JEE Mathematics Matrices and Determinants MCQs Set 07

Answer: (c) 2

Explanation:
1. Add all columns into the first one: the determinant is \( (\cos x+2\sin x)\begin{vmatrix}1&\sin x&\sin x\\1&\cos x&\sin x\\1&\sin x&\cos x\end{vmatrix} \). Subtracting row 1 from rows 2 and 3 gives \( (\cos x+2\sin x)(\cos x-\sin x)^2 \).
2. \( \cos x=\sin x \) gives \( x=\frac\pi4 \) in the interval.
3. \( \tan x=-\frac12 \) gives \( x=-\tan^{-1}\frac12\approx-0.46 \), which lies in \( \left[-\frac\pi4,\frac\pi4\right] \) since \( \frac\pi4\approx0.785 \). So there are 2 distinct roots.

Teacher's Note:
1. Factorise the determinant as \( (\cos x+2\sin x)(\cos x-\sin x)^2 \).

Answer: (d) \(-25\)

Explanation:
1. The characteristic equation is \( \lambda^2+3\lambda-1=0 \) (trace \(-3\), determinant \(-1\)), so \( A^2=-3A+I \).
2. \( A^{2016}-2A^{2015}-A^{2014}=A^{2014}\left(A^2-2A-I\right)=A^{2014}(-5A) \).
3. The determinant is \( (-1)^{2014}\cdot(-5)^2\cdot(-1)=-25 \).

Teacher's Note:
1. Use the Cayley-Hamilton theorem to simplify \( A^2-2A-I \).

Answer: (c) \( A^2+I=A\left(A^2-I\right) \)

Explanation:
1. \( A^2=-I,\ A^3=-A,\ A^4=I \).
2. (a) \( A^4-I=0 \) and \( A^2+I=0 \): true. (b) \( A^3-I=-A-I \) and \( A(A-I)=A^2-A=-I-A \): true.
3. (c) \( A^2+I=0 \) but \( A\left(A^2-I\right)=A(-2I)=-2A\ne0 \): false. (d) \( A^3+I=I-A \) and \( A\left(A^3-I\right)=A(-A-I)=I-A \): true. So (c) is not correct.

Teacher's Note:
1. Compute the powers of A first, then check each statement.

Answer: (a) \( \pm\frac15 \)

Explanation:
1. For a \( 3\times3 \) matrix, \( |5\operatorname{adj}A|=5^3|\operatorname{adj}A|=125|A|^2 \).
2. So \( 125|A|^2=5 \) gives \( |A|^2=\frac1{25} \).
3. Hence \( |A|=\pm\frac15 \).

Teacher's Note:
1. Use \( |kM|=k^3|M| \) and \( |\operatorname{adj}A|=|A|^2 \) for a \( 3\times3 \) matrix.

Answer: (b) 24

Explanation:
1. Expanding the determinant gives \( 24x-12 \).
2. Comparing with \( ax-12 \), the constant term \(-12\) matches.
3. So \( a=24 \).

Teacher's Note:
1. Expand and compare the coefficients of x.

Answer: (b) abc

Explanation:
1. In standard form the coefficient determinant is \( \begin{vmatrix}a-1&-1&-1\\-1&b-1&-1\\-1&-1&c-1\end{vmatrix} \).
2. Expanding gives \( abc-ab-bc-ca \).
3. A non-trivial solution needs this to be 0, so \( ab+bc+ca=abc \).

Teacher's Note:
1. A homogeneous system has a non-trivial solution exactly when the determinant of its coefficients is zero.

Answer: (a) 1

Explanation:
1. Since \( B^2=0 \), the binomial expansion gives \( (I+B)^{50}=I+50B \), because all higher powers of B vanish.
2. So \( (I+B)^{50}-50B=I \).
3. \( \det I=1 \).

Teacher's Note:
1. With \( B^2=0 \) only the first two terms of the binomial expansion survive.

