JEE Mathematics Matrices MCQs Set 12

Answer: (b) no solution

Explanation:
1. The coefficient determinant is \( 1(1-15)-4(3-10)-2(9-2)=-14+28-14=0 \), so there is no unique solution.
2. Combining \( (\frac{7}{11})Eq1 \) + \( (\frac{5}{11})Eq2 \) reproduces the left side 2x+3y+z of Eq3, but gives right side \( \frac{56}{11} \), not 5.
3. The system is inconsistent: rank A \( = \) 2 and rank of the augmented matrix \( = \) 3, so there is no solution.

Teacher's Note:
1. Compare the rank of the coefficient matrix with the rank of the augmented matrix.
2. When the coefficient rank is smaller, the system is inconsistent.

Answer: (a) -4, 1

Explanation:
1. By Cayley-Hamilton, \( A^2-(\operatorname{tr}A)A+|A|I=O \).
2. \( \operatorname{tr}A=4 \) and \( |A|=3-2=1 \), so \( A^2-4A+I=O \).
3. Hence \( a=-4 \), \( b=1 \).

Teacher's Note:
1. A 2x2 matrix satisfies its own characteristic equation.
2. Verify: \( A^2=\begin{bmatrix}11&8\\4&3\end{bmatrix} \) and \( A^2-4A+I=O \).

Answer: (a) A

Explanation:
1. Let \( J \) be the all-ones matrix. Then \( A=2J \) and \( J^2=3J \).
2. \( A^2=4J^2=12J=6A \) and \( A^3=6A^2=36A \).
3. So \( A^3-35A=36A-35A=A \).

Teacher's Note:
1. Write A as a multiple of the all-ones matrix to get repeated powers easily.
2. Each multiplication by A multiplies the matrix by 6.

Answer: (d) \( \begin{bmatrix} x^{100} & 0 & 0 \\ 0 & x^{100} & 0 \\ 0 & 0 & x^{100} \end{bmatrix} \)

Explanation:
1. \( A^2=x^2I \) because the matrix is x times a permutation matrix that swaps rows 1 and 3.
2. \( A^{100}=(A^2)^{50}=x^{100}I \).

Teacher's Note:
1. Find the first power that is a scalar matrix.
2. The even powers of A are scalar multiples of I.

Answer: (c) nA

Explanation:
1. \( A^2=\begin{bmatrix}0&0\\1&1\end{bmatrix}=A \), so A is idempotent.
2. Hence every power equals A and the sum of n terms is \( nA \).

Teacher's Note:
1. Idempotent matrices satisfy A squared equals A, so all higher powers collapse to A.
2. Always test A squared first.

Answer: (a) 256

Explanation:
1. A 2x2 matrix has 4 entries.
2. Each entry can be chosen in 4 ways independently, so the number is \( 4^4=256 \).

Teacher's Note:
1. Repetition allowed means every position is independent.
2. Without repetition the count would be 4! \( = \) 24.

Answer: (a) \( \begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix} \)

Explanation:
1. For symmetric \( A=\begin{bmatrix}p&q\\q&r\end{bmatrix} \) the quadratic form is \( px^2+2qxy+ry^2 \).
2. Comparing, \( p=3 \), \( 2q=10\Rightarrow q=5 \), \( r=5 \).

Teacher's Note:
1. The cross term xy gets coefficient 2q in a symmetric matrix, so halve it.
2. The diagonal entries are the coefficients of x squared and y squared.

Answer: (b) -1

Explanation:
1. \( (I-A)(I+A+\dots+A^{m-1})=I-A^m=I \).
2. So \( I+A+\dots+A^{m-1}=(I-A)^{-1} \).
3. Therefore \( (I-A)^n=(I-A)^{-1} \), giving \( n=-1 \).

Teacher's Note:
1. The geometric-series factorisation works for matrices as well as for numbers.
2. Nilpotency makes the series terminate.

Answer: (b) 8

Explanation:
1. A skew-symmetric matrix has three zeros on the diagonal, so all three zeros are used there and every off-diagonal entry is non-zero.
2. The upper entries \( a_{12},a_{13},a_{23} \) can each be 1 or -1, and the lower entries are then forced.
3. This gives \( 2^3=8 \) matrices, each with exactly three 1's and three -1's.

