JEE Mathematics Matrices and Determinants MCQs Set 11

Answer: (c) contains two elements

Explanation:
1. The system is \( (A-\lambda I)X=0 \) and a non-trivial solution needs \( \begin{vmatrix}2-\lambda&-2&1\\2&-3-\lambda&2\\-1&2&-\lambda\end{vmatrix}=0 \).
2. This expands to \( -(\lambda-1)^2(\lambda+3)=0 \).
3. The distinct values are \( \lambda=1 \) and \( \lambda=-3 \): two elements. Answer: (c)

Teacher's Note:
1. Write the system as \( (A-\lambda I)X=0 \).
2. Count distinct roots (repeated roots count once).

Answer: (c) 1

Explanation:
1. Write \( 1+f(m+n)=1\cdot1+\alpha^m\alpha^n+\beta^m\beta^n \). The matrix is therefore \( VV^T \) with \( V=\begin{bmatrix}1&1&1\\1&\alpha&\beta\\1&\alpha^2&\beta^2\end{bmatrix} \) (rows indexed from 0).
2. \( \det=(\det V)^2 \) and \( \det V=(\alpha-1)(\beta-1)(\beta-\alpha) \).
3. So the determinant is \( (1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2 \) and \( K=1 \). Answer: (c)

Teacher's Note:
1. Express the matrix as a product of a Vandermonde matrix and its transpose.
2. Square the Vandermonde determinant.

Answer: (b) \( I \)

Explanation:
1. \( B'=(A^{-1}A')'=A(A^{-1})' \).
2. \( BB'=A^{-1}A'\cdot A(A^{-1})' \). Since \( A'A=AA' \), this is \( A^{-1}AA'(A^{-1})'=A'(A^{-1})'=(A^{-1}A)'=I'=I \). Answer: (b)

Teacher's Note:
1. Use \( (XY)'=Y'X' \) and \( A^{-1}A=I \).
2. The condition \( AA'=A'A \) lets you swap \( A \) and \( A' \).

Answer: (d) \( \begin{bmatrix}1\\-1\\-1\end{bmatrix} \)

Explanation:
1. Adding: \( A(u_1+u_2)=\begin{bmatrix}1\\1\\0\end{bmatrix} \).
2. Let \( u_1+u_2=(p,q,r)^T \): \( p=1 \), \( 2p+q=1\Rightarrow q=-1 \), \( 3p+2q+r=0\Rightarrow r=-1 \).
3. So \( u_1+u_2=\begin{bmatrix}1\\-1\\-1\end{bmatrix} \). Answer: (d)

Teacher's Note:
1. Add the two given equations.
2. Solve the triangular system by forward substitution.

Answer: (c) 0

Explanation:
1. Subtract the two relations: \( P^3-P^2Q=Q^3-Q^2P \), so \( P^2(P-Q)=Q^2(Q-P)=-Q^2(P-Q) \).
2. Hence \( (P^2+Q^2)(P-Q)=0 \). Since \( P\ne Q \), \( P-Q \) is non-zero, and \( P^2+Q^2 \) must be singular (otherwise multiplying by its inverse would give \( P=Q \)).
3. So \( \det(P^2+Q^2)=0 \). Answer: (c)

Teacher's Note:
1. Subtract the two given equations and factor.
2. If a matrix is non-zero when multiplied by another, the other cannot be invertible.

Answer: (a) 2

Explanation:
1. Non-zero solution needs \( \begin{vmatrix}4&k&2\\k&4&1\\2&2&1\end{vmatrix}=0 \).
2. The determinant is \( -k^2+6k-8 \), so \( k^2-6k+8=0 \).
3. \( (k-2)(k-4)=0\Rightarrow k=2,4 \): two values. Answer: (a)

Teacher's Note:
1. Set the coefficient determinant to zero.
2. Count the roots of the quadratic.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Since \( A^T=A \) and \( B^T=B \): \( [A(BA)]^T=(BA)^TA^T=(A^TB^T)A=(AB)A=A(BA) \), so Assertion is true.
2. \( (AB)^T=B^TA^T=BA \), so \( AB \) is symmetric exactly when \( AB=BA \). The Reason is true.
3. The Reason is about \( AB \) itself; it does not explain why \( A(BA) \) is symmetric. So both are true but R is not the correct explanation. Answer: (b)

Teacher's Note:
1. Take transposes using \( (XY)^T=Y^TX^T \).
2. Check whether the second statement actually gives the first.

Answer: (b) \( R-\{2,-3\} \)

Explanation:
1. Only the trivial solution means the coefficient determinant is non-zero.
2. \( \begin{vmatrix}1&-k&1\\k&3&-k\\3&1&-1\end{vmatrix}=2(k-2)(k+3) \).
3. This is non-zero when \( k\ne2 \) and \( k\ne-3 \), so the set is \( R-\{2,-3\} \). Answer: (b)

Teacher's Note:
1. A homogeneous system has only the trivial solution when the determinant is non-zero.
2. Factorise the determinant.