Answer: (a) \( \begin{bmatrix}3&1&2\\3&0&2\\1&0&1\end{bmatrix} \)

Explanation:
1. Write \( AM=N \). Then \( A^{-1}=MN^{-1} \).
2. N is a permutation matrix, so \( N^{-1}=N^T=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix} \).
3. Multiplying, \( MN^T \) moves the columns of M: \( \begin{bmatrix}3&1&2\\3&0&2\\1&0&1\end{bmatrix} \).

Teacher's Note:
1. Use \( (AM)^{-1}=M^{-1}A^{-1} \) in the form \( A^{-1}=MN^{-1} \), and the inverse of a permutation matrix is its transpose.

Answer: (d) \( y=2x \)

Explanation:
1. Multiplying: \( y+2x+x=6 \) and \( 3y-x+2=8 \).
2. These give \( y+3x=6 \) and \( 3y-x=6 \).
3. From the first \( y=6-3x \), so \( 18-9x-x=6 \), \( x=\frac65 \) and \( y=\frac{12}5=2x \).

Teacher's Note:
1. Multiply the matrices to obtain two linear equations in x and y.

Answer: (c) Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I.

Explanation:
1. \( A^2=I \) gives \( (\det A)^2=1 \). If \( \det A=1 \), the eigenvalues are both 1 or both \(-1\), which forces \( A=I \) or \(-I \); this is excluded. So \( \det A=-1 \), and Statement II is true.
2. By the Cayley-Hamilton theorem \( A^2-\operatorname{Tr}(A)A+(\det A)I=O \). With \( A^2=I \) and \( \det A=-1 \) this gives \( \operatorname{Tr}(A)A=O \), so \( \operatorname{Tr}(A)=0 \) (since \( A\ne O \)). Statement I is true.
3. Statement I follows from Statement II, so II explains I.

Teacher's Note:
1. Use Cayley-Hamilton: for a \( 2\times2 \) matrix \( A^2-(\operatorname{Tr}A)A+(\det A)I=O \).

Answer: (a) 225

Explanation:
1. For a \( 3\times3 \) matrix \( |\operatorname{adj}A|=|A|^2 \), so \( b=a^2=9 \).
2. \( \frac S2 \) is a geometric series with first term \( \frac ab=\frac13 \) and ratio \( \frac a{b^2}=\frac1{27} \), so \( \frac S2=\frac{1/3}{26/27}=\frac9{26} \) and \( S=\frac9{13} \).
3. \( ab^2+a^2b+1=243+81+1=325 \), so the value is \( 325\cdot\frac9{13}=225 \).

Teacher's Note:
1. First find b from a, then sum the infinite geometric series.

Answer: (b) 8

Explanation:
1. \( |A^{-1}|=\frac12 \).
2. \( |\operatorname{adj}(B^{-1})|=|B^{-1}|^2=1 \).
3. \( |2A^{-1}|=8\cdot\frac12=4 \), so \( |\operatorname{adj}(2A^{-1})|=4^2=16 \). The product is \( \frac12\cdot1\cdot16=8 \).

Teacher's Note:
1. Use \( |XY|=|X||Y| \), \( |\operatorname{adj}M|=|M|^2 \) and \( |kM|=k^3|M| \) for \( 3\times3 \) matrices.

Answer: (b) 1

Explanation:
1. All the matrices are diagonal, so work with one diagonal entry, say with \( s=\sin^2\alpha \) and \( c=\cos^2\alpha \), where \( s+c=1 \).
2. The entry of C is \( s^3+c^3+3sc(c+s)=(s+c)^3=1 \).
3. So \( C=I \) and \( \det C=1 \).

Teacher's Note:
1. The expression is the expansion of \( (A^{-1}+B^{-1})^3 \) for commuting matrices.

Answer: (b) 6

Explanation:
1. Multiply \( (I-0.4A)(I-\alpha A)=I-\alpha A-0.4A+0.4\alpha A^2 \). Idempotent means \( A^2=A \).
2. This becomes \( I+(-\alpha-0.4+0.4\alpha)A=I \), so \( -0.6\alpha-0.4=0 \).
3. \( \alpha=-\frac23 \) and \( |9\alpha|=6 \).

