JEE Mathematics Matrices MCQs Set 06

Answer: (c) invertible for all \( t\in\mathbb R \)

Explanation:
1. Take \( e^t \) out of column 1 and \( e^{-t} \) out of columns 2 and 3: \( |A|=e^{t}e^{-t}e^{-t}\begin{vmatrix}1&\cos t&\sin t\\1&-\cos t-\sin t&-\sin t+\cos t\\1&2\sin t&-2\cos t\end{vmatrix} \).
2. Reducing with \( R_2\to R_2-R_1 \) and \( R_3\to R_3-R_1 \) and expanding, the remaining determinant equals 5 for every t.
3. So \( |A|=5e^{-t}\ne0 \) for all real t.

Teacher's Note:
1. Factor the exponentials out of the columns, then simplify with row operations.
2. An exponential is never zero, so the sign of the remaining determinant decides everything.

Answer: (b) Q, R and S only

Explanation:
1. Each term \( P_kBP_k^T \) is a symmetric matrix (B is symmetric) with the same trace as B, namely 3.
2. The sum has trace \( 6\times3=18 \), so (Q) is true, and X is symmetric, so (S) is true.
3. Direct computation gives \( X=\begin{bmatrix}6&12&12\\12&6&12\\12&12&6\end{bmatrix} \). Then \( X\begin{bmatrix}1\\1\\1\end{bmatrix}=30\begin{bmatrix}1\\1\\1\end{bmatrix} \), so \( \alpha=30 \) and (R) is true.
4. \( X-30I \) has all row sums equal to 0, so it is singular and (P) is false.

Teacher's Note:
1. Each \( P_k \) is a permutation matrix, so \( P_kBP_k^T \) just rearranges the entries of B.
2. A matrix whose rows each sum to zero maps the vector of ones to the zero vector, so it is singular.

Answer: (c) R and S only

Explanation:
1. R is similar to Q, so \( \det R=\det Q=48-4x^2 \). The determinant in (R) is \( 40-4x^2 \), so (R) is true: the difference is 8.
2. For \( x=1 \), \( \det R=44\ne0 \), so \( RX=O \) has only the zero solution and no unit vector exists: (P) is false.
3. The (1,2) entry of \( PQ-QP \) is 2 for every x, so \( PQ\ne QP \) for all x: (Q) is false.
4. For \( x=0 \), \( R=\begin{bmatrix}2&1&\frac23\\0&4&\frac43\\0&0&6\end{bmatrix} \). The second row of \( R\mathbf v=6\mathbf v \) gives \( a=\frac{2b}3 \) and the first row gives \( 2+a+\frac{2b}3=6 \), so \( b=3,\ a=2 \) and \( a+b=5 \): (S) is true.

Teacher's Note:
1. Similar matrices have equal determinants, which handles statement R at once.
2. For statement S, solve the eigenvector equations row by row from the bottom of the triangular matrix.

Answer: (b) \( -\frac{29}{16} \)

Explanation:
1. Multiplying by M gives \( M^2=\alpha M+\beta I \). By Cayley-Hamilton \( M^2=(\operatorname{tr}M)M-|M|I \), so \( \alpha=\operatorname{tr}M \) and \( \beta=-|M| \).
2. \( \alpha=\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta \). With \( u=\sin^2\theta\cos^2\theta\in\left[0,\frac14\right] \), \( \alpha=1-2u \) has minimum \( \frac12 \).
3. \( |M|=u^2+(1+\sin^2\theta)(1+\cos^2\theta)=u^2+u+2 \), so \( \beta=-(u^2+u+2) \), which is smallest at \( u=\frac14 \): \( \beta^*=-\left(\frac1{16}+\frac14+2\right)=-\frac{37}{16} \).
4. \( \alpha^*+\beta^*=\frac12-\frac{37}{16}=-\frac{29}{16} \).

Teacher's Note:
1. Compare \( M^2=\alpha M+\beta I \) with the Cayley-Hamilton relation to read off alpha and beta.
2. Use the substitution \( u=\sin^2\theta\cos^2\theta \) to turn both into simple functions on [0, 1/4].

