NDA Matrices previous year questions with solutions

Answer: (d) I, II and III

Method. conformability rule: (m × n)(n × p) is m × p

I. ZY: (3 × 2)(2 × 3) = 3 × 3; (ZY)X: (3 × 3)(3 × 3) = 3 × 3. Defined, square of order 3. True.
II. XZ: (3 × 3)(3 × 2) = 3 × 2; Y(XZ): (2 × 3)(3 × 2) = 2 × 2. Defined, square of order 2. True.
III. YZ: (2 × 3)(3 × 2) = 2 × 2; X(YZ): (3 × 3)(2 × 2), columns of X (3) ≠ rows of YZ (2). Not defined. True.
Correct: I, II and III.

Answer: (d) 768

Method. possible orders of a 4-entry matrix × fillings with repetition

A matrix with 4 entries has order m × n with mn = 4: 1 × 4, 4 × 1 or 2 × 2, i.e. 3 orders.
Each of the 4 positions can take any of 1, 2, 3, 4: 4⁴ = 256 fillings per order.
Total = 3 × 256 = 768.

Answer: (d) 1, 2 and 3

Method. order of a product: (m × n)(n × p) is m × p, defined only when the inner orders match

X: 3 × 3, Y: 2 × 3, Z: 3 × 2.
1. ZY: (3 × 2)(2 × 3) = 3 × 3; (ZY)X: (3 × 3)(3 × 3) = 3 × 3, a square matrix with 9 entries. Correct.
2. XZ: (3 × 3)(3 × 2) = 3 × 2; Y(XZ): (2 × 3)(3 × 2) = 2 × 2, a square matrix with 4 entries. Correct.
3. YZ: (2 × 3)(3 × 2) = 2 × 2; X(YZ): (3 × 3)(2 × 2) has 3 columns against 2 rows, so it is not defined. Correct.
All of 1, 2 and 3.

Answer: (d) 48

Method. a matrix with four entries can be 1 × 4, 4 × 1 or 2 × 2; each entry has 2 choices

A matrix with exactly four entries has order m × n with mn = 4: 1 × 4, 4 × 1 or 2 × 2.
For each order, every one of the 4 entries is 1 or 2, giving 2⁴ = 16 matrices.
Matrices of different orders are distinct, so the total is 3 × 16 = 48.

No order is given, so 2 × 2 is only one of the three orders with four entries; counting 2 × 2 matrices alone gives 16, option (a), which answers a narrower question than the one set.

Answer: (d) Orders of A and B need not be the same.

Method. order of a product: (n × m)(m × n) is n × n for any m

For AB to be n × n, A must have n rows and B must have n columns.
So A is n × m and B is m × n, where m is any positive integer.
Example: A of order 2 × 3 and B of order 3 × 2 give AB of order 2 × 2, yet neither is square and their orders differ.
So (a), (b), (c) are not necessary; the orders of A and B need not be the same.

Answer: (c) 4

Method. an order m × n holds exactly mn elements; count factor pairs of the number of elements

Primes less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, that is 10 elements.
A matrix with 10 elements has order m × n with mn = 10.
Possible orders: 1 × 10, 10 × 1, 2 × 5, 5 × 2.
So 4 matrices of different orders are possible.

Answer: (c) 3 × 3 and 5 × 5

Method. order of a product: (m × n)(n × p) is m × p

AB: (3 × 5)(5 × 3) gives order 3 × 3.
BA: (5 × 3)(3 × 5) gives order 5 × 5.
So the orders are 3 × 3 and 5 × 5.

Answer: (c) AB exists but BA does not exist

Method. conformability: AB exists only when columns of A = rows of B

A is of order 3 × 2 and B is of order 2 × 2.
AB: columns of A = 2 = rows of B, so AB exists (order 3 × 2).
BA: columns of B = 2 but rows of A = 3, so BA does not exist.
Hence AB exists but BA does not.

Answer: (a) 3 × 5 matrix

Method. (m × n)(n × p) = m × p

A is 2 × 3, so B must have 3 rows for AB to exist.
AB is 2 × (columns of B) = 2 × 5, so B has 5 columns.
B is a 3 × 5 matrix.

Answer: (c) 3 and 8

Method. conformability conditions for AB and BA

A is x × (x + 5); B is y × (11 − y).
AB exists: columns of A = rows of B, so x + 5 = y.
BA exists: columns of B = rows of A, so 11 − y = x.
Adding: x + y = 11 with y = x + 5 gives 2x + 5 = 11, so x = 3, y = 8.

Answer: (b) 1 × 1

Method. order of a product: (m × n)(n × p) is m × p

Orders:  x y z  is 1 × 3, the middle matrix is 3 × 3,  x y z  is 3 × 1.
(1 × 3)(3 × 3) is 1 × 3; then (1 × 3)(3 × 1) is 1 × 1.
So the product is of order 1 × 1.

Answer: (c) Both I and II

Method. cancellation law holds only for a non-singular factor

II. If A is non-singular, AB = AC ⇒ A^-1AB = A^-1AC ⇒ B = C. True.
I. If A is singular, cancellation can fail: A =  1 0 0 0 , B =  0 0 0 1 , C =  0 0 0 2  give AB = AC = O but B ≠ C. True.
Both I and II are correct.

