JEE Mathematics Matrices and Determinants MCQs Set 09

Answer: (a) 2

Explanation:
1. The characteristic polynomial of A is \( x^3-(\operatorname{tr}A)x^2+(\text{sum of principal minors})x-|A| \).
2. Here \( \operatorname{tr}A=6 \), the sum of principal 2 x 2 minors is \( 2+(-1)+6=7 \) and \( |A|=-2 \).
3. So the polynomial is \( x^3-6x^2+7x+2 \), and k = 2 by Cayley-Hamilton.

Teacher's Note:
1. A matrix satisfies its own characteristic equation.
2. Compute the trace, minors and determinant.

Answer: (b) a + d = 0 and k = |A|

Explanation:
1. By Cayley-Hamilton \( A^2=(a+d)A-|A|I \).
2. So \( (a+d)A=(|A|-k)I \).
3. Since \( bc\ne0 \), A is not a scalar matrix, so \( a+d=0 \) and then \( k=|A| \).

Teacher's Note:
1. Use A^2 - (trace)A + det I = O.
2. A matrix with a nonzero off-diagonal entry is not a scalar multiple of I.

Answer: (a) p = 4, mu = 2 and the solutions are \( \left(5-4\lambda,\ \frac{\lambda-1}{2},\ \lambda\right) \)

Explanation:
1. The determinant of the coefficients is \( (p-2)(\mu-3)+2=\mu p-2\mu-3p+8 \) and must vanish for infinitely many solutions.
2. With p = 4 and mu = 2 the second and third equations become identical: \( x+4y+2z=3 \).
3. Subtracting from the first: \( -2y+z=1 \), so \( y=\frac{\lambda-1}{2} \) for \( z=\lambda \), and \( x=5-4\lambda \). Check: \( 5-4\lambda+\lambda-1+3\lambda=4 \).

Teacher's Note:
1. Infinitely many solutions need the determinant to be 0 and consistent equations.
2. Substitute the given triplets back into the equations.

Answer: (b) \( \{-1\} \)

Explanation:
1. The determinant is \( (\lambda+1)\cos2\alpha \) (expand along the first column).
2. For infinitely many solutions the determinant must be 0 for every \( \alpha \), but \( \cos2\alpha \) is not identically 0.
3. So \( \lambda+1=0 \), i.e. \( \lambda=-1 \).

Teacher's Note:
1. Compute the determinant of the coefficient matrix.
2. The factor cos 2 alpha cannot vanish for all alpha.

Answer: (d) 33

Explanation:
1. A non-trivial solution exists when the determinant is 0.
2. The determinant is \( 3(-4k-9)-k(-4-6)-2(3-2k)=2k-33 \).
3. So \( 2k=33 \).

Teacher's Note:
1. Set the coefficient determinant equal to 0.
2. Expand along the first row.

Answer: (b) \( A^{N+1}=B^N(B+(N+1)C) \)

Explanation:
1. Since B and C commute, the binomial theorem applies: \( (B+C)^{N+1}=B^{N+1}+(N+1)B^NC+\dots \).
2. All terms with \( C^2 \) or higher vanish.
3. So \( A^{N+1}=B^{N+1}+(N+1)B^NC=B^N(B+(N+1)C) \).

Teacher's Note:
1. The binomial theorem needs commuting matrices.
2. C^2 = O kills all but two terms.

Answer: (c) a symmetric matrix

Explanation:
1. Using \( A'=-A \): \( (A^n)'=(A')^n=(-A)^n=A^n \) since n is even.
2. So \( A^n \) is symmetric.

Teacher's Note:
1. The transpose of a power is the power of the transpose.
2. (-1)^n = 1 for even n.

Answer: (c) 0

Explanation:
1. Transposing, \( A=(A^T)^2-I=(A^2-I)^2-I \), so A commutes with \( A^T \) and A is a normal real matrix.
2. An eigenvalue satisfies \( \bar\lambda=\lambda^2-1 \). For a complex eigenvalue \( a+bi \) (\( b\ne0 \)) this forces \( a=-\frac12 \) and \( b^2=-\frac14 \), which is impossible.
3. So all eigenvalues are real, and a normal matrix with real eigenvalues is symmetric. Hence no non-symmetric matrix exists: the number is 0.

