JEE Mathematics Matrices and Determinants MCQs Set 08

Answer: (a) -1

Explanation:
1. Compare the entries: \( x-1=-2 \) gives \( x=-1 \).
2. Check the others: \( x^2+x=0 \) holds, \( 3-x+1=5 \) holds and \( 2+x=1 \) holds for x = -1.

Teacher's Note:
1. Equate corresponding entries of the two sides.
2. Verify the value in every position.

Answer: (b) AB does not exist

Explanation:
1. A has order 3 x 1 and B has order 3 x 3.
2. For AB the number of columns of A (1) must equal the number of rows of B (3). They are different, so AB is not defined.

Teacher's Note:
1. Check the orders first.
2. Columns of the first must equal rows of the second.

Answer: (a) an odd multiple of \( \frac\pi2 \)

Explanation:
1. Each entry of AB has the factor \( \cos\theta\cos\phi+\sin\theta\sin\phi=\cos(\theta-\phi) \).
2. Indeed \( AB=\cos(\theta-\phi)\begin{pmatrix}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{pmatrix} \).
3. So AB = O requires \( \cos(\theta-\phi)=0 \), i.e. an odd multiple of pi/2.

Teacher's Note:
1. Multiply the matrices and factor out the common term.
2. cos(x) = 0 at odd multiples of pi/2.

Answer: (c) \( I\cos\theta+J\sin\theta \)

Explanation:
1. Compute \( I\cos\theta+J\sin\theta=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} \).
2. This equals B.

Teacher's Note:
1. Multiply each matrix by the scalar and add.
2. Compare entry by entry.

Answer: (c) \( i>j \)

Explanation:
1. In an upper triangular matrix all entries below the main diagonal vanish.
2. Entries below the diagonal have row index greater than column index: \( i>j \).

Teacher's Note:
1. Picture the matrix with zeros under the diagonal.
2. Row number exceeds column number below the diagonal.

Answer: (d) diag(-4, 3, 18)

Explanation:
1. Squaring a diagonal matrix squares its entries: \( A^2=\operatorname{diag}(4,1,9) \).
2. Multiply entries: \( \operatorname{diag}(4\cdot(-1),\ 1\cdot3,\ 9\cdot2)=\operatorname{diag}(-4,3,18) \).

Teacher's Note:
1. Diagonal matrices multiply entry by entry.
2. Square A first.

Answer: (b) 0

Explanation:
1. Skew-symmetry means \( a_{ii}=-a_{ii} \) for each diagonal entry.
2. So every diagonal entry is 0 and the trace is 0.

Teacher's Note:
1. Look at the diagonal entries of A = -A^T.
2. Each diagonal entry equals its own negative.

Answer: (d) F, T, T, F

Explanation:
1. (i) is false: a symmetric matrix can have any diagonal, for instance the identity.
2. (ii) is true since \( a_{ii}=-a_{ii} \).
3. (iii) is true: \( (AA')'=AA' \).
4. (iv) is false: expanding gives \( A^2+AA'+A'A+(A')^2 \), and \( A'A\ne AA' \) in general.

Teacher's Note:
1. Test each statement with a small example.
2. Matrix multiplication is not commutative.

Answer: (d) -81

Explanation:
1. For a 3 x 3 matrix \( |kM|=k^3|M| \), so \( |3AB|=27|A||B| \).
2. This is \( 27\times(-1)\times3=-81 \).

Teacher's Note:
1. A scalar multiple of a 3 x 3 matrix multiplies the determinant by k^3.
2. The determinant of a product is the product of determinants.

Answer: (b) \( 16\sqrt2 \)

Explanation:
1. Apply \( R_2\to R_2-R_3 \): the second row becomes \( (4\sqrt2,\ 4\sqrt2,\ 0) \).
2. Expand along the new second row: \( \Delta=-4\sqrt2\begin{vmatrix}2&1\\2-2\sqrt2&1\end{vmatrix}+4\sqrt2\begin{vmatrix}-1&1\\3-2\sqrt2&1\end{vmatrix} \).
3. These minors are \( 2-(2-2\sqrt2)=2\sqrt2 \) and \( -1-(3-2\sqrt2)=2\sqrt2-4 \), so \( \Delta=-4\sqrt2\cdot2\sqrt2+4\sqrt2(2\sqrt2-4)=-16+16-16\sqrt2=-16\sqrt2 \).
4. Hence \( |\Delta|=16\sqrt2 \).

Teacher's Note:
1. Use row operations R2 - R3 to simplify first.
2. R2 - R3 gives (4 sqrt 2, 4 sqrt 2, 0).

Answer: (a) 0

Explanation:
1. Use \( (u+u^{-1})^2-(u-u^{-1})^2=4 \).
2. Apply \( C_1\to C_1-C_2 \): the first column becomes (4, 4, 4), proportional to the third column (1, 1, 1).
3. A determinant with two proportional columns is 0.

Teacher's Note:
1. Subtract the second column from the first.
2. The result is a column of 4's.