Answer: (d) 1

Method. evaluate the 1 × 1 product and solve the quadratic

 1 2 3 4 5 6 7 8 9  1 1 x  =  3 + 3x 9 + 6x 15 + 9x . x 1 1  3 + 3x 9 + 6x 15 + 9x  = [3x² + 3x + 9 + 6x + 15 + 9x] = [3x² + 18x + 24].
3x² + 18x + 24 = 45 ⇒ x² + 6x − 7 = 0 ⇒ (x + 7)(x − 1) = 0.
x = 1 or x = -7; among the options, x = 1.

Shortcut. Test the options: x = 1 gives [1 1 1]·(6, 15, 24)^T = 45.

Answer: (c) B is a null matrix if A is a non-singular matrix

Method. zero divisors among matrices; cancelling a non-singular factor

If A is non-singular, A^-1 exists: AB = O ⇒ A^-1AB = A^-1O ⇒ B = O. So (c) is correct.
Counter-example to (a), (b): A =  1 0 0 0 , B =  0 0 0 1  give AB = O with neither A nor B null.
Counter-example to (d): A = I, B = O gives AB = O with A non-singular.

Answer: (c) I and III

Method. Q = vv^T with v = (a, b, c)^T, and Pv = 0

Take v =  a b c ; then Q = vv^T.
Pv =  0·a + cb − bc -ca + 0·b + ac ba − ab + 0·c  =  0 0 0 .
PQ = (Pv)v^T = O, the null matrix. I is true.
P^T = -P, so QP = v(v^TP) = v(P^Tv)^T = -v(Pv)^T = O. Hence QP is not I: II is false.
PQ = QP = O: III is true.
Correct: I and III.

Answer: (c) Both 1 and 2

Method. cancellation by A^-1; testing with the unit column vectors

1. If A is non-singular, pre-multiply AB = AC by A^-1: A^-1AB = A^-1AC, i.e. B = C. Correct.
2. Take X = e_j, the column with 1 in row j and 0 elsewhere; Be_j is the j-th column of B and Ce_j the j-th column of C.
BX = CX for every X gives every column of B equal to the corresponding column of C, so B = C. Correct.
Both 1 and 2.

Answer: (c) 2 and 3

Method. compute the products and use the distributive law

A and B are 1 × 2, C is 2 × 1.
CA =  m² mn -m² -mn , CB =  -mn -m² mn m² ; these differ in general (e.g. m = 1, n = 0), so 1 is not correct.
AC = [m·m + n(-m)] = [m² - mn], BC = [(-n)m + (-m)(-m)] = [m² - mn], so AC = BC: 2 is correct.
C(A + B) = CA + CB is the distributive law, valid since A + B and the products are defined: 3 is correct.
So 2 and 3.

Answer: (b) 2 only

Method. zero divisors exist among matrices; for square matrices AB = I forces BA = I

1. Counter-example: A =  1 0 0 0 , B =  0 0 0 1 ; AB = O though neither is null. Incorrect.
2. AB = I gives |A||B| = 1, so A is non-singular; then B = A^-1I = A^-1, hence BA = A^-1A = I = AB. Correct.
So 2 only.

Answer: (c) Both 1 and 2

Method. substitute AB = A and BA = B inside products

1. A² = A·A = (AB)A = A(BA) = AB = A. Correct.
Similarly B² = (BA)B = B(AB) = BA = B.
2. AB² = A(B²) = AB = A, and A²B = A(AB) = A·A = A² = A.
So AB² = A²B. Correct.
Both 1 and 2.

Shortcut. Since A is non-singular, AB = A gives B = I, and then BA = B gives A = I; with A = B = I both statements hold at once.

Answer: (b) -14

Method. matrix product gives two linear equations; then a 2 × 2 determinant

AB =  2(x + y) − y 2(2x) − (x − y)  =  2x + y 3x + y .
AB = C gives 2x + y = 3 and 3x + y = 2.
Subtracting: x = -1, then y = 3 − 2(-1) = 5.
A =  4 5 -2 -6 , so |A| = 4(-6) − 5(-2) = -24 + 10 = -14.

Answer: (c)  ax + hy + gz hx + by + fz gx + fy + cz 

Method. row-by-column multiplication; (3 × 3)(3 × 1) gives 3 × 1

A is 3 × 3 and B is 3 × 1, so AB is a 3 × 1 column matrix (this rules out (d), a row matrix).
Row 1 · B = ax + hy + gz; row 2 · B = hx + by + fz; row 3 · B = gx + fy + cz.
AB =  ax + hy + gz hx + by + fz gx + fy + cz .

Answer: (b) 2 only

Method. matrix multiplication is not commutative in general

1. (A + B)(A − B) = A² - AB + BA − B², which equals A² - B² only if AB = BA. False in general (e.g. A =  1 0 0 0 , B =  0 1 0 0 : AB = B, BA = O).
2. (A − I)(I + A) = A + A² - I − A = A² - I (A and I commute). So (A − I)(I + A) = O ⇔ A² = I. True.
So 2 only.

