NDA Determinants previous year questions with solutions

Answer: (c) 2500

Method. |M_k| = 2k − 1, the sum of the first 50 odd numbers

|M_k| = k² - (k − 1)² = 2k − 1.
50Σk=1 (2k − 1) = 1 + 3 + … + 99 = 50² = 2500.

Answer: (a) 0

Method. expansion of Δ along the first row and along the first column

Along R₁: Δ = aA + bB + cC, so bB + cC = Δ − aA.
Along C₁: Δ = aA + dD + gG, so dD + gG = Δ − aA.
bB + cC − dD − gG = (Δ − aA) − (Δ − aA) = 0.

Answer: (b) -6

Method. expand along the first row using i² = -1

Expand along R₁:
2[0.1 − (i − 1)(-1 − i)] - (3 + i)[(3 − i)·1 − (i − 1)(-1)] + (-1)[(3 − i)(-1 − i) − 0·(-1)].
(i − 1)(-1 − i) = -(i − 1)(i + 1) = -(i² - 1) = 2, so the first term is 2(0 − 2) = -4.
(3 − i) + (i − 1) = 2, so the second term is -(3 + i)(2) = -6 − 2i.
(3 − i)(-1 − i) = -3 − 3i + i + i² = -4 − 2i, so the third term is 4 + 2i.
Sum: -4 − 6 − 2i + 4 + 2i = -6. So A = -6, B = 0 and A + B = -6.

Shortcut. The matrix is Hermitian (a_ji = a̅_i̅j̅), so its determinant is real: B = 0.

Answer: (a) -10000

Method. expand D_n as a polynomial in n, then use Σn, Σn², Σn³

Expand along C₁: D_n = n(40.70 −⁠ 50.60) − n²(20.70 − 30.60) + n³(20.50 − 30.40)
= n(-200) − n²(-400) + n³(-200) = -200n + 400n² - 200n³.
For n = 1 to 4: Σn = 10, Σn² = 30, Σn³ = 100.
Σ D_n = -200(10) + 400(30) − 200(100) = -2000 + 12000 − 20000 = -10000.

Shortcut. D_n = -200n(n − 1)²: D₁, D₂, D₃, D₄ = 0, -400, -2400, -7200; sum -10000.

Answer: (b) an even number but not necessarily 0

Method. the six terms of the expansion are each ±1

A 3 × 3 determinant is a sum of 6 terms, each a product of three entries, so each term is +1 or -1.
A sum of six ±1's is even (it lies in {-6, -4, …, 6}); so the determinant is always even.
It need not be 0:  1 1 1 1 -1 1 1 1 -1  = 1(1 − 1) − 1(-1 − 1) + 1(1 + 1) = 4.
So the value is an even number, not necessarily 0.

Answer: (d) 10000

Method. det(A_k) = 2k − 1, sum of the first 100 odd numbers

det(A_k) = (k − 1)(k + 1) − k(k − 2) = k² - 1 − k² + 2k = 2k − 1.
Sum for k = 1 to 100: 1 + 3 + 5 + … + 199 = 100² = 10000.

Answer: (a) (z − y)(z − x)(y − x)

Method. expansion of Δ along the third column with cofactors C_i3 = (-1)ⁱ⁺³M_i3

Δ = a₁₃C₁₃ + a₂₃C₂₃ + a₃₃C₃₃ = a₁₃M₁₃ - a₂₃M₂₃ + a₃₃M₃₃.
= yz(z − y) − zx(z − x) + xy(y − x).
This cubic vanishes when x = y, y = z or z = x, so Δ = k(x − y)(y − z)(z − x).
Coefficient of yz²: in Δ it is 1 (from yz·z); in (x − y)(y − z)(z − x) it is (-y)(-z)(z), also 1, so k = 1.
Δ = (x − y)(y − z)(z − x) = (z − y)(z − x)(y − x), since (z − y)(y − x) = (y − z)(x − y).

Answer: (c) Δ

Method. expansion of a determinant along a row

a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ is the expansion of Δ along its first row, so it equals Δ.

Answer: (a) 0

Method. elements of one row times cofactors of another row sum to zero

a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ is the expansion along R₁ of the determinant obtained by replacing R₁ with R₂.
That determinant has two equal rows, so the sum is 0.

Answer: (d) -Δ

Method. transpose, then count row and column interchanges

Transpose: value =  a₂₁ a₂₃ a₂₂ a₃₁ a₃₃ a₃₂ a₁₁ a₁₃ a₁₂ .
Interchange C₂ and C₃ (sign changes once):  a₂₁ a₂₂ a₂₃ a₃₁ a₃₂ a₃₃ a₁₁ a₁₂ a₁₃  with a minus sign.
Rows are now R₂, R₃, R₁ of A, a cyclic shift (two interchanges, no sign change), which equals Δ.
Value = -Δ.

Answer: (c) 317

Method. maximise ps − qr: largest possible ps, smallest possible qr

Primes less than 20: 2, 3, 5, 7, 11, 13, 17, 19.
det = ps − qr. Take ps as large as possible: p, s = 19, 17 gives 323.
Take qr as small as possible from the rest: q, r = 2, 3 gives 6.
Maximum = 323 − 6 = 317.