Teacher's Note:
1. Use \( A^2=A \) to reduce the product to a multiple of A.

Answer: (b) 52

Explanation:
1. Write \( P=I+N \) with \( N=\begin{bmatrix}0&0&0\\2&0&0\\4&2&0\end{bmatrix} \). Then \( N^2=\begin{bmatrix}0&0&0\\0&0&0\\4&0&0\end{bmatrix} \) and \( N^3=O \).
2. \( P^{50}=I+50N+\binom{50}2N^2 \), so \( Q=50N+1225N^2 \).
3. \( q_{21}=100,\ q_{32}=100,\ q_{31}=200+4900=5100 \). So \( \frac{5100+100}{100}=52 \).

Teacher's Note:
1. Write P as \( I+N \) with N nilpotent, then use the binomial theorem.

Answer: (b) 2

Explanation:
1. Split the third column: the determinant is \( x^3\begin{vmatrix}1&1&1\\2&4&1\\3&9&1\end{vmatrix}+x^6\begin{vmatrix}1&1&1\\2&4&8\\3&9&27\end{vmatrix}=2x^3+12x^6 \).
2. Setting \( 12x^6+2x^3=10 \) gives \( 6x^6+x^3-5=0 \), so \( x^3=\frac56 \) or \( x^3=-1 \).
3. Each value of \( x^3 \) gives exactly one real x, so there are 2 distinct real values.

Teacher's Note:
1. Split the determinant using the additive property of the third column.

Answer: (b) 9

Explanation:
1. The first condition gives the second column of M: \( (-1,2,3)^T \).
2. The second gives \( Me_1-Me_2=(1,1,-1)^T \), so \( Me_1=(0,3,2)^T \). The third gives \( Me_1+Me_2+Me_3=(0,0,12)^T \), so \( Me_3=(1,-5,7)^T \).
3. \( M=\begin{bmatrix}0&-1&1\\3&2&-5\\2&3&7\end{bmatrix} \), with diagonal sum \( 0+2+7=9 \).

Teacher's Note:
1. Express the standard basis vectors through the given vectors to find the columns of M.

Answer: (b) 1

Explanation:
1. The homogeneous system \( -x+cy+bz=0,\ cx-y+az=0,\ bx+ay-z=0 \) has a non-zero solution, so the determinant of coefficients is zero.
2. \( \begin{vmatrix}-1&c&b\\c&-1&a\\b&a&-1\end{vmatrix}=a^2+b^2+c^2+2abc-1 \).
3. Setting it to 0 gives \( a^2+b^2+c^2+2abc=1 \).

Teacher's Note:
1. A non-zero solution of a homogeneous system needs a zero determinant.

Answer: (b) 4

Explanation:
1. Apply \( C_3\to C_3-C_2 \) and \( C_2\to C_2-C_1 \). Then apply \( R_3\to R_3-R_2,\ R_2\to R_2-R_1 \) twice and finally \( R_3\to R_3-R_2 \).
2. The determinant becomes \( \begin{vmatrix}x&y&z\\0&x&x+y\\0&0&x\end{vmatrix}=x^3 \).
3. \( x^3=64 \) gives \( x=4 \).

Teacher's Note:
1. Row and column operations reduce the determinant to a triangular one.

Answer: (a) 0

Explanation:
1. \( A^2=\begin{bmatrix}3&0\\0&3\end{bmatrix}=3I \).
2. So \( A^8+A^6+A^4+A^2+I=(81+27+9+3+1)I=121I \).
3. \( 121V=\begin{bmatrix}0\\11\end{bmatrix} \) gives \( V=\begin{bmatrix}0\\\frac1{11}\end{bmatrix} \), and the product of all its elements is 0.

Teacher's Note:
1. Compute \( A^2 \) first; it is a scalar matrix.