Teacher's Note:
1. Take the transpose of the given relation twice.
2. A real normal matrix with real eigenvalues is symmetric.

Answer: (c) \( nA-(n-1)I \)

Explanation:
1. The relation says \( (A-I)^2=O \). Write \( A=I+N \) with \( N^2=O \).
2. Then \( A^n=(I+N)^n=I+nN=I+n(A-I)=nA-(n-1)I \).

Teacher's Note:
1. Put N = A - I and note N^2 = O.
2. Binomial expansion stops after two terms.

Answer: (d) \( \begin{pmatrix}1&2005\\0&1\end{pmatrix} \)

Explanation:
1. P is orthogonal: \( P^TP=PP^T=I \).
2. Then \( Q^{2005}=PA^{2005}P^T \), so \( x=P^TPA^{2005}P^TP=A^{2005} \).
3. Since \( A^n=\begin{pmatrix}1&n\\0&1\end{pmatrix} \), \( x=\begin{pmatrix}1&2005\\0&1\end{pmatrix} \).

Teacher's Note:
1. Check that P^T P = I.
2. Powers of Q telescope: Q^n = P A^n P^T.

Answer: (d) independent of \( \phi \)

Explanation:
1. Expanding along the third column: \( \Delta=\cos\theta\,(\sin\theta\cos\theta\cos^2\phi+\sin\theta\cos\theta\sin^2\phi)+\sin\theta\,(\sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi) \).
2. This simplifies to \( \sin\theta\cos^2\theta+\sin^3\theta=\sin\theta \).
3. So \( \Delta=\sin\theta \), which does not involve \( \phi \).

Teacher's Note:
1. Expand along the third column.
2. Use cos^2 phi + sin^2 phi = 1.

Answer: (d) \( (a-b)^2(b-c)^2(c-a)^2 \)

Explanation:
1. The matrix equals \( VV^T \) with \( V=\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix} \) (each entry is \( 1+xy+x^2y^2 \)).
2. So \( \Delta=|V|^2 \).
3. The Vandermonde determinant is \( |V|=(b-a)(c-a)(c-b) \), hence \( \Delta=(a-b)^2(b-c)^2(c-a)^2 \).

Teacher's Note:
1. Write each entry as a sum of products of two vectors.
2. The determinant of a product is the product of determinants.

Answer: (a) \( 1+a^2+b^2+c^2 \)

Explanation:
1. Write the matrix as \( I+vv^T \) with \( v=(a,b,c)^T \).
2. Then \( \det(I+vv^T)=1+v^Tv=1+a^2+b^2+c^2 \).

Teacher's Note:
1. Notice the pattern: identity plus an outer product.
2. det(I + v v^T) = 1 + v^T v.

Answer: (b) \( 0 \)

Explanation:
1. Apply \( R_3\to R_3-R_2 \) and \( R_2\to R_2-R_1 \): the new rows have entries \( a^5-a^3,\ a^6-a^4,\ a^7-a^5 \) and \( a^7-a^5,\ a^8-a^6,\ a^9-a^7 \) with no x.
2. The second is \( a^2 \) times the first, so the two rows are proportional.
3. A determinant with two proportional rows is 0.

Teacher's Note:
1. Subtract consecutive rows to remove x.
2. The resulting rows differ by the factor a^2.

Answer: (b) A is non-singular

Explanation:
1. If A is non-singular, multiplying by \( A^{-1} \) gives \( B=C \).
2. Singular, symmetric or merely square matrices do not guarantee a left inverse.

Teacher's Note:
1. Cancellation needs an inverse.
2. A square matrix may still be singular.

Answer: (a) 16

Explanation:
1. Direct multiplication gives \( A^2=\begin{pmatrix}4&-14+2\sqrt3&-14-2\sqrt3\\0&-4&0\\0&0&-4\end{pmatrix} \).
2. Squaring again, the diagonal entries are \( 16,16,16 \) and the off-diagonal entries in the first row are \( 4(-14\pm2\sqrt3)-4(-14\pm2\sqrt3)=0 \).
3. So \( A^4=16I \) and \( \lambda=16 \).

Teacher's Note:
1. A^2 turns out to be upper triangular with diagonal 4, -4, -4.
2. Square twice.