Answer: (d) no solution

Explanation:
1. Add the first two equations: \( 3x_1+5x_2+2x_3=6 \).
2. The third equation says the same left side equals 1.
3. \( 6\ne1 \), a contradiction, so there is no solution. Answer: (d)

Teacher's Note:
1. Look for a relation between the left-hand sides of the equations.
2. Compare the right-hand sides under the same relation.

Answer: (d) at least 7

Explanation:
1. The identity matrix with one extra 1 at any of the 6 off-diagonal positions is triangular with determinant 1, giving 6 non-singular matrices already.
2. Counting all \( \binom94=126 \) placements by their determinant gives exactly 36 non-singular matrices.
3. 36 is at least 7. Answer: (d)

Teacher's Note:
1. Count placements of four 1's among nine positions and test the determinant.
2. Only the lower bound is needed to pick the option.

Answer: (c) A is true, but R is false.

Explanation:
1. By the Cayley-Hamilton theorem \( A^2-\mathrm{tr}(A)A+|A|I=0 \). With \( A^2=I \): \( \mathrm{tr}(A)A=(1+|A|)I \).
2. Since \( A \) has non-zero off-diagonal entries it is not a scalar matrix, so both sides must vanish: \( \mathrm{tr}(A)=0 \) and \( |A|=-1 \).
3. So the Assertion is true and the Reason is false. Answer: (c)

Teacher's Note:
1. Use the characteristic equation of a \( 2\times2 \) matrix.
2. A matrix with non-zero off-diagonal entries cannot equal a multiple of \( I \).

Answer: (c) any odd integer

Explanation:
1. Transpose the second determinant and swap two columns: it becomes \( (-1)^{n+2} \) times the first determinant. (The first determinant equals \( 4b(a+c)\ne0 \).)
2. The sum is \( \left[1+(-1)^{n+2}\right]\cdot4b(a+c)=0 \), so \( 1+(-1)^{n+2}=0 \).
3. \( (-1)^{n}=-1 \) means \( n \) is odd. Answer: (c)

Teacher's Note:
1. Transpose the second determinant and compare it with the first.
2. The sum vanishes only when the sign factor is \( -1 \).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For a matrix of order \( n \), \( \mathrm{adj}(\mathrm{adj}A)=|A|^{n-2}A \). With \( n=2 \) this is \( A \), so the Assertion is true.
2. \( |\mathrm{adj}A|=|A|^{n-1}=|A| \) for \( n=2 \), so the Reason is true.
3. The Reason is a statement about determinants and does not prove the Assertion about the matrix. So both are true but R is not the correct explanation. Answer: (b)

Teacher's Note:
1. Recall the formulae for \( \mathrm{adj}(\mathrm{adj}A) \) and \( |\mathrm{adj}A| \) in terms of the order \( n \).
2. Substitute \( n=2 \).

Answer: (d) If \( \det(A)=\pm1 \), then \( A^{-1} \) exists and all its entries are integers

Explanation:
1. If \( \det A=\pm1\ne0 \), \( A^{-1} \) exists, so (a) is false.
2. \( A^{-1}=\frac1{\det A}\mathrm{adj}A=\pm\mathrm{adj}A \), and every entry of \( \mathrm{adj}A \) is a cofactor, hence an integer. So (b) is false and (d) is true.
3. (c) is false: for example \( A=\mathrm{diag}(2,1) \) has \( \det\ne\pm1 \) but \( A^{-1} \) has the integer entry 1. Answer: (d)

Teacher's Note:
1. Write the inverse as adjoint divided by determinant.
2. Cofactors of an integer matrix are integers.

Answer: (b) divisible by both \( x \) and \( y \)

Explanation:
1. Apply \( R_2\to R_2-R_1 \) and \( R_3\to R_3-R_1 \): \( D=\begin{vmatrix}1&1&1\\0&x&0\\0&0&y\end{vmatrix} \).
2. This is triangular, so \( D=xy \).
3. \( xy \) is divisible by both \( x \) and \( y \). Answer: (b)

Teacher's Note:
1. Subtract the first row from the other two.
2. The determinant of a triangular matrix is the product of the diagonal entries.

Answer: (c) \( \frac15 \)

Explanation:
1. \( A \) is upper triangular, so \( |A|=5\cdot\alpha\cdot5=25\alpha \).
2. \( |A^2|=|A|^2=625\alpha^2=25 \), so \( \alpha^2=\frac1{25} \).
3. \( |\alpha|=\frac15 \). Answer: (c)

Teacher's Note:
1. The determinant of a triangular matrix is the product of its diagonal entries.
2. Use \( |A^2|=|A|^2 \).

Answer: (a) \( AB=BA \)

Explanation:
1. Expand: \( (A-B)(A+B)=A^2+AB-BA-B^2 \).
2. Equating with \( A^2-B^2 \) gives \( AB-BA=0 \).
3. So \( AB=BA \) always, and the other options need not hold. Answer: (a)

Teacher's Note:
1. Multiply out the right side without assuming commutativity.
2. Cancel the common terms.