Answer: (a) 0

Method. rows in AP make the determinant zero

Entries 2(i + j) give  4 6 8 6 8 10 8 10 12 .
R₃ - R₂ = (2, 2, 2) = R₂ - R₁.
R₃ → R₃ - R₂, R₂ → R₂ - R₁ gives two equal rows (2, 2, 2), so the determinant is 0.

Answer: (d) 0

Method. interchanging two rows changes the sign, so the determinants cancel in pairs

The 2 × 2 determinants using 2, 4, 6, 8 once each number 4! = 24.
Interchanging the two rows of one of them gives another arrangement in the list, with value of opposite sign.
The 24 determinants therefore fall into 12 pairs, each pair summing to 0, so the total is 0.

Answer: (d) -4i

Method. reduce the powers of i (i² = -1, i⁴ = 1), then expand

i² = -1, i³ = -i, i⁴ = 1, i⁶ = -1, i⁸ = 1, i⁹ = i, i¹² = 1, i¹⁵ = -i.
Determinant =  i -1 -i 1 -1 1 i 1 -i .
Expand along R₁: i[(-1)(-i) − 1 · 1] - (-1)[1 · (-i) − 1 · i] + (-i)[1 · 1 − (-1)i]
= i(i − 1) + (-2i) − i(1 + i)
= (-1 − i) − 2i − (i − 1) = -4i.

Answer: (c) 24

Method. take common factors out of rows, then expand

The determinant is  1 2 6 2 6 24 6 24 120 .
Take 2 out of R₂ and 6 out of R₃: 12 1 2 6 1 3 12 1 4 20 .
R₂ - R₁, R₃ - R₁: 12 1 2 6 0 1 6 0 2 14  = 12(1 · 14 − 6 · 2) = 12 · 2 = 24.

Answer: (b) cos θ

Method. difference of squares and cos² t − sin² t = cos 2t

Value = cos⁴ θ2 − sin⁴ θ2
= (cos² θ2 − sin² θ2)(cos² θ2 + sin² θ2)
= cos θ · 1 = cos θ.

Answer: (b) B has as many elements as C

Method. swapping two rows changes the sign of a determinant

Interchanging R₁ and R₂ turns a 0-1 matrix into another 0-1 matrix and changes the sign of the determinant.
So it sends every matrix of B (det = 1) to a matrix of C(det = -1), and vice versa; doing it twice gives back the original matrix.
This pairing is one-to-one, so B and C have the same number of elements.
(C is not empty:  0 1 0 1 0 0 0 0 1  has det -1; and A ≠ B ∪ C, since the zero matrix has det 0.)

Shortcut. Counting all 512 matrices gives 84 with det 1 and 84 with det -1.

Answer: (b) 0

Method. the three rows add to the zero row

R₁ → R₁ + R₂ + R₃: the first row becomes (a − b + b − c + c − a, …) = (0, 0, 0).
So the left-hand determinant is 0 for all a, b, c, p, q, r, x, y, z.
Since  a b c p q r x y z  is not zero in general, 0 = k × (that determinant) needs k = 0.

Answer: (c)  p q r a b c l m n 

Method. interchanging two rows changes the sign; a cyclic shift of rows is two interchanges

Let Δ have rows R₁ = (a, b, c), R₂ = (l, m, n), R₃ = (p, q, r).
(a) rows R₁, R₃, R₂: one interchange → -Δ.
(b) rows R₂, R₁, R₃: one interchange → -Δ.
(c) rows R₃, R₁, R₂: R₁ ↔ R₃, then R₂ ↔ R₃ — two interchanges → +Δ.
(d) is the transpose of the determinant in (a), so it equals -Δ.
Only (c) equals Δ.

Answer: (a) Both Statement–⁠I and Statement–⁠II are correct and Statement–⁠II explains Statement–⁠I

Method. multiplying one row by m multiplies the determinant by m; mX scales all n rows

Statement–⁠II is a standard property of determinants: correct.
mX is obtained from X by multiplying each of its n rows by m. Applying Statement–⁠II once per row gives det(mX) = m·m·…·m det(X) = mⁿ det(X), so Statement–⁠I is correct.
Statement–⁠I follows by applying Statement–⁠II n times, so Statement–⁠II explains Statement–⁠I.

Answer: (c) 6

Method. Pascal's rule C(n, r) + C(n, r −⁠ 1) = C(n + 1, r) on C₁ + C₂

C₁ + C₂ = (C(10, 4), C(12, 6), C(m + 1, 7)).
If C₃ equals this column the determinant is 0 for every m. With n = 6: C₃ = (C(10, 4), C(12, 6), C(m + 1, 7)). Exactly that.
So for n = 6, C₁ + C₂ - C₃ = 0 and the determinant vanishes for every m > 6.
For n = 4, 5, 7 the third column is not C₁ + C₂ (e.g. C(12, n) ≠ C(12, 6)), and the determinant is non-zero (e.g. at m = n + 1).

No correct option. This question is defective as printed in UPSC's paper.

Method. C₁ - 2C₂ + C₃ reduces the determinant

As printed (row 3 = x + a, x + b, x + 3): C₁ → C₁ - 2C₂ + C₃ gives C₁ = (0, 0, a − 2b + 3).
Δ = (a − 2b + 3) x + 2 x + 3 x + 3 x + 4  = (a − 2b + 3)(-1) = 2b − a − 3.
With 2b = a + c this is c − 3, which depends on c, so no option is always right.
If the (3, 3) entry is x + c: C₁ becomes (0, 0, a − 2b + c) = (0, 0, 0), so Δ = 0, option (b).