Answer: (b)  -1 -1 -1 2 

Method. multiply, then use α + β = -1, αβ = 1, α² + α = -1

For 1 + x + x² = 0: α + β = -1, αβ = 1, and α² + α = -1, β² + β = -1. 1 β α α  α β 1 β  =  α + β β + β² α² + α αβ + αβ 
=  -1 -1 -1 2 .

Answer: (a)  2 + 3i 5 7 2 − 3i 

Method. scalar multiplication with k = 12i = -i2

k = 12i = -i2 (since 1i = -i).
(4i − 6)·-i2 = -4i² + 6i2 = 4 + 6i2 = 2 + 3i.
10i·-i2 = 5; 14i·-i2 = 7.
(6 + 4i)·-i2 = -6i − 4i²2 = 4 − 6i2 = 2 − 3i.
kA =  2 + 3i 5 7 2 − 3i .

Answer: (d)  -4 -8 8 12 

Method. solve AB = C for x, y, then square A

AB =  3(x + y) − 2y 3x − 2(x − y)  =  3x + y x + 2y .
AB = C gives 3x + y = 4 and x + 2y = -2, so x = 2, y = -2.
Then A =  0 -2 2 4 .
A² =  0.0 + (-2)·2 0·(-2) + (-2)·4 2.0 + 4.2 2·(-2) + 4.4  =  -4 -8 8 12 .

Answer: (d)  1 0 0 1 

Method. substitute θ = π and square

cos π = -1, sin π = 0, so f(π) =  -1 0 0 -1  = -I.
{f(π)}² = (-I)² = I =  1 0 0 1 .

Shortcut. f(θ)f(φ) = f(θ + φ), so {f(π)}² = f(2π) = I.

Answer: (d) 5I₃

Method. direct squaring of A

A² =  1 2 2 2 1 2 2 2 1  1 2 2 2 1 2 2 2 1  =  9 8 8 8 9 8 8 8 9 
(diagonal: 1 + 4 + 4 = 9; off-diagonal: 2 + 2 + 4 = 8).
4A =  4 8 8 8 4 8 8 8 4 .
A² - 4A =  5 0 0 0 5 0 0 0 5  = 5I₃.

Answer: (b) 2 and 3 only

Method. powers of a diagonal matrix

Aⁿ =  2ⁿ 0 0 0 3ⁿ 0 0 0 4ⁿ  for every positive integer n.
|Aⁿ| = 2ⁿ·3ⁿ·4ⁿ = 24ⁿ ≠ 0, so Aⁿ is never singular: statement 1 is false.
Aⁿ is diagonal: statement 2 is true.
Every diagonal matrix equals its transpose, so Aⁿ is symmetric: statement 3 is true.
Correct: 2 and 3 only.

Answer: (a) Null matrix of order 3

Method. powers of the identity matrix

A = I₃, so A² = A³ = I₃.
23A³ - 19A² - 4A = (23 − 19 − 4)I₃ = 0·I₃.
This is the null matrix of order 3.

Answer: (a) -2I

Method. powers of  1 a 0 1 : Aⁿ =  1 na 0 1 

A² =  1 a 0 1  1 a 0 1  =  1 2a 0 1 ; by induction Aⁿ =  1 na 0 1 .
A¹⁰⁰ - A⁵⁰ - 2A²⁵ =  1 − 1 − 2 (100 − 50 − 50)a 0 1 − 1 − 2 .
=  -2 0 0 -2  = -2I.

Answer: (d) 32

Method. expand (mI + nA)² using A² = -4I and compare coefficients of I and A

A² =  0 2 -2 0  0 2 -2 0  =  -4 0 0 -4  = -4I.
Since I and A commute, (mI + nA)² = m²I + 2mnA + n²A² = (m² - 4n²)I + 2mnA.
Setting this equal to A: m² - 4n² = 0 and 2mn = 1.
With m, n > 0: m = 2n, so 4n² = 1, n = 12, m = 1.
m + n = 1 + 12 = 32.

Answer: (c) 1 and 3 only

Method. J = all-ones 3 × 3 matrix satisfies J² = 3J

1. All rows of A are equal, so |A| = 0 and A^-1 does not exist: correct.
Each entry of A² is 1.1 + 1.1 + 1.1 = 3, so A² = 3A.
3. 3A = A²: correct.
2. A³ = A·A² = A·3A = 3A² = 9A ≠ A: incorrect.
So 1 and 3 only.

Answer: (a) a null matrix

Method. powers of a matrix satisfying A² = kA

A² =  1 -1 -1 1  1 -1 -1 1  =  2 -2 -2 2  = 2A.
A³ = A·A² = A·2A = 2A² = 4A.
A³ - 2A² = 4A − 2(2A) = 4A − 4A = O, the null matrix.

Shortcut. A³ - 2A² = A²(A − 2I); with A² = 2A this is 2A(A − 2I) = 2(A² - 2A) = O.

Answer: (b) Involutory matrix

Method. involutory means A² = I

A² =  0 1 1 0  0 1 1 0  =  1 0 0 1  = I, so A is involutory.
|A| = 0 − 1 = -1 ≠ 0, so not singular; A² = I, so every power of A is A or I and never O: A is not nilpotent; A² = I ≠ A, so A is not idempotent.