Teacher's Note:
1. Count the zeros first: the diagonal already supplies three.
2. Each independent upper entry gives two choices.

Answer: (d) \( \frac12(n^2-n) \)

Explanation:
1. The diagonal entries are 0.
2. The entries above the diagonal are free: \( \binom n2=\frac{n(n-1)}{2} \).
3. Entries below the diagonal are their negatives, so they are not new independent values.

Teacher's Note:
1. Different entries means independent values.
2. Only the entries above the diagonal can be chosen freely.

Answer: (d) \( A^{-1}B^nA \)

Explanation:
1. \( (A^{-1}BA)^2=A^{-1}B(AA^{-1})BA=A^{-1}B^2A \).
2. By induction, \( (A^{-1}BA)^n=A^{-1}B^nA \).

Teacher's Note:
1. Adjacent A and A inverse cancel in the middle of the product.
2. This is called a similarity transformation.

Answer: (c) \( \frac1{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix} \)

Explanation:
1. \( (A-aI)(A-cI)=\begin{bmatrix}0&b\\0&c-a\end{bmatrix}\begin{bmatrix}a-c&b\\0&0\end{bmatrix}=O \).
2. So the expression equals \( A^{-1}=\frac1{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix} \).

Teacher's Note:
1. The product of the factors (A minus each eigenvalue) is zero by Cayley-Hamilton.
2. The inverse of a 2x2 matrix swaps the diagonal and negates the off-diagonal.

Answer: (d) only I and IV

Explanation:
1. Composing rotations about the z-axis adds the angles, so I is true and II is false.
2. The inverse of the rotation by \( \alpha \) is the rotation by \( -\alpha \), so IV is true.
3. \( -A_\alpha \) has diagonal entry -1 in the last place, so III is false.

Teacher's Note:
1. Multiply the two 2x2 rotation blocks and use the addition formulas for sine and cosine.
2. The inverse of a rotation undoes it.

Answer: (c) \( \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \)

Explanation:
1. \( (BA)^{-1}=C \) means \( A^{-1}B^{-1}=C \).
2. Multiplying on the right by \( B \) gives \( A^{-1}=CB \).
3. \( CB=\begin{bmatrix}-3&-5&-5\\0&9&2\\2&14&6\end{bmatrix} \).

Teacher's Note:
1. Use the reversal law for the inverse of a product, then multiply both sides by B.
2. Multiply row by column carefully.

Answer: (a) 1

Explanation:
1. The coefficient determinant is \( (k+1)(k+3)-8k=k^2-4k+3=(k-1)(k-3) \).
2. For \( k=1 \) the equations are \( 2x+8y=4 \) and \( x+4y=2 \), which are the same line (infinitely many solutions).
3. For \( k=3 \) they are \( x+2y=3 \) and \( 3x+6y=8 \), which are parallel and different, so there is no solution.
4. Only one value of k works.

Teacher's Note:
1. Set the coefficient determinant to zero, then test each root for consistency.
2. A zero determinant gives either no solution or infinitely many.

Answer: (c) 3, 10

Explanation:
1. Subtract the second equation from the third: \( (\lambda-3)z=k-10 \).
2. The system is inconsistent when \( \lambda=3 \) and \( k\ne10 \).

Teacher's Note:
1. Inconsistency needs a zero coefficient on the left and a non-zero constant on the right.
2. At k \( = \) 10 the system would instead have infinitely many solutions.

Answer: (c) \( \{m: m>30\text{ or }m<-15/2\} \)

Explanation:
1. Solving gives \( x=\frac{25m}{2m+15} \) and \( y=\frac{2(m-30)}{2m+15} \).
2. If \( 2m+15>0 \), both are positive only when \( m>30 \).
3. If \( 2m+15<0 \), we need \( m<0 \) and \( m<30 \), so \( m<-\frac{15}{2} \).

Teacher's Note:
1. Solve by Cramer's rule, then analyse the sign of numerator and denominator.
2. Consider both signs of the common denominator.

Answer: (d) -2

Explanation:
1. Subtract (k+1) times the third equation from the second: \( y=2 \), hence \( x=-1 \).
2. Substitute in the first with \( u=k+2 \): \( -(u-1)^3+2u^3-(u+1)^3=-6u \).
3. The first equation then requires \( -6u=0 \), so \( u=0 \) and \( k=-2 \).