The (3, 3) entry is printed x + 3, so c never enters and the value is 2b − a − 3 = c − 3, which no option gives in general; with the evident intended entry x + c the determinant is 0, option (b).

Answer: (c) always negative

Method. transpose Δ₂; its rows are a cyclic shift of Δ₁'s rows

Δ₂ = Δ₂' (transpose) =  1 q r 1 r p 1 p q : its rows are R₂, R₃, R₁ of Δ₁, a cyclic shift (two swaps), so Δ₂ = Δ₁.
Δ₁ = 1(qp − r²) − p(p − r) + q(r − q) = -(p² + q² + r² - pq − qr − rp).
p² + q² + r² - pq − qr − rp = 12[(p − q)² + (q − r)² + (r − p)²] > 0 since p, q, r are not all equal.
Hence Δ₁ + Δ₂ = -[(p − q)² + (q − r)² + (r − p)²] < 0: always negative.

Answer: (d) -72Δ

Method. transpose, then split rows and pull out scalar factors

Transposing does not change a determinant, so the value is  3d + 5g 3e + 5h 3f + 5i 4a + 7g 4b + 7h 4c + 7i 6g 6h 6i .
Take 6 out of R₃: R₃ = (g, h, i). Now R₁ → R₁ - 5R₃ and R₂ → R₂ - 7R₃ remove the g, h, i parts.
Value = 6 3d 3e 3f 4a 4b 4c g h i  = 6 × 3 × 4  d e f a b c g h i .
Swapping R₁ and R₂ gives Δ with a minus sign: value = 72(-Δ) = -72Δ.

Answer: (d) pqΔ

Method. a common factor of a column comes out of the determinant

Take p out of C₁ and q out of C₃: pa₁ b₁ qc₁ pa₂ b₂ qc₂ pa₃ b₃ qc₃  = pq a₁ b₁ c₁ a₂ b₂ c₂ a₃ b₃ c₃  = pqΔ.

Answer: (a) 0

Method. logs of a GP form an AP, so C₃ - C₂ = C₂ - C₁

If the GP has ratio R, ln a_{n + 1} - ln a_n = ln R = d for every n, so ln a₁, ln a₂, …, ln a₉ are in AP.
In each row the entries are t, t + d, t + 2d.
C₃ → C₃ - C₂ and C₂ → C₂ - C₁ make both these columns (d, d, d): two equal columns, so the determinant is 0.

Answer: (a) 0

Method. logs of GP terms are linear in the term number, making C₁ a combination of C₂ and C₃

Let the GP have first term A and ratio R (both positive, as u, v, w > 0).
u = AR^{p − 1}, so ln u = (ln A − ln R) + p ln R; similarly ln v = (ln A − ln R) + q ln R, ln w = (ln A − ln R) + r ln R.
So C₁ = (ln R) C₂ + (ln A − ln R) C₃.
C₁ → C₁ - (ln R) C₂ - (ln A − ln R) C₃ makes C₁ all zeros, so the determinant is 0.

Answer: (a) 0

Method. subtract successive rows, then take a, b, c out of the rows

R₃ → R₃ - R₂, then R₂ → R₂ - R₁:
R₂ = (1/⁠b, c(a + b) − bc, b³) = (1/⁠b, ca, b³); R₃ = (1/⁠c, ab + bc + ca − ca − bc, c³) = (1/⁠c, ab, c³).
Δ =  1/⁠a bc a³ 1/⁠b ca b³ 1/⁠c ab c³ . Multiply R₁, R₂, R₃ by a, b, c and divide by abc:
Δ = 1abc  1 abc a⁴ 1 abc b⁴ 1 abc c⁴  = abcabc  1 1 a⁴ 1 1 b⁴ 1 1 c⁴  = 0 (C₁ = C₂).

Answer: (d) 4

Method. standard determinant  b + c a a b c + a b c c a + b  = 4abc

Put a = r², b = p², c = q²: the determinant is  b + c a a b c + a b c c a + b .
R₁ → R₁ - R₂ - R₃: R₁ = (0, -2c, -2b). Expand along R₁:
Δ = 2c[b(a + b) − bc] - 2b[bc − c(c + a)] = 2bc(a + b − c) + 2bc(c + a − b) = 4abc.
So Δ = 4p²q²r² = 4(pqr)². The cube roots of unity are 1, ω, ω², with product ω³ = 1.
Δ = 4(1)² = 4.

Answer: (c) q + 2 > p

Method. expand the determinant in terms of xyz and x + y + z

Expand along R₁: x(yz − 1) − 1(z − 1) + 1(1 − y) = xyz − x − z + 1 + 1 − y = xyz − (x + y + z) + 2 = q − p + 2.
Determinant > 0 means q − p + 2 > 0, i.e. q + 2 > p. Option (d) q + 2 ≥ p is implied by (c) but is weaker: it also allows q + 2 = p, which the strict positivity of the determinant excludes. (c) is the complete conclusion.

Answer: (b) 2p³

Method. C₁ → C₁ + C₂ + C₃, take out the common factor, then subtract rows

With a + b + c = p, C₁ → C₁ + C₂ + C₃ makes every entry of C₁ equal to p + a + b + c = 2p:
Δ = 2p  1 a b 1 p + a b 1 a p + b .
R₂ → R₂ - R₁, R₃ → R₃ - R₁: Δ = 2p  1 a b 0 p 0 0 0 p  = 2p · p² = 2p³.