Answer: (a) 4

Method. compute A² and compare entries

A² =  1 2 2 3  1 2 2 3  =  1 + 4 2 + 6 2 + 6 4 + 9  =  5 8 8 13 .
A² - kA − I₂ =  4 − k 8 − 2k 8 − 2k 12 − 3k  = O.
So k = 4 (all entries vanish together).

Shortcut. Cayley–⁠Hamilton: A² - (trace A)A + |A|I = O, i.e. A² - 4A − I = O, so k = 4.

Answer: (b) A² = -4A

Method. compute A² and compare with A

A² =  -2 2 2 -2  -2 2 2 -2  =  4 + 4 -4 − 4 -4 − 4 4 + 4  =  8 -8 -8 8 .
-4A =  8 -8 -8 8 , so A² = -4A.

Answer: (a)  cos 3θ sin 3θ -sin 3θ cos 3θ 

Method. powers of a rotation matrix: angles add

A² =  cos² θ − sin² θ 2 sin θ cos θ -2 sin θ cos θ cos² θ − sin² θ  =  cos 2θ sin 2θ -sin 2θ cos 2θ .
A³ = A²A =  cos 2θ cos θ − sin 2θ sin θ cos 2θ sin θ + sin 2θ cos θ -(sin 2θ cos θ + cos 2θ sin θ) cos 2θ cos θ − sin 2θ sin θ 
=  cos 3θ sin 3θ -sin 3θ cos 3θ .

Shortcut. For this matrix Aⁿ =  cos nθ sin nθ -sin nθ cos nθ ; put n = 3.

Answer: (a)  1 0 0 1 

Method. A² = I, so even powers are I

A² =  0 1 1 0  0 1 1 0  =  0.0 + 1.1 0.1 + 1.0 1.0 + 0.1 1.1 + 0.0  =  1 0 0 1  = I.
A⁴ = (A²)² = I² = I =  1 0 0 1 .

Answer: (b) π/⁠3

Method. add the transpose and compare with I

A^T =  cos θ sin θ -sin θ cos θ .
A + A^T =  2cos θ 0 0 2cos θ .
A + A^T = I ⇒ 2cos θ = 1 ⇒ cos θ = 12.
The least positive θ is π/⁠3.

Answer: (b) II only

Method. transpose of a commutator of symmetric matrices

A^T = A, B^T = B.
(AB − BA)^T = B^TA^T - A^TB^T = BA − AB = -(AB − BA).
So AB − BA is skew-symmetric, not symmetric (unless it is O, i.e. A and B commute). I is false in general.
BA − AB = -(AB − BA) and (BA − AB)^T = AB − BA = -(BA − AB): skew-symmetric. II is true.
E.g. A =  1 0 0 0 , B =  0 1 1 0 : AB − BA =  0 1 -1 0 , which is not symmetric.
Correct: II only.

Answer: (b) 1

Method. columns of an orthogonal matrix are unit vectors

A orthogonal ⇒ A^TA = I₃.
The (1, 1) entry of A^TA is the sum of squares of column 1: y² + z² + x².
So x² + y² + z² = 1.
(With integers this means and then A is a permutation matrix or its negative, which is indeed orthogonal.)

Answer: (b) Identity matrix

Method. a symmetric orthogonal matrix squares to I

A^T =  x y z y z x z x y  = A, so A is symmetric.
A orthogonal ⇒ A^TA = I.
Replacing A^T by A: A² = I, the identity matrix.

Answer: (a) 1 and 2 only

Method. diagonal of a skew-symmetric matrix; odd-order skew-symmetric determinant is zero

A^T = -A gives a_ii = -a_ii, so every diagonal element is 0. Statement 1 is true.
Hence the sum of the diagonal elements is 0. Statement 2 is true.
For order 3, |A| = |A^T| = |-A| = (-1)³|A| = -|A|, so |A| = 0.
An orthogonal matrix has |A| = ±1 (A^TA = I ⇒ |A|² = 1), so A cannot be orthogonal. Statement 3 is false.
Correct: 1 and 2 only.

Answer: (a) A² is symmetric matrix with det(A²) = 0.

Method. square of a skew-symmetric matrix is symmetric; odd-order skew-symmetric matrix has zero determinant

A^T =  0 -3 -4 3 0 -5 4 5 0  = -A, so A is skew-symmetric.
(A²)^T = (A^T)² = (-A)² = A², so A² is symmetric (not skew-symmetric, its diagonal is -25, -34, -41).
|A| = 0(0 + 25) − 3(0 + 20) + 4(15 − 0) = -60 + 60 = 0.
det(A²) = |A|² = 0.
So A² is symmetric with det(A²) = 0.

Shortcut. Odd-order skew-symmetric ⇒ |A| = 0 ⇒ det(A²) = 0; and (A²)^T = A².

Answer: (b) 2 only

Method. for symmetric A, B the matrix AB − BA is skew-symmetric

(AB − BA)^T = B^TA^T - A^TB^T = BA − AB = -(AB − BA), so AB − BA is skew-symmetric.
1. A skew-symmetric matrix has every diagonal entry 0 (a_ii = -a_ii), so its diagonal entries are not nonzero. Incorrect.
2. Its off-diagonal entries pair as a_ij + a_ji = 0, so their sum is zero. Correct.
So 2 only.