Teacher's Note:
1. Solve the two simplest equations first and substitute into the third.
2. The cubic terms cancel and only a linear condition in u remains.

Answer: (a) \( [-\sqrt2,\sqrt2] \)

Explanation:
1. A non-trivial solution needs the coefficient determinant to be zero.
2. Expanding gives \( -\lambda+2\sin\alpha\cos\alpha+\cos^2\alpha-\sin^2\alpha=0 \), i.e. \( \lambda=\sin2\alpha+\cos2\alpha \).
3. As \( \alpha \) varies, \( \sin2\alpha+\cos2\alpha \) takes all values in \( [-\sqrt2,\sqrt2] \).

Teacher's Note:
1. Write the determinant condition and isolate lambda.
2. Use the amplitude of a sin plus b cos, which is the square root of a squared plus b squared.

Answer: (b) \( a=b=c\ (\ne0) \)

Explanation:
1. The coefficient determinant factors as \( -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \).
2. It is zero when \( a+b+c=0 \) or \( a=b=c \).
3. Among the options, only \( a=b=c\ (\ne0) \) qualifies.

Teacher's Note:
1. Set the determinant to zero and factor it.
2. The factor a squared plus b squared plus c squared minus ab minus bc minus ca vanishes only if \( a=b=c \) for real numbers.

Answer: (c) G.P.

Explanation:
1. The coefficient determinant equals \( -(ac-b^2)(a\alpha^2+2b\alpha+c) \).
2. It must be zero and the second factor is non-zero, so \( ac=b^2 \).
3. That means a, b, c are in G.P.

Teacher's Note:
1. Expand the determinant and factor out the given non-zero expression.
2. b squared equal to ac is the definition of a geometric progression.

Answer: (d) 0

Explanation:
1. Row 1 of A times Column 1 of A: \( 1(1)+(-3)(-1)+(-4)(1)=0 \); the other entries are zero in the same way.
2. So \( A^2=O=0\cdot I \) and \( \lambda=0 \).

Teacher's Note:
1. A is a nilpotent matrix of index 2: every row of A is a multiple of (1, -3, -4) and that row dotted with any column is zero.
2. Compute one row-column product and then check the pattern.

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

Explanation:
1. Collect the coefficients of \( \lambda^2 \), \( \lambda \) and the constant terms entry by entry.
2. \( B=\begin{bmatrix}-2&1\\3&0\end{bmatrix} \) and \( C=\begin{bmatrix}1&-2\\1&1\end{bmatrix} \).
3. \( B+C=\begin{bmatrix}-1&-1\\4&1\end{bmatrix} \).

Teacher's Note:
1. Treat each entry as a polynomial in lambda and read off the coefficients.
2. Check: setting lambda \( = \) 1 on both sides gives matching entries.

Answer: (b) \( \begin{bmatrix} 2 & 9/2 & 8 \\ 9/2 & 8 & 25/2 \end{bmatrix} \)

Explanation:
1. \( a_{11}=\frac{4}{2}=2 \), \( a_{12}=\frac92 \), \( a_{13}=\frac{16}2=8 \).
2. \( a_{21}=\frac92 \), \( a_{22}=8 \), \( a_{23}=\frac{25}2 \).
3. So \( A=\begin{bmatrix}2&\frac{9}{2}&8\\9/2&8&\frac{25}{2}\end{bmatrix} \).

Teacher's Note:
1. Fill row by row; the matrix has 2 rows and 3 columns.
2. Note that a_12 and a_21 are equal because the rule is symmetric in i and j.

Answer: (a) \( 3\times2 \)

Explanation:
1. The row matrix has order 1x3, so B must have 3 rows.
2. The product has order 1x2, so B must have 2 columns.
3. Hence B is 3x2.

Teacher's Note:
1. Columns of the first matrix must equal rows of the second.
2. The number of columns of the product equals the number of columns of B.

Answer: (a) \( 1\times13 \) or \( 13\times1 \)

Explanation:
1. The number of elements equals rows times columns, so \( mn=13 \).
2. 13 is prime, so the only factorisations are \( 1\times13 \) and \( 13\times1 \).

Teacher's Note:
1. List the factor pairs of the number of elements.
2. A prime number of elements gives only a row matrix or a column matrix.