Answer: (b) 0

Method. for real A, B, C the condition forces A = B = C = 0

A, B, C are real and A² + B² + C² = 0, so A = B = C = 0 and cos A = cos B = cos C = 1.
The determinant becomes  1 1 1 1 1 1 1 1 1  = 0 (identical rows).

Answer: (b) 0

Method. tan(B + C) = tan(π − A) = -tan A, then expand

In a triangle, tan(B + C) = -tan A.
Expand along R₁: cos C[0·cos C − sin B tan(B + C)] - sin B[tan A cos C − 0] + 0
= cos C sin B tan A − sin B tan A cos C = 0.

Answer: (a) Zero

Method. expand, then the cosine rule a² = b² + c² - 2bc cos A

Expand along R₁: a²(1 − cos² A) − b sin A(b sin A − c sin A cos A) + c sin A(b sin A cos A − c sin A)
= sin² A(a² - b² + bc cos A + bc cos A − c²) = sin² A(a² - b² - c² + 2bc cos A).
By the cosine rule the bracket is 0, so the determinant is zero.

Answer: (d) 0

Method. sine rule turns the condition into a³ + b³ + c³ = 3abc; circulant determinant

By the sine rule sin A = a/⁠(2R), sin B = b/⁠(2R), sin C = c/⁠(2R). Multiplying the condition by (2R)³: a³ + b³ + c³ = 3abc.
Expand the determinant along R₁: a(cb − a²) − b(b² - ac) + c(ab − c²) = 3abc − a³ - b³ - c³.
So the determinant = 3abc − (a³ + b³ + c³) = 3abc − 3abc = 0.

Answer: (b) 12

Method. take abc out, then the Vandermonde determinant

Take a, b, c out of C₁, C₂, C₃: value = abc 1 1 1 a b c a² b² c² .
The Vandermonde determinant  1 1 1 a b c a² b² c²  = (a − b)(b − c)(c − a).
Value = 6 × 2 = 12.

Answer: (d) 1, 2 and 3

Method. circulant determinant = -(a + b + c)(a² + b² + c² - ab − bc − ca)

C₁ → C₁ + C₂ + C₃ puts a + b + c in every entry of C₁; taking it out and expanding gives
Δ = -(a + b + c)(a² + b² + c² - ab − bc − ca) = -(a³ + b³ + c³ - 3abc).
1: a + b + c = 0 makes the first factor zero, so Δ = 0.
3: makes the second factor zero, so Δ = 0.
2: a³ + b³ + c³ = 3abc is exactly Δ = 0.
All three conditions make the determinant vanish.

Answer: (a) -4

Method. take common factors out of columns, then expand

Take a from C₁ and b from C₂: Δ = ab 1 -1 a − b − c -1 1 -a + b − c -1 -1 -a − b + c .
R₂ → R₂ + R₁, R₃ → R₃ + R₁: Δ = ab 1 -1 a − b − c 0 0 -2c 0 -2 -2b .
Expand along C₁: Δ = ab[0·(-2b) − (-2c)(-2)] = ab(-4c) = -4abc.
So -4abc − kabc = 0 with abc ≠ 0, giving k = -4.

Answer: (b) 0

Method. direct expansion along the first row

Expand along R₁: cos C[0·cos C − (-tan A)(sin B)] - tan A[sin B cos C − (-tan A)·0] + 0
= cos C tan A sin B − tan A sin B cos C
= 0.

Answer: (a) 0

Method. expand, then use xy + yz + zx = -1 for these x, y, z

Expand along R₁: 1(1 + yz) + x(1 + y) + x(z − 1) = 1 + xy + yz + zx.
xy = ab(b − c)(c − a), yz = bc(c − a)(a − b), zx = ca(a − b)(b − c).
Over the common denominator (a − b)(b − c)(c − a): numerator = ab(a − b) + bc(b − c) + ca(c − a) = -(a − b)(b − c)(c − a).
So xy + yz + zx = -1 and the determinant = 1 − 1 = 0.

Answer: (b) -64

Method. circulant determinant = -(a³ + b³ + c³ - 3abc)

Expand: value = a(cb − a²) − b(b² - ac) + c(ab − c²) = -(a³ + b³ + c³ - 3abc).
a³ + b³ + c³ - 3abc = (a + b + c)[(a + b + c)² - 3(ab + bc + ca)] = 4(16 − 0) = 64.
So the value is -64.

Answer: (a) D < 0

Method. circulant determinant = -(p³ + q³ + r³ - 3pqr)

Expand: D = p(rq − p²) − q(q² - pr) + r(pq − r²) = 3pqr − p³ - q³ - r³.
So D = -(p + q + r)(p² + q² + r² - pq − qr − rp) = -12(p + q + r)[(p − q)² + (q − r)² + (r − p)²].
p + q + r > 0, and the bracket is > 0 because p, q, r are distinct.
Hence D < 0.

Answer: (a) 0

Method. C₁ + C₂ + C₃ gives a common factor x + a + b + c

C₁ → C₁ + C₂ + C₃: every entry of C₁ becomes x + a + b + c.
So the determinant = (x + a + b + c) 1 b c 1 x + b c 1 b x + c  = 0, since x + a + b + c = 0.