Answer: (b) P, R and S only

Explanation:
1. The (1,1) entry of \( \operatorname{adj}M \) is the cofactor \( 2-3b=-1 \), so \( b=1 \). The (1,3) entry is the cofactor \( 3-2a=-1 \), so \( a=2 \). Hence (P) \( a+b=3 \) is true.
2. \( |M|=0(2-3)-1(1-9)+2(1-6)=8-10=-2 \).
3. (Q) \( \det(\operatorname{adj}M^2)=|M^2|^2=|M|^4=16\ne81 \), so it is false.
4. (R) \( (\operatorname{adj}M)^{-1}=\frac{M}{|M|} \) and \( \operatorname{adj}M^{-1}=\frac{M}{|M|} \), so the sum is \( \frac{2M}{-2}=-M \): true.
5. (S) \( \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=M^{-1}\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}1\\-1\\1\end{bmatrix} \), so \( \alpha-\beta+\gamma=3 \): true.

Teacher's Note:
1. Use \( A^{-1}=\frac{\operatorname{adj}A}{|A|} \) to turn adjoint information into the determinant and the inverse.
2. Match single entries of the adjoint with cofactors to find the unknowns a and b.

Answer: (c) \( \begin{bmatrix}4&-2\\-1&-4\end{bmatrix} \)

Explanation:
1. With \( S=A+B \), the symmetric part is \( A=\frac12(S+S^T)=\begin{bmatrix}2&4\\4&-1\end{bmatrix} \) and the skew-symmetric part is \( B=\frac12(S-S^T)=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \).
2. \( AB=\begin{bmatrix}2\cdot0+4\cdot1&2(-1)+4\cdot0\\4\cdot0+(-1)\cdot1&4(-1)+0\end{bmatrix}=\begin{bmatrix}4&-2\\-1&-4\end{bmatrix} \).

Teacher's Note:
1. The decomposition of a matrix into symmetric and skew-symmetric parts is unique.
2. Use \( \frac12(S+S^T) \) and \( \frac12(S-S^T) \) to find the two parts.

Answer: (c) \( A^3 \)

Explanation:
1. Since \( \alpha^3=1 \), \( \alpha^4=\alpha \). Squaring A and using \( 1+\alpha+\alpha^2=0 \) gives \( A^2=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \).
2. Then \( A^4=(A^2)^2=I \).
3. \( A^{31}=(A^4)^7A^3=A^3 \).

Teacher's Note:
1. Look for the smallest power of A that equals the identity, then reduce the exponent.
2. Here \( A^4=I \), and 31 leaves remainder 3 when divided by 4.

Answer: (b) \( A-6I \)

Explanation:
1. \( \operatorname{tr}A=6 \) and \( |A|=8-18=-10 \). By Cayley-Hamilton, \( A^2-6A-10I=O \).
2. Hence \( A^2-6A=10I \). Multiplying by \( A^{-1} \): \( A-6I=10A^{-1} \).

Teacher's Note:
1. For a \( 2\times2 \) matrix the characteristic equation is \( \lambda^2-(\operatorname{tr}A)\lambda+|A|=0 \).
2. Multiply by \( A^{-1} \) to obtain the inverse.

Answer: (c) 8

Explanation:
1. \( |A|=1(9+4)-1(3-4)+2(-1-3)=13+1-8=6 \).
2. \( |B|=|A|^2 \) and \( |\operatorname{adj}B|=|B|^2=|A|^4=1296 \).
3. \( |C|=3^3|A|=162 \).
4. \( \frac{1296}{162}=8 \).

Teacher's Note:
1. For a \( 3\times3 \) matrix, \( |\operatorname{adj}A|=|A|^2 \) and \( |kA|=k^3|A| \).
2. Express everything in terms of \( |A| \) before computing.

Answer: (b) infinitely many solutions \( (x,y,z) \) satisfying \( x=2z \)

Explanation:
1. The determinant of the coefficients is \( 7(-24+4)-6(-18-2)-2(-6-4)=-140+120+20=0 \), so the homogeneous system has infinitely many solutions.
2. Adding the first two equations: \( 10x+10y=0 \), so \( y=-x \). Putting this in the third: \( x+2x-6z=0 \), so \( x=2z \).
3. The solutions are \( (2z,-2z,z) \), which satisfy \( x=2z \) (and \( y=-2z \), not \( y=2z \)).

