JEE Mathematics Matrices and Determinants MCQs Set 06

Answer: (a) \( g(x)+g(-x)=0 \)

Explanation:
1. \( e^{kx\log_ea}=a^{kx} \), so the first two columns are \( a^{-x},a^{x} \) and so on.
2. Replacing x by \(-x\) interchanges the first two columns and leaves the third column unchanged (since \( x^2,x^4 \) are even).
3. Interchanging two columns changes the sign of the determinant, so \( g(-x)=-g(x) \), i.e. \( g(x)+g(-x)=0 \).

Teacher's Note:
1. Replace x by \(-x\) and see what happens to the columns.

Answer: (a) 0

Explanation:
1. Expanding the determinant gives \( 24x-12 \).
2. So \( A=24 \) and \( B=-12 \).
3. Then \( A+2B=24-24=0 \).

Teacher's Note:
1. The determinant is linear in x after simplification; read off the coefficient and the constant term.

Answer: (c) \( \pm\frac1{\sqrt2},\pm\frac1{\sqrt6},\pm\frac1{\sqrt3} \)

Explanation:
1. For an orthogonal matrix the columns (and rows) are unit vectors. The first column gives \( 2a^2=1 \), so \( a=\pm\frac1{\sqrt2} \).
2. The second column gives \( 4b^2+b^2+b^2=6b^2=1 \), so \( b=\pm\frac1{\sqrt6} \).
3. The third column gives \( c^2+c^2+c^2=3c^2=1 \), so \( c=\pm\frac1{\sqrt3} \). The columns are also mutually orthogonal.

Teacher's Note:
1. Use \( A^TA=I \): each column has unit length.

Answer: (c) 6

Explanation:
1. \( P^2=I-P \). Then \( P^3=P-P^2=2P-I \).
2. \( P^4=2P^2-P=2I-3P \) and \( P^5=2P-3P^2=5P-3I \).
3. \( P^6=5P^2-3P=5I-8P \), so \( n=6 \).

Teacher's Note:
1. Keep reducing powers of P using \( P^2=I-P \).

Answer: (a) \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \)

Explanation:
1. P is orthogonal, so \( P^TP=PP^T=I \).
2. \( Q^{n}=PA^nP^T \), so \( P^TQ^{2005}P=A^{2005} \).
3. \( A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix} \), so \( A^{2005}=\begin{bmatrix}1&2005\\0&1\end{bmatrix} \).

Teacher's Note:
1. Because P is orthogonal, the P and \( P^T \) cancel inside the power.

Answer: (b) \( X=\begin{bmatrix}2&-1&1\\1&2&0\end{bmatrix} \)

Explanation:
1. Multiply the first equation by 2 and add the second: \( 5X=\begin{bmatrix}10&-5&5\\5&10&0\end{bmatrix} \), so \( X=\begin{bmatrix}2&-1&1\\1&2&0\end{bmatrix} \).
2. Then \( Y=2X-\begin{bmatrix}3&-3&0\\3&3&2\end{bmatrix}=\begin{bmatrix}1&1&2\\-1&1&-2\end{bmatrix} \).
3. Hence \( X+Y=\begin{bmatrix}3&0&3\\0&3&-2\end{bmatrix} \) and \( X-Y=\begin{bmatrix}1&-2&-1\\2&1&2\end{bmatrix} \), so only (b) is correct.

Teacher's Note:
1. Eliminate Y by taking twice the first equation plus the second.

Answer: (c) \( \Delta \)

Explanation:
1. The elements of row 2 are \( x,y,z \).
2. Multiplying them by their own cofactors \( C_{21},C_{22},C_{23} \) is exactly the expansion of \( \Delta \) along row 2.
3. So the value is \( \Delta \).

Teacher's Note:
1. Elements multiplied by their own cofactors give the determinant.

Answer: (b) \(-\Delta\)

Explanation:
1. \( C_{12}=-M_{12},\ C_{22}=M_{22},\ C_{32}=-M_{32} \).
2. So \( qM_{12}-yM_{22}+mM_{32}=-\left(qC_{12}+yC_{22}+mC_{32}\right) \).
3. The bracket is the expansion of \( \Delta \) along column 2, so the value is \( -\Delta \).

Teacher's Note:
1. Convert minors to cofactors using the sign \( (-1)^{i+j} \).

Answer: (a) 0

Explanation:
1. \( (m+m^{-1})^2-(m-m^{-1})^2=4 \) for any m.
2. So \( R_2-R_3=(4,4,4)=4R_1 \).
3. A row operation \( R_2\to R_2-R_3 \) makes row 2 a multiple of row 1, so the determinant is 0.

Teacher's Note:
1. The difference of the squares of the sum and difference is a constant.

Answer: (d) 2

Explanation:
1. A non-zero solution requires \( \begin{vmatrix}p&b&c\\a&q&c\\a&b&r\end{vmatrix}=0 \). Expanding: \( pqr-pbc-abr-acq+2abc=0 \).
2. One can check the identity \( \Delta=p(q-b)(r-c)+q(p-a)(r-c)+r(p-a)(q-b)-2(p-a)(q-b)(r-c) \).
3. Dividing by \( (p-a)(q-b)(r-c)\ne0 \): \( \frac p{p-a}+\frac q{q-b}+\frac r{r-c}-2=0 \), so the value is 2.

Teacher's Note:
1. Use row operations to bring out the factors \( p-a,\ q-b,\ r-c \).

Answer: (c) \( A^{-1}B^nA \)

Explanation:
1. Write \( \left(A^{-1}BA\right)\left(A^{-1}BA\right)=A^{-1}B\left(AA^{-1}\right)BA=A^{-1}B^2A \).
2. Repeating the cancellation of \( AA^{-1} \) in the middle for n factors gives \( A^{-1}B^nA \).