Answer: (a)  0 1/⁠2 1/⁠2 1/⁠2 0 1/⁠2 1/⁠2 1/⁠2 0 

Method. symmetric part P = 12(A + A^T)

A^T =  0 cos² θ sin² θ sin² θ 0 cos² θ cos² θ sin² θ 0 .
Every off-diagonal entry of A + A^T is sin² θ + cos² θ = 1, and the diagonal is 0.
P = 12(A + A^T) =  0 1/⁠2 1/⁠2 1/⁠2 0 1/⁠2 1/⁠2 1/⁠2 0 .

Answer: (d) cos 2θ  0 -1/⁠2 1/⁠2 1/⁠2 0 -1/⁠2 -1/⁠2 1/⁠2 0 

Method. skew-symmetric part Q = 12(A − A^T), with cos² θ − sin² θ = cos 2θ

A^T =  0 cos² θ sin² θ sin² θ 0 cos² θ cos² θ sin² θ 0 , so A − A^T has entries:
(1,2): sin² θ − cos² θ = -cos 2θ; (1,3): cos² θ − sin² θ = cos 2θ; (2,3): sin² θ − cos² θ = -cos 2θ, with (j,i) = -(i,j) and zero diagonal.
Q = 12(A − A^T) = cos 2θ  0 -1/⁠2 1/⁠2 1/⁠2 0 -1/⁠2 -1/⁠2 1/⁠2 0 .

Shortcut. Q = A − P: the (1,2) entry is sin² θ − 12 = -12cos 2θ, which matches only (d).

Answer: (a) 14

Method. determinant with zero diagonal = a₁₂a₂₃a₃₁ + a₁₃a₂₁a₃₂; then a³ + b³ with a + b = 1

With zero diagonal, |A| = a₁₂a₂₃a₃₁ + a₁₃a₂₁a₃₂ = sin⁶ θ + cos⁶ θ.
sin⁶ θ + cos⁶ θ = (sin² θ + cos² θ)³ - 3sin² θ cos² θ(sin² θ + cos² θ) = 1 − 3sin² θ cos² θ.
= 1 − 34sin² 2θ.
sin² 2θ ≤ 1 (equality at θ = π/⁠4), so the minimum of |A| is 1 − 34 = 14.

Answer: (c) (A̅)^T + A is hermitian

Method. conjugate transpose: (A̅)^T + A always equals its own conjugate transpose

A̅ =  1 + i -i i 1 + i , so (A̅)^T =  1 + i i -i 1 + i .
(A̅)^T ≠ A (diagonal 1 + i vs 1 − i) and (A̅)^T ≠ -A, so A is neither hermitian nor skew-hermitian.
(A̅)^T + A =  2 2i -2i 2 : real diagonal, and the (2,1) entry -2i is the conjugate of the (1,2) entry 2i.
So (A̅)^T + A is hermitian (and not skew-hermitian, since its diagonal is not purely imaginary).

Shortcut. For any square A, the conjugate transpose of (A̅)^T + A is A + (A̅)^T, the same matrix, so it is always hermitian.

Answer: (b) A′ = A^-1

Method. definition of an orthogonal matrix

A is orthogonal when AA′ = A'A = I.
That says A′ is the inverse of A, i.e. A′ = A^-1.

Answer: (c) 1 and 3 only

Method. transpose of a sum and the reversal law for transposes

1. Transpose distributes over addition: (A + B + C)' = A′ + B′ + C′. True.
2. (AB)' = B'A′, which in general is not A'B′. False.
3. (ABC)' = ((AB)C)' = C′(AB)' = C'B'A′. True.
So 1 and 3 only.

Answer: (b) Identity matrix

Method. product of a rotation-type matrix with its transpose

A^T =  cos α -sin α sin α cos α .
AA^T =  cos² α + sin² α -cos α sin α + sin α cos α -sin α cos α + cos α sin α sin² α + cos² α 
=  1 0 0 1  = I.
So AA^T is the identity matrix.

Shortcut. Rows of A are unit vectors and perpendicular to each other, so A is orthogonal and AA^T = I.

Answer: (c) 2A

Method. adj(adjA) = |A|^n-2A

For a square matrix of order n with |A| ≠ 0, adj(adjA) = |A|^n-2A.
Here n = 3 and |A| = 2: adj(adjA) = 2^3-2A = 2A.

Answer: (d) 64

Method. A(adjA) = |A|I

A(adjA) = |A|I₃ for every square matrix of order 3.
Given A(adjA) =  64 0 0 0 64 0 0 0 64  = 64I₃.
So |A| = 64.
(The trap: |adjA| = |A|² = 4096 would be asked separately; the product A(adjA) carries |A| itself.)

Answer: (c) Both I and II

Method. (A^-1)^T = (A^T)^-1; |adjA| = |A|^n-1

I. If A is symmetric and invertible, (A^-1)^T = (A^T)^-1 = A^-1, so A^-1 is symmetric. True.
II. For an n × n matrix, |adjA| = |A|^n-1. If A is singular, |A| = 0, and since n > 1, n − 1 ≥ 1, so |adjA| = 0^n-1 = 0: adjA is singular. True.
Both I and II are correct.