Teacher's Note:
1. A homogeneous system always has the trivial solution; it has more exactly when the determinant is zero.
2. Use two equations to eliminate a variable, then check the third.

Answer: (d) (P) is false and (Q) is true

Explanation:
1. The invertible 0-1 matrices are \( \begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\1&1\end{bmatrix},\begin{bmatrix}1&1\\0&1\end{bmatrix} \) (determinant 1, trace 2) and \( \begin{bmatrix}0&1\\1&0\end{bmatrix},\begin{bmatrix}0&1\\1&1\end{bmatrix},\begin{bmatrix}1&1\\1&0\end{bmatrix} \) (determinant -1).
2. (P) is false: \( \begin{bmatrix}1&0\\1&1\end{bmatrix}\ne I_2 \) has determinant 1.
3. (Q) is true: every such matrix with determinant 1 has trace 2.

Teacher's Note:
1. Enumerate all sixteen 0-1 matrices and keep the six with non-zero determinant.
2. One counter-example is enough to disprove a statement.

Answer: (c) contains exactly two elements

Explanation:
1. \( |P|=1(-3+36)-2(2+4)+1(-18-3)=33-12-21=0 \).
2. The first two rows are not proportional, so the rank is 2 and the solutions of \( PX=O \) form a line through the origin.
3. A line through the origin meets the unit sphere in exactly two points.

Teacher's Note:
1. A singular homogeneous system has a non-trivial solution set; its dimension is 3 minus the rank.
2. A one-dimensional solution space meets the unit sphere in two antipodal points.

Answer: (d) \( \frac13 \)

Explanation:
1. \( A^TA=I \) gives \( a^2+b^2+c^2=1 \) and \( ab+bc+ca=0 \), so \( (a+b+c)^2=1 \) and \( a+b+c=\pm1 \).
2. \( a^3+b^3+c^3-3abc=(a+b+c)(1-0) \), so \( 2-3abc=a+b+c \).
3. For \( a+b+c=1 \): \( abc=\frac13 \). (For \( a+b+c=-1 \): \( abc=1 \), which is not among the options.)

Teacher's Note:
1. The product \( A^TA \) gives two symmetric conditions on a, b, c.
2. The factorisation of \( a^3+b^3+c^3-3abc \) connects the given cubic sum to abc.

Answer: (b) 10

Explanation:
1. \( A^2=\begin{bmatrix}x^2+1&x\\x&1\end{bmatrix} \).
2. \( A^4=(A^2)^2 \) has \( a_{11}=(x^2+1)^2+x^2=x^4+3x^2+1 \) and \( a_{22}=x^2+1 \).
3. \( x^4+3x^2+1=109 \) gives \( (x^2-9)(x^2+12)=0 \), so \( x^2=9 \).
4. \( a_{22}=9+1=10 \).

Teacher's Note:
1. Square twice rather than multiplying four times.
2. Only \( x^2 \) matters, and it is determined by the quartic.

Answer: (c) \( \left(3,\frac1{81}\right) \)

Explanation:
1. \( |\operatorname{adj}A|=2(0+4)+1(1-2)+1(2-0)=9 \), and \( |\operatorname{adj}A|=|A|^2 \), so \( |\lambda|=3 \).
2. \( B=\operatorname{adj}(\operatorname{adj}A)=|A|A \), so \( |B|=|A|^3|A|=|A|^4=81 \).
3. \( |(B^{-1})^T|=\frac1{|B|}=\frac1{81} \).

Teacher's Note:
1. Remember \( \operatorname{adj}(\operatorname{adj}A)=|A|^{n-2}A \) for an \( n\times n \) matrix.
2. The determinant of a transpose or inverse is easy to handle once \( |B| \) is known.