Answer: (a) x − 3

Method. factor theorem: a value of x that makes two columns equal gives a factor

C₃ = (3, 9, 27) = (3, 3², 3³) and C₁ = (x, x², x³).
Put x = 3: C₁ becomes (3, 9, 27) = C₃, so the determinant is 0 for every y.
As a polynomial in x the expansion vanishes at x = 3, so (x − 3) is a factor.
The others are not factors: x = 2, y = 3 makes y − 3 = 0 and x − y ≠ 0 yet gives -10638 ≠ 0; x = 6, y = 2 makes x − 3y = 0 yet gives -216 ≠ 0; and at x = y the value is not identically 0.

Answer: (d) (z − x)² (x + y + z)

Method. expand along R₁ and compare with the expanded options

Expand along R₁:
x[x(x + y) − z(z + x)] - y[z(x + y) − y(z + x)] + (y + z)[z² - xy]
= x(x² + xy − z² - zx) − y(zx − xy) + (y + z)(z² - xy)
= x³ + x²y − xz² - x²z − xyz + xy² + yz² - xy² + z³ - xyz
= x³ + z³ + x²y − x²z − xz² + yz² - 2xyz.
(z − x)²(x + y + z) = (z² - 2zx + x²)(x + y + z) = x³ + z³ + x²y − x²z − xz² + yz² - 2xyz — the same.
So the determinant is (z − x)²(x + y + z).

Shortcut. Test x = 1, y = z = 0: the determinant is  1 0 0 0 1 1 0 0 1  = 1; only (d) gives 1 (a, b, c give 0).

Answer: (b) (α − β)(β − γ)(γ − α)

Method. column operations reduce it to the Vandermonde determinant

C₁ → C₁ + C₂ + C₃: each row gives (1 − α) + (α − α²) + α² = 1.
C₂ → C₂ + C₃: each row gives α − α² + α² = α.
The determinant becomes  1 α α² 1 β β² 1 γ γ² .
R₂ - R₁, R₃ - R₁ and take out (β − α), (γ − α): (β − α)(γ − α) 1 β + α 1 γ + α  = (β − α)(γ − α)(γ − β).
(β − α)(γ − β) = (α − β)(β − γ), so the value is (α − β)(β − γ)(γ − α).

Answer: (a) 0

Method. expand, then use x³ + y³ + z³ = 3xyz when x + y + z = 0

Expand along R₁:
pa(qa · ra − pb · pc) − qb(qc · qa − pb · rb) + rc(qc · pc − ra · rb)
= pqr a³ - p³abc − q³abc + pqr b³ + pqr c³ - r³abc
= pqr(a³ + b³ + c³) − abc(p³ + q³ + r³).
Since a + b + c = 0, a³ + b³ + c³ = 3abc; since p + q + r = 0, p³ + q³ + r³ = 3pqr.
Value = pqr · 3abc − abc · 3pqr = 0.

Answer: (c) x²y²z²

Method. subtract the first row from the others to make the determinant triangular

R₂ → R₂ - R₁, R₃ → R₃ - R₁: 1 1 1 0 xyz 0 0 0 xyz .
This is upper triangular, so its value is the product of the diagonal entries: 1 · xyz · xyz = x²y²z².

Answer: (b) One

Method. row operations reduce Δ to (k − 1)³

R₁ → R₁ - R₂, R₂ → R₂ - R₃:
Δ =  k² - 1 k − 1 0 2k − 2 k − 1 0 3 3 1 .
Expand along C₃: Δ = (k² - 1)(k − 1) − (k − 1)(2k − 2) = (k − 1)[(k + 1)(k − 1) − 2(k − 1)] = (k − 1)²(k − 1) = (k − 1)³.
I. k = 1/⁠2 > 0 gives Δ = -1/⁠8 < 0: not correct.
II. k < 0 gives k − 1 < 0, so Δ < 0: correct.
III. k = 0 gives Δ = -1 ≠ 0: not correct.
One statement is correct.

Answer: (a) x = 0

Method. C₁ → C₁ + C₂ + C₃ with 1 + ω + ω² = 0

C₁ → C₁ + C₂ + C₃: every entry of C₁ becomes x + 1 + ω + ω² = x. Take x out:
Δ = x  1 ω ω² 1 x + ω² 1 1 1 x + ω .
R₂ → R₂ - R₁, R₃ → R₃ - R₁ and expand along C₁:
Δ = x[(x + ω² - ω)(x + ω − ω²) − (1 − ω)(1 − ω²)] = x[x² - (ω² - ω)² - (1 − ω − ω² + ω³)].
(ω² - ω)² = ω⁴ - 2ω³ + ω² = ω + ω² - 2 = -3 and 1 − ω − ω² + 1 = 3.
Δ = x[x² + 3 − 3] = x³ = 0, so x = 0.

Shortcut. Put x = 0: every row sums to 1 + ω + ω² = 0, so the determinant vanishes; x = 0 is a root.

Answer: (b) 0

Method. the constant term is the value at x = 0

Put x = 0: e =  0 -1 3 1 0 -4 -3 4 0 .
= 0 − (-1)(0 − 12) + 3(4 − 0) = -12 + 12 = 0.
(The matrix is skew-symmetric of order 3, so this had to be 0.)