Answer: (c) there exist infinitely many \( B \)'s such that \( AB=BA \)

Explanation:
1. \( AB=\begin{bmatrix}a&2b\\3a&4b\end{bmatrix} \) and \( BA=\begin{bmatrix}a&2a\\3b&4b\end{bmatrix} \).
2. They are equal exactly when \( 2b=2a \) and \( 3a=3b \), i.e. \( a=b \).
3. Every natural number \( a=b \) works, so there are infinitely many such \( B \). Answer: (c)

Teacher's Note:
1. Compute both products.
2. Compare corresponding entries.

Answer: (d) \( -2 \)

Explanation:
1. The coefficient determinant is \( (\alpha-1)^2(\alpha+2) \), which is zero for \( \alpha=1 \) or \( \alpha=-2 \).
2. For \( \alpha=1 \) all three equations become \( x+y+z=0 \), so there are infinitely many solutions.
3. For \( \alpha=-2 \), adding the three equations gives \( 0=3(\alpha-1)=-9 \), a contradiction, so there is no solution. Answer: (d)

Teacher's Note:
1. Find where the determinant vanishes, then test each value.
2. Add the three equations for \( \alpha=-2 \).

Answer: (c) 0

Explanation:
1. If the terms are in GP, their logarithms are in AP: \( \log a_{n+k}=a+kd \).
2. Then \( \Delta=\begin{vmatrix}a&a+d&a+2d\\a+3d&a+4d&a+5d\\a+6d&a+7d&a+8d\end{vmatrix} \). Applying \( C_2\to C_2-C_1 \), \( C_3\to C_3-C_1 \) gives columns 2 and 3 proportional (\( d \) and \( 2d \)).
3. So \( \Delta=0 \). Answer: (c)

Teacher's Note:
1. Logarithms of a GP form an AP.
2. Column operations then give proportional columns.

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

Explanation:
1. \( A \) has non-zero entries, so (a) is false. It is not diagonal, so \( A\ne-I \) and (b) is false.
2. Expanding along row 1: \( |A|=-1\cdot\begin{vmatrix}0&-1\\-1&0\end{vmatrix}=-1\cdot(0-1)=1\ne0 \), so \( A^{-1} \) exists and (c) is false.
3. \( A^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I \). Answer: (d)

Teacher's Note:
1. Test each statement against the matrix.
2. Square the matrix by direct multiplication.

Answer: (d) 5

Explanation:
1. \( |A|=1(1+3)+1(2+3)+1(2-1)=10 \).
2. Computing the adjoint gives \( 10A^{-1}=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix} \), so the (2,3) entry is 5.
3. So \( \alpha=5 \). Answer: (d) (Check: \( A\cdot10B \) with \( \alpha=5 \) gives \( 10I \).)

Teacher's Note:
1. \( B=A^{-1} \) means \( 10A^{-1} \) is the given matrix.
2. Use \( A\cdot(10B)=10I \) to find \( \alpha \).

Answer: (c) are in HP

Explanation:
1. A non-zero solution needs \( \begin{vmatrix}1&2a&a\\1&3b&b\\1&4c&c\end{vmatrix}=0 \).
2. Expanding gives \( -ab+2ac-bc=0 \), i.e. \( 2ac=ab+bc \).
3. Dividing by \( abc \): \( \frac2b=\frac1c+\frac1a \), so \( a,b,c \) are in HP. Answer: (c)

Teacher's Note:
1. Set the determinant of the coefficient matrix to zero.
2. Divide the relation by \( abc \) to see reciprocals.

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

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

Teacher's Note:
1. Multiply the matrix by itself.
2. Match entries.

Answer: (a) 0

Explanation:
1. Multiply column 1 by \( \omega^n \): \( (\omega^n,\omega^{2n},\omega^{3n})=(\omega^n,\omega^{2n},1) \), which is column 2.
2. So column 2 is a multiple of column 1 and the determinant is 0. Answer: (a)

Teacher's Note:
1. Look for proportional columns.
2. Use \( \omega^{3n}=1 \).

Answer: (d) 0

Explanation:
1. Let the GP have first term \( A \) and ratio \( R \): \( \log l=\log A+(p-1)\log R \), and similarly for \( m,n \).
2. Apply \( C_1\to C_1-\log R\cdot C_2-(\log A-\log R)C_3 \): the first column becomes zero.
3. A zero column gives determinant 0. Answer: (d)

Teacher's Note:
1. Write each logarithm as a linear function of its index.
2. Use a column operation to produce a zero column.

Answer: (a) 0

Explanation:
1. Apply \( R_1\to R_1+R_3 \). Using \( 1+\omega+\omega^2=0 \) the new first row is \( (1-i,\ -1,\ \omega^2-1) \).
2. This is identical to row 2, so the determinant is 0. Answer: (a)

Teacher's Note:
1. Add the third row to the first.
2. Two identical rows make a determinant vanish.