Answer: (d) \( a^2-b^2=\frac12 \)

Explanation:
1. By induction \( A^n=\begin{bmatrix}\cos n\theta&i\sin n\theta\\i\sin n\theta&\cos n\theta\end{bmatrix} \), so \( a=d=\cos\frac{5\pi}{24} \) and \( b=c=i\sin\frac{5\pi}{24} \).
2. (a) \( a^2-d^2=0 \) is true. (b) \( a^2-c^2=\cos^2+\sin^2=1 \) is true.
3. (c) \( a^2+b^2=\cos^2\frac{5\pi}{24}-\sin^2\frac{5\pi}{24}=\cos\frac{5\pi}{12}\in[0,1] \) is true.
4. (d) \( a^2-b^2=\cos^2+\sin^2=1\ne\frac12 \), so (d) is not true.

Teacher's Note:
1. Because \( i^2=-1 \), the square of b is negative.
2. Check each statement against the closed form of \( A^5 \).

Answer: (a) 2

Explanation:
1. Put the vectors as columns: \( AX=B \) with \( X=\begin{bmatrix}1&0&0\\1&2&0\\1&1&1\end{bmatrix} \) and \( B=\begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix} \).
2. \( |X|=2 \) and \( |B|=4 \).
3. \( |A|=\frac{|B|}{|X|}=2 \).

Teacher's Note:
1. Combine the three equations into one matrix equation \( AX=B \).
2. Both X and B are triangular, so their determinants are easy.

Answer: (b) lies in (1, 2)

Explanation:
1. A is the rotation by \( -\theta \) and behaves like the complex number \( e^{-i\theta} \); \( A^4 \) behaves like \( e^{-4i\theta} \). The determinant of such a matrix is the squared modulus of the corresponding complex number.
2. \( \det B=|e^{-i\theta}+e^{-4i\theta}|^2=2+2\cos3\theta=2+2\cos\frac{3\pi}5 \).
3. \( \cos\frac{3\pi}5\approx-0.309 \), so \( \det B\approx1.38 \), which lies in (1, 2).

Teacher's Note:
1. Matrices of the form [[c, s], [-s, c]] add and multiply like complex numbers.
2. The determinant equals the squared modulus of the corresponding complex number.

Answer: (a) \( \frac19 \)

Explanation:
1. \( b_{ij}=3^{i-1}\,a_{ji}\,3^{j-1} \), so \( B=DA^TD \) with \( D=\operatorname{diag}(1,3,9) \).
2. \( |D|=27 \), so \( |B|=|D|^2|A^T|=729|A| \).
3. \( 729|A|=81 \) gives \( |A|=\frac19 \).

Teacher's Note:
1. Scaling row i by \( 3^{i-1} \) and column j by \( 3^{j-1} \) is multiplication by a diagonal matrix on each side.
2. The transpose does not change the determinant.

Answer: (b) Q, R and S only

Explanation:
1. For a \( 3\times3 \) matrix \( \operatorname{adj}(\operatorname{adj}M)=|M|M \), so \( M^{-1}=|M|M \), that is \( |M|M^2=I \).
2. Taking determinants, \( |M|^3|M|^2=1 \), so \( |M|^5=1 \) and \( |M|=1 \) (real). (Q) is true.
3. Then \( M^2=I \), so (R) is true, and \( \operatorname{adj}(M^2)=\operatorname{adj}I=I \), so (S) is true.
4. (P) is not always true: \( M=\operatorname{diag}(1,-1,-1) \) satisfies the condition but \( M\ne I \).

Teacher's Note:
1. Use \( \operatorname{adj}(\operatorname{adj}M)=|M|^{n-2}M \) with n = 3.
2. To disprove "always true", give a single counter-example.

Answer: (b) 5

Explanation:
1. For a \( 2\times2 \) matrix with eigenvalues \( \lambda_1,\lambda_2 \), the trace is \( t=\lambda_1+\lambda_2 \) and the determinant is \( d=\lambda_1\lambda_2 \).
2. \( \operatorname{tr}A^3=\lambda_1^3+\lambda_2^3=t^3-3td \).
3. \( 27-9d=-18 \), so \( d=5 \).

Teacher's Note:
1. The trace of \( A^n \) is the sum of the nth powers of the eigenvalues.
2. The identity \( \lambda_1^3+\lambda_2^3=(\lambda_1+\lambda_2)^3-3\lambda_1\lambda_2(\lambda_1+\lambda_2) \) does the work.