Answer: (a) None

Method. reversal laws for inverse and transpose

I. (AB)^-1 = B^-1A^-1, which differs from A^-1B^-1 unless A and B commute. False in general.
II. (BA)(AB)^-1 = BAB^-1A^-1, which equals I only when AB = BA. False in general.
III. (AB)^T = B^TA^T, not A^TB^T in general. False.
E.g. A =  1 1 0 1 , B =  1 0 1 1  (both non-singular, AB ≠ BA) break all three.
None is correct.

Answer: (c) Statement-I is correct but Statement-II is not correct

Method. a column equal to the sum of two others makes the determinant zero

Column 1 = Column 2 + Column 3: 71 = 23 + 48, 57 = 28 + 29, 65 = 17 + 48.
C₁ → C₁ - C₂ - C₃ makes column 1 all zeros, so |M| = 0.
M is singular, so M^-1 does not exist: Statement-I is correct.
Statement-II (M is non-singular) is not correct.

Answer: (d)  1 0 0 0 1 0 0 0 1 

Method. A(adjA) = |A|I

|A| = 3[(-3)(1) − (4)(-1)] - (-3)[(2)(1) − (4)(0)] + 4[(2)(-1) − (-3)(0)]
= 3(1) + 3(2) + 4(-2) = 3 + 6 − 8 = 1.
A(adjA) = |A|I₃ = I₃ =  1 0 0 0 1 0 0 0 1 .

Answer: (a)  1 -1 0 -2 3 -4 -2 3 -3 

Method. A^-1 = adjA|A| with |A| = 1

From the previous item |A| = 1, so A^-1 = adjA.
Cofactors: C₁₁ = -3 + 4 = 1, C₁₂ = -(2 − 0) = -2, C₁₃ = -2 − 0 = -2;
C₂₁ = -(-3 + 4) = -1, C₂₂ = 3 − 0 = 3, C₂₃ = -(-3 − 0) = 3;
C₃₁ = -12 + 12 = 0, C₃₂ = -(12 − 8) = -4, C₃₃ = -9 + 6 = -3.
adjA = transpose of the cofactor matrix =  1 -1 0 -2 3 -4 -2 3 -3 .
So A^-1 =  1 -1 0 -2 3 -4 -2 3 -3 .

Shortcut. |A| = 1, so A^-1 = adjA has integer entries, which rules out (b) and (d); of the integral options (a) and (c) = 2(a), only (a) gives A·(a) = I (A·(c) = 2I).

Answer: (b) Only one statement

Method. reversal law adj(AB) = (adjB)(adjA) and A(adjA) = |A|I

For non-singular A, B: adj(AB) = (adjB)(adjA), the order reverses.
1. (adjA)(adjB) ≠ (adjB)(adjA) in general, so statement 1 is false.
2. adj(AB) = (adjB)(adjA) and adj(BA) = (adjA)(adjB); these differ in general (AB ≠ BA in general), so statement 2 is false.
3. For any square matrix M, M(adjM) = |M|I_n; with M = AB, (AB)adj(AB) − |AB|I_n = O. Statement 3 is true.
Only one statement is correct.

Answer: (a) 1 and 2 only

Method. adj(A^T) = (adjA)^T; cancelling a non-singular factor

1. The cofactors of A^T are the cofactors of A transposed, so adj(A^T) = (adjA)^T; hence A(adjA^T) = A(adjA)^T. True.
2. A² = A ⇒ A^-1A² = A^-1A ⇒ A = I_n. True.
3. A³ = A ⇒ A^-1A³ = A^-1A ⇒ A² = I_n, which does not force A = I_n: e.g. A = -I_n is non-singular and (-I_n)³ = -I_n = A. False.
Correct: 1 and 2 only.

Answer: (c) 1 and 3

Method. the matrix is orthogonal with determinant 1

2 sin² θ − 1 = -cos 2θ and 2 sin θ cos θ = sin 2θ, so A =  sin 2θ -cos 2θ 0 cos 2θ sin 2θ 0 0 0 1 .
|A| = 1·(sin² 2θ + cos² 2θ) = 1.
A^TA: columns 1 and 2 have length 1 (sin² 2θ + cos² 2θ = 1) and are perpendicular (-sin 2θ cos 2θ + cos 2θ sin 2θ = 0); column 3 is (0, 0, 1). So A^TA = I and A^-1 = A^T. Statement 3 is true.
A^-1 = adjA|A| = adjA since |A| = 1. Statement 1 is true.
The (3, 3) entry is 1 ≠ 0, so A is not skew-symmetric. Statement 2 is false.
Correct: 1 and 3.

Answer: (d) 1, 2 and 3

Method. A is symmetric with A² = I, so A^-1 = A and adj A = |A|A^-1 = -A

|A| = 1 ×  cos θ sin θ sin θ -cos θ  = -cos² θ − sin² θ = -1.
A²: the lower 2 × 2 block squares to  cos² θ + sin² θ cos θ sin θ − sin θ cos θ sin θ cos θ − cos θ sin θ sin² θ + cos² θ  = I, so A² = I and A^-1 = A.
adj A = |A| A^-1 = -A.
1. A + adj A = A − A = O. 2. A^-1 + adj A = A − A = O. 3. A − A^-1 = A − A = O.
All of 1, 2 and 3 are correct.

