Class 12th Determinants NCERT Solutions Chapter 4 Misc

Concept used. Expand the determinant along R₁ and simplify; if all θ-dependent terms cancel, the result is independent of θ. Use the Pythagorean identity sin²θ + cos²θ = 1.

1. Expand along R₁: Δ = x vmatrix -x & 1 1 & x vmatrix - sinθ vmatrix -sinθ & 1 cosθ & x vmatrix + cosθ vmatrix -sinθ & -x cosθ & 1 vmatrix.
2. Evaluate each minor. vmatrix -x & 1 1 & x vmatrix = -x· x - 1· 1 = -x² - 1. vmatrix -sinθ & 1 cosθ & x vmatrix = -xsinθ - cosθ. vmatrix -sinθ & -x cosθ & 1 vmatrix = -sinθ - (-xcosθ) = -sinθ + xcosθ.
3. Substitute into the expansion: Δ = x(-x²-1) - sinθ(-xsinθ - cosθ) + cosθ(-sinθ + xcosθ).
4. Expand term by term: x(-x²-1) = -x³ - x. -sinθ(-xsinθ - cosθ) = xsin²θ + sinθ. cosθ(-sinθ + xcosθ) = -sinθ + xcos²θ.
5. Add the three pieces: Δ = -x³ - x + xsin²θ + sinθ - sinθ + xcos²θ. The two sinθ terms cancel: Δ = -x³ - x + x(sin²θ + cos²θ).
6. Use sin²θ + cos²θ = 1: Δ = -x³ - x + x· 1 = -x³.
7. The right side has no θ, so Δ is independent of θ.

Δ = -x³, independent of θ.

Concept used. Expand along the row or column with the most zeros. Here R₂ has a zero. Use sin²+cos²=1 to collapse the result.

1. Expand along R₂ (third entry is zero): Δ = -(-sinβ)vmatrix cosβ & -sinα sinβ & cosα vmatrix + cosβvmatrix cosβ & -sinα sinβ & cosα vmatrix - 0. Simplify the first sign: -(-sinβ) = sinβ.
2. First minor: vmatrix cosβ & -sinα sinβ & cosα vmatrix = (cosβ)(cosα) - (-sinα)(sinβ). = cos²α sinβ + sin²α sinβ = sinβ(cos²α + sin²α) = sinβ.
3. Second minor: vmatrix cosβ & -sinα sinβ & cosα vmatrix = (cosβ)(cosα) - (-sinα)(sinβ). = cos²α cosβ + sin²α cosβ = cosβ(cos²α + sin²α) = cosβ.
4. Substitute back: Δ = sinββ + cosββ = sin²β + cos²β = 1.

Δ = 1.

Concept used. (AB)^-1 = B^-1A^-1. We must first compute B^-1, then multiply B^-1 with the given A^-1.

1. Compute |B| along R₁: |B| = 1(3+0) - 2(-1-0) + (-2)(2-0) = 3 + 2 - 4 = 1.
2. Compute cofactors of B. Row 1: B₁₁ = +(3-0) = 3, B₁₂ = -(-1-0) = 1, B₁₃ = +(2-0) = 2. Row 2: B₂₁ = -(2-4) = 2, B₂₂ = +(1-0) = 1, B₂₃ = -(-2-0) = 2. Row 3: B₃₁ = +(0+6) = 6, B₃₂ = -(0-2) = 2, B₃₃ = +(3+2) = 5.
3. Adjoint of B = transpose of cofactor matrix: adj(B) = pmatrix 3 & 2 & 6 1 & 1 & 2 2 & 2 & 5 pmatrix.
4. B^-1 = 1|B|adj(B) = adj(B) = pmatrix 3 & 2 & 6 1 & 1 & 2 2 & 2 & 5 pmatrix.
5. Compute (AB)^-1 = B^-1A^-1: B^-1A^-1 = pmatrix 3 & 2 & 6 1 & 1 & 2 2 & 2 & 5 pmatrixpmatrix 3 & -1 & 1 -15 & 6 & -5 5 & -2 & 2 pmatrix. Row 1: 3(3)+2(-15)+6(5) = 9-30+30 = 9;  3(-1)+2(6)+6(-2) = -3+12-12 = -3;  3(1)+2(-5)+6(2) = 3-10+12 = 5. Row 2: 1(3)+1(-15)+2(5) = 3-15+10 = -2;  1(-1)+1(6)+2(-2) = -1+6-4 = 1;  1(1)+1(-5)+2(2) = 1-5+4 = 0. Row 3: 2(3)+2(-15)+5(5) = 6-30+25 = 1;  2(-1)+2(6)+5(-2) = -2+12-10 = 0;  2(1)+2(-5)+5(2) = 2-10+10 = 2. (AB)^-1 = pmatrix 9 & -3 & 5 -2 & 1 & 0 1 & 0 & 2 pmatrix.