Answer: (b) a − b + c

Method. expand along the first column, then sum the roots

C₁ has only one non-zero entry, x + b in position (3, 1), with sign (+).
Δ = (x + b) x − a x − b 0 x − c  = (x + b)(x − a)(x − c).
Roots: -b, a, c. Sum = a − b + c.

Answer: (b) a, b, c are in GP

Method. C₃ → C₃ - αC₁ - C₂ factors the determinant

C₃ → C₃ - αC₁ - C₂: rows 1 and 2 of C₃ become 0; row 3 becomes -(aα² + 2bα + c).
Expand along C₃: ∆ = -(aα² + 2bα + c)(ac − b²).
A quadratic in α cannot vanish for every α > 0 unless a = b = c = 0, so ∆ = 0 for every α > 0 forces ac − b² = 0.
b² = ac: a, b, c are in GP.

Answer: (c) 7x² + 8x + 2 = 0

Method. use the factorisation ∆ = -(aα² + 2bα + c)(ac − b²)

As in the previous item, ∆(a, b, c, α) = -(aα² + 2bα + c)(ac − b²).
For a = 7, b = 4, c = 2: ac − b² = 14 − 16 = -2 ≠ 0.
So ∆ = 2(7α² + 8α + 2) = 0 gives 7α² + 8α + 2 = 0: α is a root of 7x² + 8x + 2 = 0.

Answer: (b) 0

Method. put x = 0 directly

At x = 0: R₂ = (2 sin 0, 0, 0) = (0, 0, 0).
A row of zeros gives f(0) = 0.

Answer: (b) 0

Method. simplify f(x) = x²(tan x − cos x)

Expand along R₁: f(x) = cos x(x² - 2x²) − x(2 sin x − 2x tan x) + 1(2x sin x − x² tan x)
= -x² cos x − 2x sin x + 2x² tan x + 2x sin x − x² tan x = x²(tan x − cos x).
f(x)x = x(tan x − cos x) → 0 × (0 − 1) = 0 as x → 0.

Answer: (a) -1

Method. simplify f(x) = x²(tan x − cos x)

Expand along R₁: f(x) = cos x(x² - 2x²) − x(2 sin x − 2x tan x) + 1(2x sin x − x² tan x)
= -x² cos x − 2x sin x + 2x² tan x + 2x sin x − x² tan x = x²(tan x − cos x).
f(x)x² = tan x − cos x → 0 − 1 = -1 as x → 0.

Answer: (d) Any real value

Method. two proportional rows

1 − 2sin² θ = cos 2θ and 2cos² θ − 1 = cos 2θ, so R₂ = (cos 2θ, cos 2θ, 3).
R₁ = (2cos 2θ, 2cos 2θ, 6) = 2R₂.
Two proportional rows make the determinant 0 whatever k is, so the matrix is singular for any real k.

Answer: (c) -1 or 1

Method. expand along the row with two zeros

Expand along R₂ = (0, 0, 1): the only term is from the (2, 3) entry, with sign (-1)^{2 + 3} = -1.
Value = -1 ·  x 1 1 x  = -(x² - 1) = 1 − x².
1 − x² = 0 gives x = -1 or 1.

Answer: (a) 0

Method. at x = -1, 0, 1 a whole row or column of the determinant vanishes

f(0): R₂ = (0, 0, 0), so f(0) = 0.
f(1): R₃ = (3 · 1 · 0, 2 · 0 · (-1), 1 · 2 · 0) = (0, 0, 0), so f(1) = 0.
f(-1): C₃ = (-1 + 1, (-1)(0), (-1)(0)(-2)) = (0, 0, 0), so f(-1) = 0.
f(-1) + f(0) + f(1) = 0.

Answer: (d) Infinite

Method. a matrix is singular when its determinant is 0; here the determinant vanishes identically

Expand along R₁:
0 · (0 · (-1) − (-5)k) − k[(-k)(-1) − (-5)(-k)] + 4[(-k)k − 0 · (-k)]
= -k(k − 5k) + 4(-k²) = 4k² - 4k² = 0.
The determinant is 0 for every k, so the matrix is singular for infinitely many values of k.

No correct option. This question is defective as printed in UPSC's paper.

Method. C₁ + C₂ + C₃ brings out the factor (a + b + c − x)

C₁ → C₁ + C₂ + C₃: every entry of C₁ becomes a + b + c − x = -x.
So the determinant = -x  1 c b 1 b − x a 1 a c − x .
R₂ - R₁, R₃ - R₁ and expanding: -x[(b − c − x)(c − b − x) − (a − c)(a − b)] = -x[x² - (a² + b² + c² - ab − bc − ca)].
So x = 0 is a root, and x² = a² + b² + c² - (ab + bc + ca).
From (a + b + c)² = 0: ab + bc + ca = -a² + b² + c²2, so x² = 3(a² + b² + c²)2.
Roots: x = 0 and x = ± ⁠√(3(a²⁠ + b² + c²)2); both option (b) and option (d) are solutions.

Both option (b), x = ⁠√(3(a²⁠ + b² + c²)2), and option (d), x = 0, satisfy the equation, so the item has two correct options and no single printable answer.

Answer: (b) -16

Method. a square matrix has no inverse exactly when its determinant is 0

det = 2 · x − 4 · (-8) = 2x + 32.
No inverse ⇔ 2x + 32 = 0 ⇔ x = -16.