Teacher's Note:
1. Neighbouring A and \( A^{-1} \) cancel in the middle of the product.

Answer: (d) 512

Explanation:
1. A \( 3\times3 \) matrix has 9 entries.
2. Each entry can be chosen in 2 ways (2 or 0).
3. The total number is \( 2^9=512 \).

Teacher's Note:
1. Count by the multiplication principle.

Answer: (a) Skew symmetric matrix

Explanation:
1. \( \left(AB^T-BA^T\right)^T=(AB^T)^T-(BA^T)^T=BA^T-AB^T \).
2. This equals \( -\left(AB^T-BA^T\right) \).
3. A matrix whose transpose is its negative is skew symmetric.

Teacher's Note:
1. Take the transpose using \( (XY)^T=Y^TX^T \).

Answer: (a) 0

Explanation:
1. Expanding along the first row: \( f(t)=\cos t\left(t^2-2t^2\right)-t\left(2t\sin t-2t\sin t\right)+\left(2t\sin t-t\sin t\right)=t\sin t-t^2\cos t \).
2. \( \frac{f(t)}{t^2}=\frac{\sin t-t\cos t}t \). Using \( \sin t-t\cos t=\frac{t^3}3+\dots \) this is about \( \frac{t^2}3 \).
3. So the limit is 0.

Teacher's Note:
1. Expand the determinant and use series for sine and cosine.

Answer: (a) \( A^2+B^2 \)

Explanation:
1. \( B=-A^{-1}BA \) gives, on multiplying by A on the left, \( AB=-BA \).
2. \( (A+B)^2=A^2+AB+BA+B^2 \).
3. Since \( AB+BA=0 \), \( (A+B)^2=A^2+B^2 \).

Teacher's Note:
1. Multiply the given relation by A to obtain \( AB=-BA \).

Answer: (c) \( (A+B)^2=A+B \)

Explanation:
1. \( (A+B)^2=A^2+AB+BA+B^2=A+0+0+B=A+B \).
2. \( (A-B)^2=A^2-AB-BA+B^2=A+B \), which is neither \( B-A \) nor \( A-B \) in general.
3. So only (c) is correct.

Teacher's Note:
1. Expand the squares and use the given relations.

Answer: (c) Exactly one real value of t

Explanation:
1. Non-trivial solutions need the determinant of the coefficients to be zero.
2. Expanding: \( \begin{vmatrix}t&t+1&t-1\\t+1&t&t+2\\t-1&t+2&t\end{vmatrix}=-4(2t+1) \).
3. This vanishes only for \( t=-\frac12 \), exactly one real value.

Teacher's Note:
1. Set the determinant of the coefficient matrix equal to zero.

Answer: (d) 500

Explanation:
1. \( |A|=1(1+1)-1(-1+1)+1(1+1)=2-0+2=4 \).
2. For a \( 3\times3 \) matrix \( |5A|=5^3|A| \).
3. So \( |5A|=125\times4=500 \).

Teacher's Note:
1. Multiplying a \( 3\times3 \) matrix by a scalar multiplies the determinant by the cube of the scalar.

Answer: (a) \( (\det A)^2 \)

Explanation:
1. From \( A\,\operatorname{adj}A=|A|I \), taking determinants gives \( |A|\,|\operatorname{adj}A|=|A|^3 \).
2. Since A is non-singular, \( |\operatorname{adj}A|=|A|^2 \).

Teacher's Note:
1. Take determinants of both sides of \( A\,\operatorname{adj}A=|A|I \).

Answer: (a) 2

Explanation:
1. A non-zero solution requires \( \begin{vmatrix}4&k&2\\k&4&1\\2&2&1\end{vmatrix}=0 \).
2. Expanding: \( 4(4-2)-k(k-2)+2(2k-8)=-k^2+6k-8 \).
3. \( k^2-6k+8=0 \) gives \( k=2 \) or \( 4 \), so there are 2 values.

Teacher's Note:
1. Equate the determinant of the coefficients to zero.

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

Explanation:
1. Let \( u_1=(a,b,c)^T \). Then \( a=1,\ 2a+b=0,\ 3a+2b+c=0 \), so \( b=-2,\ c=1 \).
2. Let \( u_2=(p,q,r)^T \). Then \( p=0,\ 2p+q=1,\ 3p+2q+r=0 \), so \( q=1,\ r=-2 \).
3. \( u_1+u_2=(1,-1,-1)^T \).

Teacher's Note:
1. Because A is lower triangular, solve the equations by forward substitution.

Answer: (c) 0

Explanation:
1. \( \left(P^2+Q^2\right)(P-Q)=P^3-P^2Q+Q^2P-Q^3=0 \).
2. If \( \left|P^2+Q^2\right|\ne0 \), then \( P^2+Q^2 \) is invertible and \( P-Q=0 \), contradicting \( P\ne Q \).
3. Hence \( \left|P^2+Q^2\right|=0 \).

Teacher's Note:
1. Factor \( (P^2+Q^2)(P-Q) \) and use the fact that \( P\ne Q \).

Answer: (d) \( a=8,\ b=15 \)

Explanation:
1. Subtract equation 1 and equation 2 from equation 3: \( 0x_1+0x_2+(a-8)x_3=b-15 \).
2. For infinitely many solutions this equation must hold for every \( x_3 \), so \( a-8=0 \) and \( b-15=0 \).
3. Hence \( a=8,\ b=15 \).

Teacher's Note:
1. Reduce the third equation using the first two and look at the final row.