Answer: (d) 2I

Method. A(adj A) = |A| I

Expand |A| along column 3 (entries 0, 0, 1):
|A| = 1 ×  2sin θ cos θ -2cos θ sin θ  = 2sin² θ + 2cos² θ = 2.
A(adj A) = |A| I = 2I.

Answer: (b) 1 and 2 only

Method. A(adj A) = (adj A)A = |A| I

For any square A, A(adj A) = (adj A)A = |A| I.
1. AB = A(adj A) = (adj A)A = BA: correct.
2. AB = |A| I is a scalar matrix: correct.
3. A is non-singular, so |A| ≠ 0 and AB = |A| I is never the null matrix: incorrect.
So 1 and 2 only.

Answer: (a)  1 2 3 4 

Method. A = (A^-1)^-1; inverse of a 2 × 2 matrix by swapping and negating

Let M = A^-1 =  -2 1 32 -12 ; then A = M^-1.
|M| = (-2)(-12) − (1)(32) = 1 − 32 = -12.
adj M =  -12 -1 -32 -2 .
A = 1|M| adj M = -2 -12 -1 -32 -2  =  1 2 3 4 .

Shortcut. Check each option against AA^-1 = I: (a) gives  1 2 3 4  -2 1 32 -12  =  1 0 0 1 .

Answer: (a)  0 0 0 0 0 0 -2 -1 8 

Method. adj B = transpose of the cofactor matrix (exists for every square matrix)

Cofactors of B =  3 2 0 2 4 0 1 1 0 :
C₁₁ = 4.0 − 0.1 = 0, C₁₂ = -(2.0 − 0.1) = 0, C₁₃ = 2.1 − 4.1 = -2;
C₂₁ = -(2.0 − 0.1) = 0, C₂₂ = 3.0 − 0.1 = 0, C₂₃ = -(3.1 − 2.1) = -1;
C₃₁ = 2.0 − 0.4 = 0, C₃₂ = -(3.0 − 0.2) = 0, C₃₃ = 3.4 − 2.2 = 8.
Cofactor matrix  0 0 -2 0 0 -1 0 0 8 ; its transpose is adj B =  0 0 0 0 0 0 -2 -1 8 .
(The adjoint exists even though |B| = 0; only the inverse fails to exist.)

Answer: (b) is equal to A

Method. I·I = I, so I^-1 = I

For the identity matrix A = I₃, AA = I₃·I₃ = I₃.
So A is its own inverse: A^-1 = A.

Answer: (a) 1 and 2 only

Method. properties of the inverse; (λA)^-1 = λ^-1A^-1

1. A A^-1 = I shows A is the inverse of A^-1, so (A^-1)^-1 = A. True.
2. det(A)·det(A^-1) = det(I) = 1, so det(A^-1) = 1det A. True.
3. (λA)(1λA^-1) = I, so (λA)^-1 = 1λA^-1, not λA^-1 (e.g. λ = 2 gives 12A^-1). False.
So 1 and 2 only.

Answer: (a)  cos θ -sin θ -sin θ cos θ 

Method. simplify with cos(-θ) = cos θ, sin(-θ) = -sin θ, then adj  a b c d  =  d -b -c a 

cos(-θ) = cos θ and -sin(-θ) = sin θ, so the matrix is  cos θ sin θ sin θ cos θ .
For a 2 × 2 matrix, swap the diagonal entries and change the sign of the off-diagonal ones:
adj =  cos θ -sin θ -sin θ cos θ .

Answer: (a) B^-1A^-1

Method. reversal law for inverses

(AB)(B^-1A^-1) = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I.
So (AB)^-1 = B^-1A^-1.

Answer: (b) kl

Method. A·adj A = |A| I

For any square matrix, A(adj A) = |A| l.
Here B = adj A and |A| = k, so AB = kl.

Answer: (a)  cos θ -sin θ 0 sin θ cos θ 0 0 0 1 

Method. A is orthogonal, so A^-1 = A^T

Rows of A: (cos θ, sin θ, 0), (-sin θ, cos θ, 0), (0, 0, 1) are unit vectors, mutually perpendicular, so AA^T = I.
Hence A^-1 = A^T =  cos θ -sin θ 0 sin θ cos θ 0 0 0 1 .
(Check: |A| = cos² θ + sin² θ = 1, and the (1, 1) entry of AA^T is cos² θ + sin² θ = 1, the (1, 2) entry is -cos θ sin θ + sin θ cos θ = 0.)

Answer: (c) null matrix whose order is same as that of A

Method. adj(A^T) = (adj A)^T

The cofactor of the (i, j) entry of A^T equals the cofactor of the (j, i) entry of A, so the cofactor matrix of A^T is the transpose of the cofactor matrix of A.
Hence adj(A^T) = (adj A)^T, and adj A^T - (adj A)^T = O, the null matrix of the same order as A.

