JEE Mathematics Matrices MCQs Set 15

Answer: (d) 1 and 3

Explanation:
1. Expanding the given equality: \( -AB+BA=AB-BA \) gives \( AB=BA \).
2. Then \( (AB)^T=B^TA^T=-BA=-AB \), so \( (-1)^kAB=-AB \) requires \( k \) odd (for \( AB\ne0 \)): \( k=1,3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( A^2=I \)

Explanation:
1. Multiplying, \( A^2=I_3 \) (each row picks up a product of two entries equal to -1).
2. So \( A^{-1}=A \) exists, \( A \) is not zero and \( A\ne-I_3 \) because the off-diagonal entries are non-zero.

Teacher's Note:
1. Check A squared by direct row-by-column multiplication.
2. A matrix with A squared equal to I is its own inverse.

Answer: (b) Both are true and Statement 2 is a correct explanation of Statement 1

Explanation:
1. The coefficient determinant is 0. Adding equations 1 and 2 gives \( -y+z=k-1 \) while equation 3 minus equation 1 gives \( -y+z=2 \); so solutions exist only for \( k=3 \) (Statement 1 true).
2. The determinant in Statement 2 equals \( 3-k \), non-zero for \( k\ne3 \): the augmented system has rank 3 against coefficient rank 2, which is exactly why there is no solution.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( 3\omega(\omega-1) \)

Explanation:
1. Using \( 1+\omega+\omega^2=0 \): \( -1-\omega^2=\omega \) and \( \omega^4=\omega \), so the matrix is \( \begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix} \).
2. Expanding: \( (\omega^2-\omega^4)-(\omega-\omega^2)+(\omega^2-\omega)=3\omega^2-3\omega=3\omega(\omega-1) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \sqrt{10}+2 \)

Explanation:
1. A homogeneous system has non-trivial solutions iff the determinant is zero.
2. \( \begin{vmatrix}1&1&-2\\\lambda&-2&1\\1&3&-\lambda\end{vmatrix}=\lambda^2-4\lambda-6=0 \).
3. So \( \lambda=2\pm\sqrt{10} \); with \( \lambda>0 \): \( \lambda=2+\sqrt{10} \).

Teacher's Note:
1. Set the coefficient determinant equal to zero.
2. Reject the negative root.

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

Explanation:
1. A unique solution means \( \det A\ne0 \). Checking the 12 matrices (by enumeration) gives exactly 6 invertible ones, which is at least 4 and less than 7.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) the equations represent the whole of three-dimensional space

Explanation:
1. The second condition gives \( a=b=c \) and the first gives \( 3a=0 \), so \( a=b=c=0 \).
2. All three equations become \( 0=0 \): every point satisfies them.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. \( \begin{vmatrix}1&-k&-1\\k&-1&-1\\1&1&-1\end{vmatrix}=1-k^2 \).
2. Non-zero solutions need the determinant to be zero: \( k=\pm1 \).

Teacher's Note:
1. A homogeneous system has non-trivial solutions iff its determinant vanishes.
2. Expand along the first row.

Answer: (b) the equations represent the line \( x=y=z \)

Explanation:
1. The determinant is \( -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0 \), so there are non-trivial solutions, and \( (1,1,1) \) satisfies every equation since \( a+b+c=0 \).
2. The planes are not identical (not all coefficients equal), so they meet in the line \( x=y=z \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 3

Explanation:
1. \( AU=\begin{bmatrix}1&2&2\\0&3&3\\0&0&1\end{bmatrix} \), so \( |A||U|=1\cdot3\cdot1=3 \).
2. Since \( |A|=1 \), \( |U|=3 \).

Teacher's Note:
1. Combine the three column equations into AU = M.
2. The determinant of a triangular matrix is the product of its diagonal.

Answer: (d) \( \alpha=a^2+b^2,\ \beta=2ab \)

Explanation:
1. Multiply: \( A^2=\begin{bmatrix}a^2+b^2&2ab\\2ab&a^2+b^2\end{bmatrix} \).
2. So \( \alpha=a^2+b^2 \) and \( \beta=2ab \).

Teacher's Note:
1. Row-by-column multiplication.
2. The diagonal entries are sums of squares.

Answer: (a) 1

Explanation:
1. Infinite solutions need \( (k+1)(k+3)-8k=k^2-4k+3=0 \), so \( k=1 \) or \( 3 \).
2. For \( k=1 \): \( 2x+8y=4 \) and \( x+4y=2 \) are the same line: infinite solutions.
3. For \( k=3 \): \( x+2y=3 \) and \( 3x+6y=8 \) are inconsistent. So only one value.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. The characteristic polynomial is \( \lambda^3-6\lambda^2+11\lambda-6 \) (eigenvalues 1, 2, 3).
2. By Cayley-Hamilton, \( A^3-6A^2+11A-6I=0 \), so \( A^{-1}=\frac16(A^2-6A+11I) \).
3. Thus \( (c,d)=(-6,11) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 12

Explanation:
1. The independent entries are 3 diagonal (\( d \) ones) and 3 off-diagonal pairs (\( o \) ones counted twice): \( d+2o=5 \).
2. Cases \( (d,o)=(3,1) \): \( \binom31=3 \); \( (1,2) \): \( \binom32\binom31=9 \). Total 12.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) more than 2

Explanation:
1. Among the 12 matrices, 6 are invertible (consistent) and 6 are singular. Comparing ranks of \( A \) and the augmented matrix, exactly 4 singular matrices give an inconsistent system, which is more than 2.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) the equations represent identical planes

Explanation:
1. The condition \( a^2+b^2+c^2=ab+bc+ca \) means \( (a-b)^2+(b-c)^2+(c-a)^2=0 \), so \( a=b=c\ne0 \).
2. All three equations reduce to \( x+y+z=0 \): identical planes.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 0

Explanation:
1. A linear system has no solution, exactly one, or infinitely many solutions over the reals.
2. If two distinct solutions existed, every point of the line joining them would be a solution, so exactly two is impossible: the number is 0.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) the equations represent planes meeting only at a single point

Explanation:
1. The determinant is \( -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ne0 \).
2. So only the trivial solution exists: the planes meet only at the origin.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.