(AB)^-1 = pmatrix 9 & -3 & 5 -2 & 1 & 0 1 & 0 & 2 pmatrix.

Concept used. Two identities.
(I) [adjA]^-1 = adj(A^-1) when |A| ≠ 0.
(II) (A^-1)^-1 = A for every invertible A, by the uniqueness of inverses.

Part (i). The identity follows from A adj(A) = |A| I. Dividing by |A|: A·adj(A)|A| = I, so A^-1=adj(A)|A|. Then adj(A^-1) = |A^-1| (A^-1)^-1 = 1|A| A = A|A|. Also, [adj(A)]^-1 = 1|adj(A)|adj(adj(A)). For a 3× 3 matrix one shows adj(adjA) = |A| A (a standard identity), and |adjA| = |A|². So [adj(A)]^-1 = |A|· A|A|² = A|A| = adj(A^-1).

We now verify by computation.

1. Compute |A| along R₁: |A| = 1(15-1) - (-2)(-10-1) + 1(-2-3) = 14 - 22 - 5 = -13.
2. Compute adj(A). Cofactors. Row 1: A₁₁=+(15-1)=14, A₁₂=-(-10-1)=11, A₁₃=+(-2-3)=-5. Row 2: A₂₁=-(-10-1)=11, A₂₂=+(5-1)=4, A₂₃=-(1+2)=-3. Row 3: A₃₁=+(-2-3)=-5, A₃₂=-(1+2)=-3, A₃₃=+(3-4)=-1.
3. Adjoint: adj(A) = pmatrix 14 & 11 & -5 11 & 4 & -3 -5 & -3 & -1 pmatrix.
4. A^-1 = 1-13adj(A) = -113pmatrix 14 & 11 & -5 11 & 4 & -3 -5 & -3 & -1 pmatrix.
5. adj(A^-1) = A|A| = A-13 = -113A = -113pmatrix 1 & -2 & 1 -2 & 3 & 1 1 & 1 & 5 pmatrix.
6. Meanwhile [adj(A)]^-1: since |adj(A)|=|A|²=169 and adj(adjA) = |A| A = -13 A, [adj(A)]^-1 = -13 A169 = -A13 = -113A.
7. Comparing steps 5 and 6: [adj(A)]^-1 = -113A = adj(A^-1).

Part (ii). For any invertible A, A· A^-1 = I and A^-1· A = I. Reading the second equation, A is a matrix whose product with A^-1 on the right gives I; that means A is the inverse of A^-1, i.e. (A^-1)^-1=A.

Both (i) and (ii) verified.

Concept used. Use the column operation C₁ → C₁ + C₂ + C₃ (which does not change the determinant). When all three entries in C₁ become the common quantity 2(x+y), pull it out and reduce.