Answer: (b) \( 2\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} \)

Explanation:
1. Use the multilinearity of the determinant in the columns: expand each column as a sum of two columns.
2. Terms with a repeated column vanish; the surviving terms are the two cyclic arrangements \( |b\ c\ a| \) and \( |c\ a\ b| \), each equal to \( |a\ b\ c| \) (a cyclic shift of columns is an even permutation).
3. So the determinant is \( 2|a\ b\ c| \).

Teacher's Note:
1. Add all three columns.
2. Use the standard result: this determinant equals twice the determinant of (a, b, c).

Answer: (b) 2

Explanation:
1. Apply \( C_3\to C_3-C_2 \) and \( C_2\to C_2-C_1 \): the determinant becomes \( \begin{vmatrix}x&y&z\\2x&3x+2y&2x+3y+2z\\3x&4x+3y&2x+4y+3z\end{vmatrix} \).
2. Reducing further (row and column operations) gives \( \Delta=-2x^3 \).
3. So \( -2x^3=-16 \) and x = 2.

Teacher's Note:
1. Use column operations C2 - C1 and C3 - C2 repeatedly.
2. The determinant reduces to a multiple of x^3.

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

Explanation:
1. Expanding along the first row and simplifying with the addition formulas, the determinant equals \( 1+\cos2\phi=2\cos^2\phi \).
2. It contains no \( \theta \), so it is independent of \( \theta \) but does depend on \( \phi \).

Teacher's Note:
1. Expand along the first row and use cos(theta + phi) formulas.
2. The value 1 + cos 2 phi has no theta.

Answer: (b) Both statements are true, but Statement 2 is not a correct explanation of Statement 1

Explanation:
1. Interchanging two rows changes the sign of a determinant, so determinants equal to 1 and -1 are paired off: \( n(B)=n(C) \) (true).
2. B and C have no common element, so \( B\cap C=\varnothing \), and the empty set is a subset of A (true).
3. The second statement does not explain the first.

Teacher's Note:
1. A row swap pairs +1 with -1 determinants.
2. The empty set is a subset of every set.

Answer: (b) \( \begin{pmatrix}1&-2\\-2&1\end{pmatrix} \)

Explanation:
1. For a 2 x 2 matrix \( \begin{pmatrix}a&b\\c&d\end{pmatrix} \) the adjoint is \( \begin{pmatrix}d&-b\\-c&a\end{pmatrix} \).
2. Here it is \( \begin{pmatrix}1&-2\\-2&1\end{pmatrix} \).

Teacher's Note:
1. Swap the diagonal entries and change the sign of the off-diagonal ones.
2. This is the transpose of the cofactor matrix.

Answer: (c) AB is non-singular

Explanation:
1. The determinant of a product is the product of the determinants: \( |AB|=|A||B|\ne0 \).
2. So AB is non-singular. (The inverse is actually \( B^{-1}A^{-1} \), not \( A^{-1}B^{-1} \).)

Teacher's Note:
1. |AB| = |A||B|.
2. The order of the factors reverses in the inverse of a product.

Answer: (d) \( \frac12\begin{pmatrix}2&4\\3&-5\end{pmatrix} \)

Explanation:
1. Since B is invertible, \( X=B^{-1}A \).
2. With \( B^{-1}=\begin{pmatrix}1&0\\0&\frac12\end{pmatrix} \), the second row of A is halved: \( X=\begin{pmatrix}1&2\\\frac32&-\frac52\end{pmatrix}=\frac12\begin{pmatrix}2&4\\3&-5\end{pmatrix} \).

Teacher's Note:
1. Pre-multiply both sides by the inverse of B.
2. The inverse of a diagonal matrix has reciprocal entries.

Answer: (b) \( \begin{pmatrix}-1&2&-3\\-2&0&3\\3&-3&1\end{pmatrix} \)

Explanation:
1. Since \( \operatorname{adj}(A^{-1})=|A^{-1}|\,(A^{-1})^{-1}=\frac{A}{|A|} \).
2. Therefore \( |A|\operatorname{adj}(A^{-1})=A \).

Teacher's Note:
1. Use adj(M) = |M| M^(-1) with M = A^(-1).
2. The determinant of the inverse is 1/|A|.

Answer: (d) T, F, T

Explanation:
1. S1: |A| = 1 and A is orthogonal, so \( \operatorname{adj}A=|A|A^{-1}=A^{-1}=A' \) (true).
2. S2: the inverse of diag(a, b, c) is diag(1/a, 1/b, 1/c) (false).
3. S3: \( \det(B^{-1}AB)=\frac{1}{|B|}|A||B|=|A| \) (true).

Teacher's Note:
1. Use adj A = |A| A^(-1).
2. Rotation matrices satisfy A^(-1) = A^T.

Answer: (c) All four are true

Explanation:
1. For order n, the determinant of the cofactor matrix is \( D^{n-1} \), which is \( D^2 \) for n = 3 (so (i) is true).
2. If D = 0 then \( \Delta=0 \) ((ii) true).
3. If D = 27, \( \Delta=729=9^3=27^2 \), a perfect cube and a perfect square ((iii), (iv) true).

Teacher's Note:
1. The cofactor determinant is D^(n-1).
2. 729 = 27^2 = 9^3.