Answer: (a)  a^-1 0 0 0 b^-1 0 0 0 c^-1 

Method. inverse of a diagonal matrix: reciprocal of each diagonal entry

 a 0 0 0 b 0 0 0 c  a^-1 0 0 0 b^-1 0 0 0 c^-1  =  1 0 0 0 1 0 0 0 1  = I (a, b, c ≠ 0).
So A^-1 =  a^-1 0 0 0 b^-1 0 0 0 c^-1 .
(For options b, c, d the product with A is 1abcI, 1abcA and 1abcA², none equal to I in general.)

Answer: (b)  1 6 -2 -2 1 4 6 -3 1 

Method. adj A = transpose of the cofactor matrix

A =  1 0 2 2 1 0 0 3 1 .
Cofactors: C₁₁ = 1.1 − 0.3 = 1, C₁₂ = -(2.1 − 0.0) = -2, C₁₃ = 2.3 − 1.0 = 6;
C₂₁ = -(0.1 − 2.3) = 6, C₂₂ = 1.1 − 2.0 = 1, C₂₃ = -(1.3 − 0.0) = -3;
C₃₁ = 0.0 − 2.1 = -2, C₃₂ = -(1.0 − 2.2) = 4, C₃₃ = 1.1 − 0.2 = 1.
adj A = (cofactor matrix)^T =  1 6 -2 -2 1 4 6 -3 1 .

Shortcut. Check one entry only: the (1, 1) entry of adj A is C₁₁ = 1, and only option (b) has it.

Answer: (c) is inconsistent

Method. compare coefficient ratios of two linear equations

First: 2x − 3y = 5.
Second: -10x + 15y = -50; divide by -5: 2x − 3y = 10.
The left sides are identical but 5 ≠ 10, so no (x, y) satisfies both (parallel lines).
Ratios: 2-10 = -315 = -15 but -550 = -110 ≠ -15.
The system is inconsistent.

Answer: (c) no solution

Method. compare the equations as multiples of the first: same left side, different right side means inconsistency

Equation 2 is 2 × (x + 2y + z = 4), so it adds nothing.
Equation 1 multiplied by 3 gives 3x + 6y + 3z = 12.
Equation 3 says 3x + 6y + 3z = 10, and 12 ≠ 10.
The equations contradict each other, so the system has no solution.

Answer: (b) a + b = 1

Method. linearity: A(aX₁ + bX₂) = aAX₁ + bAX₂

AX₁ = B and AX₂ = B.
A(aX₁ + bX₂) = aAX₁ + bAX₂ = aB + bB = (a + b)B.
For this to equal B we need (a + b)B = B, i.e. (a + b − 1)B = O.
For a non-homogeneous system (B ≠ O) this gives a + b = 1.

The question does not say B ≠ O. If B = O, every aX₁ + bX₂ solves AX = O and nothing is forced on a, b; the question is answerable only for a non-homogeneous system, and then a + b = 1 is the one condition that must hold.

Answer: (b) 21 ± ⁠√16110⁠

Method. three lines a_ix + b_iy + c_i = 0 are concurrent (consistent) when  a_i b_i c_i  = 0

For the system to be consistent the three lines must pass through one point, so 2k² 3 -1 7 -2 3 6k 1 1  = 0.
Expanding along row 1: 2k²(-2 − 3) − 3(7 − 18k) − 1(7 + 12k) = 0.
-10k² - 21 + 54k − 7 − 12k = 0, i.e. 5k² - 21k + 14 = 0.
k = 21 ± ⁠√(441⁠ −⁠ 280)10 = 21 ± ⁠√16110.⁠
(The last two lines meet in a single point for both values, since 6k ≠ -72, so the system is indeed consistent.)

Answer: (b) is consistent, with a unique solution

Method. non-zero coefficient determinant gives a unique solution

 2 1 -3 3 -2 2 5 -3 -1  = 2(2 + 6) − 1(-3 − 10) − 3(-9 + 10) = 16 + 13 − 3 = 26 ≠ 0.
So the system is consistent with a unique solution: x = 1, y = -3, z = -2 (check: 2 − 3 + 6 = 5, 3 + 6 − 4 = 5, 5 + 9 + 2 = 16).
The solution has y ≠ 0, z ≠ 0, so it does not lie on the x-axis; (d) fails.

Answer: (d) -2

Method. coefficient determinant zero, then test consistency

 k 1 1 1 k 1 1 1 k  = k³ - 3k + 2 = (k − 1)²(k + 2), zero for k = 1 or k = -2.
k = 0 or k = -1: determinant 2 or 4, non-zero, so a unique solution.
k = 1: all three equations become x + y + z = 1, infinitely many solutions.
k = -2: -2x + y + z = 1, x − 2y + z = -2, x + y − 2z = 4. Adding all three: 0 = 1 − 2 + 4 = 3, impossible.
So there is no solution when k = -2.

Answer: (a) have the unique solution

Method. non-zero coefficient determinant gives a unique solution

Coefficient determinant:  1 2 3 2 1 3 5 5 9 
= 1(9 − 15) − 2(18 − 15) + 3(10 − 5) = -6 − 6 + 15 = 3 ≠ 0.
So the system has a unique solution (it is x = 0, y = -1, z = 1; check: 0 − 2 + 3 = 1, 0 − 1 + 3 = 2, 0 − 5 + 9 = 4).