Answer: (d) 0

Method. test the options; at x = 0 the determinant is skew-symmetric of odd order

Put x = 0:  0 -a -b a 0 -c b c 0 .
This is skew-symmetric (entry (j, i) = - entry (i, j)) and of odd order 3, so its value is 0.
Direct check: 0 − (-a)(0 + bc) + (-b)(ac − 0) = abc − abc = 0.
So x = 0 satisfies the equation.

Answer: (d) xy is one of the cube roots of -1

Method. expand, then factor x³ + y³ = 0

Expand along R₁: x(x · x − y · 0) − y(0 · x − y · y) + 0 = x³ + y³.
So x³ + y³ = 0, i.e. x³ = -y³.
y ≠ 0 here (if y = 0 then x = 0 too and xy is meaningless), so (xy)³ = -1.
Hence xy is one of the cube roots of -1.

Answer: (b) 16

Method. |adj A| = |A|^{n − 1} for order n

For order 3, |adj A| = |A|² = 4² = 16.

Answer: (c) -1/⁠2

Method. |A^-1| = 1/⁠|A|

|A| = (-1)(-4) − (2)(3) = 4 − 6 = -2.
|A^-1| = 1/⁠|A| = -1/⁠2.

Answer: (d) p² - q²

Method. product of determinants = determinant of the product matrix

 a b c d   e f g h  =  ae + bg af + bh ce + dg cf + dh  =  p q q p .
Since |X||Y| = |XY|, the product =  p q q p  = p² - q².

Answer: (d) 1

Method. |AB^-1| = |A|/⁠|B|

|A| = 1.1 − (-tan θ)(tan θ) = 1 + tan² θ = sec² θ.
|B| = 1.1 − (tan θ)(-tan θ) = sec² θ.
det(C) = |A| |B^-1| = |A|/⁠|B| = sec² θ / sec² θ = 1.

Answer: (d) 4

Method. |AA^T| = |A||A^T| = |A|²

|AA^T| = |A| |A^T| = |A| |A| = (-2)² = 4.

Answer: (d) 512

Method. |adj M| = |M|^{n − 1} for order n

M = 2I₃, so |M| = 2³ = 8.
|adj M| = |M|^{3 − 1} = 8² = 64.
|M| |adj M| = 8 × 64 = 512.

Answer: (b) II only

Method. |2M| = 2²|M| for a 2 × 2 matrix; M + M^T = O for anti-symmetric M

I. M symmetric: M + M^T = 2M, so |M + M^T| = |2M| = 4|M|, while |M| + |M^T| = 2|M|. These agree only when |M| = 0; e.g. M = I gives 4 ≠ 2. Not correct.
II. M anti-symmetric: M^T = -M, so M + M^T = O and |O| = 0. Correct.
II only.

Answer: (c) Three

Method. take determinants: |M|³ = |M|

|M³| = |M|³, so |M|³ = |M|, i.e. |M|(|M| − 1)(|M| + 1) = 0, giving |M| = 0, 1 or -1.
Each occurs: M = O gives 0, M = I gives 1, M =  1 0 0 -1  gives -1 (each satisfies M³ = M).
Three values are possible.

Answer: (a) 12

Method. |A^-1| = 1/⁠|A|

|A| = (-4)(2) − (-5)(2) = -8 + 10 = 2.
|A^-1| = 1/⁠|A| = 1/⁠2.

Answer: (b) 1

Method. |Aⁿ| = |A|ⁿ

|A| = cos² θ + sin² θ = 1.
|A⁴| = |A|⁴ = 1.

Answer: (c) A

Method. adj A = |A| A^-1; A is orthogonal with |A| = 1

For  p q r s , adj =  s -q -r p , so adj A =  cos θ -sin θ sin θ cos θ  = A^T.
A A^T =  cos² θ + sin² θ 0 0 sin² θ + cos² θ  = I, so (A^T)^-1 = A.
[adj A]^-1 = (A^T)^-1 = A.

Answer: (c) Two

Method. |AB| = |A||B|, |A^-1| = 1/⁠|A|, |A^T| = |A|

I. |M²| = |M·M| = |M||M| = |M|²: correct.
II. MM^-1 = I gives |M||M^-1| = 1, so |M^-1| = 1/⁠|M|, which equals |M| only when |M| = ±1: not correct in general.
III. A determinant is unchanged by transposition, |M^T| = |M|: correct.
Two statements are correct.

Answer: (b) 0

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

P^T = -P, so |P| = |P^T| = |-P| = (-1)³|P| = -|P|.
Hence 2|P| = 0, i.e. det(P) = 0.

Answer: (d) 1

Method. |kM| = k³|M|, |adj M| = |M|² for order 3

|3A| = 27 × 12⁠√2,⁠ so |adj(3A)| = |3A|² = 7298.
|2B adj(3A)| = 2³|B||adj(3A)| = 8 × 1729 × 7298 = 1.

Answer: (a) 15

Method. form I + AA′ and evaluate

AA′ =  1 2 3 2 4 6 3 6 9 , so I + AA′ =  2 2 3 2 5 6 3 6 10 .
det = 2(50 − 36) − 2(20 − 18) + 3(12 − 15) = 28 − 4 − 9 = 15.

