JEE Mathematics Matrices MCQs Set 16

Answer: (b) -7

Explanation:
1. \( A^2=\begin{bmatrix}1&0\\-1\cdot1+7\cdot(-1)&49\end{bmatrix}=\begin{bmatrix}1&0\\-8&49\end{bmatrix} \).
2. \( 8A=\begin{bmatrix}8&0\\-8&56\end{bmatrix} \).
3. \( A^2-8A=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}=-7I \), so \( k=-7 \).

Teacher's Note:
1. Compute A squared and subtract 8A; the result should be a scalar multiple of I.

Answer: (a) \( \tfrac25 \)

Explanation:
1. A matrix is singular when its determinant is zero.
2. \( \lambda(2+3)-7(8-6)-2(-4-2)=5\lambda-14+12=5\lambda-2 \).
3. Setting this to zero gives \( \lambda=\tfrac25 \).

Teacher's Note:
1. Singular means determinant zero.
2. Expand along the first row.

Answer: (b) \( \begin{bmatrix}a^2+bc&ab+bd\\ac+dc&bc+d^2\end{bmatrix} \)

Explanation:
1. \( A^2=A\cdot A \) uses row-by-column multiplication.
2. Row 1 by column 1: \( a\cdot a+b\cdot c=a^2+bc \); row 1 by column 2: \( ab+bd \).
3. Row 2 by column 1: \( ca+dc \); row 2 by column 2: \( cb+d^2 \).

Teacher's Note:
1. Matrix squaring is not entrywise squaring.

Answer: (d) none of these

Explanation:
1. \( A^2=\begin{bmatrix}\alpha^2&0\\\alpha+1&1\end{bmatrix} \).
2. Equating with B: \( \alpha^2=1 \) and \( \alpha+1=5 \).
3. The first gives \( \alpha=\pm1 \) and the second gives \( \alpha=4 \); no value satisfies both, so none of the options is correct.

Teacher's Note:
1. Every entry must match, so check all the equations together.

Answer: (b) \( \lambda\ne8 \)

Explanation:
1. \( |A|=0(-\lambda+6)+4(-2+3)+1(4-\lambda)=4+4-\lambda=8-\lambda \).
2. The inverse exists exactly when \( |A|\ne0 \).
3. So \( \lambda\ne8 \).

Teacher's Note:
1. A matrix is invertible exactly when its determinant is nonzero.

Answer: (c) 1

Explanation:
1. \( |A|=1(0-0)+1(0-0)+1(3-2)=1 \).
2. For a 3 by 3 matrix \( |\operatorname{adj}A|=|A|^{2} \).
3. So \( |\operatorname{adj}A|=1 \).

Teacher's Note:
1. Use the result that the determinant of adj A equals the determinant of A to the power n - 1.

Answer: (b) I

Explanation:
1. Row 1 of \( A^2 \): \( (16-3-12,\ -4+0+4,\ -16+4+12)=(1,0,0) \).
2. Row 2: \( (12-12,\ -3+0+4,\ -12+12)=(0,1,0) \). Row 3: \( (12-3-9,\ -3+0+3,\ -12+4+9)=(0,0,1) \).
3. So \( A^2=I \), an involutory matrix.

Teacher's Note:
1. Multiply row by column carefully; an involutory matrix squares to the identity.

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

Explanation:
1. Using \( \omega^3=1 \) and \( 1+\omega+\omega^2=0 \): \( AC=\begin{bmatrix}1+\omega^2+\omega^4\\\omega+\omega^3+\omega^2\\\omega^2+\omega+\omega^3\end{bmatrix}=\begin{bmatrix}1+\omega^2+\omega\\\omega+1+\omega^2\\\omega^2+\omega+1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} \).
2. \( BC=\begin{bmatrix}\omega+\omega^3+\omega^2\\\omega^2+\omega+\omega^3\\\omega+\omega^3+\omega^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} \) since each entry equals \( 1+\omega+\omega^2 \).
3. Hence \( (A+B)C=AC+BC=\begin{bmatrix}0\\0\\0\end{bmatrix} \).

Teacher's Note:
1. Reduce every power of omega with omega cubed equal to 1 and use 1 + omega + omega squared = 0.

Answer: (a) \( \lambda\ne\tfrac{18}{11} \)

Explanation:
1. The rank is 3 exactly when the determinant is nonzero.
2. \( 1(2+12)-2(\lambda-8)+3(-3\lambda-4)=14-2\lambda+16-9\lambda-12=18-11\lambda \).
3. So the rank is 3 when \( \lambda\ne\tfrac{18}{11} \).

Teacher's Note:
1. For a square matrix, full rank means a nonzero determinant.

Answer: (b) 3

Explanation:
1. The matrix has 3 rows, so its rank cannot exceed 3.
2. The last three columns form the identity matrix \( I_3 \), whose determinant is 1.
3. So there is a nonzero 3 by 3 minor and the rank is 3.

Teacher's Note:
1. The rank cannot exceed the smaller of the number of rows and columns.

Answer: (d) none of these

Explanation:
1. The row times the matrix gives \( [1+2x+15,\ 3+5x+3,\ 2+x+2]=[2x+16,\ 5x+6,\ x+4] \).
2. Multiplying by the column: \( (2x+16)+2(5x+6)+x(x+4)=x^2+16x+28 \).
3. \( x^2+16x+28=(x+2)(x+14)=0 \) gives \( x=-2 \) or \( x=-14 \). Since x can be either, no single option lists the full answer.

Teacher's Note:
1. Multiply from left to right and then solve the resulting quadratic.
2. Check whether both roots are allowed by the question.

Answer: (a) -5

Explanation:
1. The product is \( \begin{bmatrix}2(x+y)-y\\4x-(x-y)\end{bmatrix}=\begin{bmatrix}2x+y\\3x+y\end{bmatrix} \).
2. Equating: \( 2x+y=3 \) and \( 3x+y=2 \), so \( x=-1,\ y=5 \).
3. \( xy=-5 \).

Teacher's Note:
1. Multiply the matrices and equate corresponding entries.