and -4.
2. Hence the minimum is -4.

Teacher's Note:
1. A 3 x 3 determinant with entries +-1 is always divisible by 4.
2. The maximum possible in absolute value is 4.

Answer: (c) 2

Explanation:
1. Rows of an orthogonal matrix are mutually perpendicular unit vectors. Row 2 with row 3: \( 2a+2-2b=0 \) so \( a=b-1 \).
2. Row 1 with row 3: \( a+4+2b=0 \) so \( a=-4-2b \).
3. Equating: \( b-1=-4-2b \) gives \( b=-1,\ a=-2 \). Check the length: \( 4+4+1=9 \) works.
4. \( ab=2 \).

Teacher's Note:
1. Use perpendicularity of rows, then confirm the length of the third row.
2. Dividing by 3 makes the first row a unit vector.

Answer: (d) 3

Explanation:
1. \( \begin{bmatrix}1&p\\0&1\end{bmatrix}\begin{bmatrix}1&q\\0&1\end{bmatrix}=\begin{bmatrix}1&p+q\\0&1\end{bmatrix} \), so the top-right entry of the product is \( 1+2+\cdots+n=\frac{n(n+1)}2 \).
2. \( \frac{n(n+1)}2=378 \) gives \( n(n+1)=756=27\times28 \), so n = 27.
3. \( a^3=27 \) gives a = 3.

Teacher's Note:
1. Upper unit triangular matrices of this type add their off-diagonal entries.
2. Solve the quadratic in n by factorising 756.

Answer: (c) 4

Explanation:
1. From the recurrence, \( A^n-A^{n-2}=A^2-I \). Adding the relations for n = 4, 6, ..., 50 (24 steps) gives \( A^{50}-A^2=24(A^2-I) \).
2. So \( A^{50}=25A^2-24I \) and k = 25.
3. \( \left[\frac{25}6\right]=[4.17]=4 \).

Teacher's Note:
1. The recurrence telescopes in steps of 2.
2. Count the number of steps carefully: from 2 to 50 is 24 steps.

Answer: (d) 2

Explanation:
1. \( A^n=2^{n-1}A \) by induction.
2. \( A^n-I=\begin{bmatrix}2^{n-1}-1&2^{n-1}\\2^{n-1}&2^{n-1}-1\end{bmatrix} \).
3. \( \det=(2^{n-1}-1)^2-2^{2n-2}=1-2^n \). So \( \lambda=2 \).

Teacher's Note:
1. A^2 = 2A for this matrix, which gives the power pattern.
2. The determinant is a difference of squares.

Answer: (b) 7

Explanation:
1. \( B^{-1}AB=A^2 \). Applying this repeatedly, \( B^{-3}AB^3=A^8 \).
2. Since \( B^3=I \), the left side is A, so \( A=A^8 \) and \( A^7=I \).
3. Because 7 is prime and \( A\ne I \), the order of A is exactly 7, so the least k greater than 1 is 7.

Teacher's Note:
1. Conjugation by B squares A each time.
2. A prime order has no smaller divisor other than 1.