Shortcut. det(I + AA′) = 1 + A'A = 1 + (1 + 4 + 9) = 15 for a column vector A.

Answer: (a) 16

Method. det(kM) = k³det M, det(A^-1) = 1/⁠det A, det is multiplicative

det(2A^-1BC) = 2³ det(A^-1) det(BC) = 8 × 1det A × 2 det A = 16.

Answer: (a) -1

Method. a skew-symmetric matrix of odd order has determinant 0

det A = det A′ = det(-A) = (-1)³ det A, so det A = 0.
Hence det(A²), det(A³), det(A⁴) = 0, and so det(4A⁴) = det(3A³) = det(2A²) = det A = 0.
det(-I) = (-1)³ = -1.
Value = 0 − 0 + 0 − 0 − 1 = -1.

Answer: (d) 625

Method. |adj(adj A)| = |A|^(n-1)2 for order n

|A| = 2(3 − 0) − (-1)(-1 − 0) + 0 = 6 − 1 = 5.
|adj A| = |A|² = 25, and |adj(adj A)| = |adj A|² = 625.

Answer: (c) 81

Method. cofactor matrix C = (adj A)', so |C| = |adj A| = |A|²

|C| = |adj A| = |A|^3-1 = 3² = 9.
|C²| = |C|² = 81.

Answer: (b) 13

Method. |kA| = kⁿ|A| and |adj B| = |B|^n-1 for order n = 3

|3A| = 3³·4 = 2²·3³.
|adj(3A)| = |3A|² = 2⁴·3⁶.
|2 adj(3A)| = 2³·2⁴·3⁶ = 2⁷·3⁶.
α = 7, β = 6, so α + β = 13.

Answer: (d) 8

Method. det(kI) = kⁿ for order n

B = I′ = I, so C = I + I = 2I of order 3.
det C = 2³ det I = 8.

Answer: (d) A and B need not be unit matrices.

Method. det (AB) = det A · det B = det (BA) for all square matrices of the same order

det (AB) = det A · det B = det B · det A = det (BA) holds for every pair of 2 × 2 matrices.
So the given condition puts no restriction on A and B; they need not be unit matrices (e.g. A = 2I, B = 3I).

Answer: (a) 1 only

Method. A(adj A) = (adj A)A = |A| I, and |adj A| = |A|^{n − 1}

Statement 1: A(adj A) = (adj A)A = |A| I for every square matrix, so it is correct.
Statement 2: for order 3, |adj A| = |A|^{3 − 1} = |A|², which equals |A| only when |A| = 1 (A non-singular, so |A| ≠ 0).
E.g. A = 2I: |A| = 8, adj A = 4I, |adj A| = 64 ≠ 8. So statement 2 is not correct in general.
Only statement 1 is correct.

Answer: (d) 4

Method. det (A^T) = det A

A determinant is unchanged when rows and columns are interchanged, so det (A^T) = det A = 4.

Answer: (b) det (- A) = (- 1)ⁿ det A

Method. taking a factor k out of each of n rows: det (kA) = kⁿ det A

-A is A with every one of its n rows multiplied by -1.
Each row contributes a factor -1, so det (-A) = (-1)ⁿ det A.

Answer: (a) 96

Method. det (kA) = k³ det A for order 3, and det (AB) = det A · det B

det (2AB) = 2³ det (AB) = 8 · det A · det B = 8 · 4 · 3 = 96.

Answer: (c) 36

Method. det (kA) = k³ det A for order 3, and det (B^-1) = 1det B

det (3AB^-1) = 3³ · det A · det (B^-1) = 27 · 4 · 13 = 36.

Answer: (c) ± 3

Method. det of a power: det (A³) = (det A)³

det A = α² - 4.
det (A³) = (det A)³ = (α² - 4)³ = 125 = 5³.
So α² - 4 = 5 (the only real cube root of 125), α² = 9, α = ± 3.

Answer: (b) det (A)

Method. product rule det (PQ) = det P · det Q

det (B^-1AB) = det (B^-1) · det A · det B.
det (B^-1) = 1det B (B is non-singular, so det B ≠ 0).
So det (B^-1AB) = 1det B · det A · det B = det (A).

Answer: (b) 1

Method. x = 1, y = z = 0 satisfies all three equations, and the solution is unique

The coefficient determinant is the Vandermonde determinant  1 a a² 1 b b² 1 c c²  = (a − b)(b − c)(c − a) ≠ 0 for distinct a, b, c, so the system has exactly one solution.
x = 1, y = 0, z = 0 satisfies each equation (1 + 0 + 0 = 1).
So x + y + z = 1.

Answer: (b) -3

Method. Cramer's rule: D₂ is D with its second column replaced by the constants

D₂ =  1 7 1 1 16 3 1 22 4 .
Expand along R₁: 1(64 − 66) − 7(4 − 3) + 1(22 − 16) = -2 − 7 + 6 = -3.
Check: D = 1(8 − 9) − 1(4 − 3) + 1(3 − 2) = -1, and the system gives x = 1, y = 3, z = 3; y = D₂/⁠D = -3/⁠(-1) = 3.

The stem says y is obtained by 'dividing D by D₂'; by Cramer's rule it is D₂ divided by D. D₂ as the Cramer determinant is -3 (option b); a literal reading y = D/⁠D₂ would give D₂ = -1/⁠3, which is not an option.