1. Apply C₁ → C₁ + C₂ + C₃. New C₁ entries: x + y + (x+y) = 2(x+y), y + (x+y) + x = 2(x+y), (x+y) + x + y = 2(x+y). So Δ = vmatrix 2(x+y) & y & x+y 2(x+y) & x+y & x 2(x+y) & x & y vmatrix.
2. Pull 2(x+y) out of C₁: Δ = 2(x+y)vmatrix 1 & y & x+y 1 & x+y & x 1 & x & y vmatrix.
3. Apply R₂ → R₂ - R₁ and R₃ → R₃ - R₁ to introduce zeros in C₁: Δ = 2(x+y)vmatrix 1 & y & x+y 0 & x & -y 0 & x-y & -x vmatrix.
4. Expand along C₁ (only the top entry survives): Δ = 2(x+y)· 1·vmatrix x & -y x-y & -x vmatrix.
5. Compute the 2× 2 minor: vmatrix x & -y x-y & -x vmatrix = (x)(-x) - (-y)(x-y) = -x² + y(x-y) = -x² + xy - y².
6. Multiply: Δ = 2(x+y)(-x² + xy - y²) = -2(x+y)(x² - xy + y²).
7. Recognise (x+y)(x²-xy+y²) = x³ + y³: Δ = -2(x³ + y³).

Δ = -2(x³ + y³).

Concept used. Row operations preserve the determinant. Subtract the first row from the others to introduce zeros.

1. Apply R₂ → R₂ - R₁ and R₃ → R₃ - R₁: Δ = vmatrix 1 & x & y 0 & y & 0 0 & 0 & x vmatrix. Row 2: (1-1, (x+y)-x, y-y) = (0, y, 0).   Row 3: (1-1, x-x, (x+y)-y) = (0, 0, x).
2. Expand along C₁ (only a₁₁=1 is non-zero): Δ = 1·vmatrix y & 0 0 & x vmatrix = y· x - 0· 0 = xy.

Δ = xy.

Concept used. Introduce u = 1/x, v = 1/y, w = 1/z. The system becomes linear in u,v,w; solve by the matrix method, then invert.

1. Substitute: 2u + 3v + 10w = 4, 4u - 6v + 5w = 1, 6u + 9v - 20w = 2.
2. Write in matrix form A U = B, with U = (u,v,w)^T: A = pmatrix 2 & 3 & 10 4 & -6 & 5 6 & 9 & -20 pmatrix, B = pmatrix 4 1 2 pmatrix.
3. Compute |A| along R₁: |A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36) = 2(75) - 3(-110) + 10(72). = 150 + 330 + 720 = 1200.
4. Cofactors. Row 1: A₁₁ = +(120 - 45) = 75, A₁₂ = -(-80-30) = 110, A₁₃ = +(36+36) = 72. Row 2: A₂₁ = -(-60-90) = 150, A₂₂ = +(-40-60) = -100, A₂₃ = -(18-18) = 0. Row 3: A₃₁ = +(15+60) = 75, A₃₂ = -(10-40) = 30, A₃₃ = +(-12-12) = -24.
5. Adjoint = transpose of cofactor matrix: adj(A) = pmatrix 75 & 150 & 75 110 & -100 & 30 72 & 0 & -24 pmatrix.
6. Compute adj(A)B: row 1: 75(4) + 150(1) + 75(2) = 300 + 150 + 150 = 600. row 2: 110(4) + (-100)(1) + 30(2) = 440 - 100 + 60 = 400. row 3: 72(4) + 0(1) + (-24)(2) = 288 + 0 - 48 = 240.
7. Divide by |A|=1200: U = 11200pmatrix 600 400 240 pmatrix = pmatrix 1/2 1/3 1/5 pmatrix.
8. Recover original variables: u = 1/x = 1/2 ⇒ x = 2, v = 1/y = 1/3 ⇒ y = 3, w = 1/z = 1/5 ⇒ z = 5.

x = 2, y = 3, z = 5.

Concept used. The inverse of a diagonal matrix with non-zero diagonal entries is the diagonal matrix of reciprocals: diag(a,b,c)^-1 = diag(1/a,1/b,1/c).

1. A = diag(x,y,z), |A| = xyz ≠ 0, so A^-1 exists.
2. Build A^-1 directly: A^-1 = pmatrix 1/x & 0 & 0 0 & 1/y & 0 0 & 0 & 1/z pmatrix = pmatrix x^-1 & 0 & 0 0 & y^-1 & 0 0 & 0 & z^-1 pmatrix.
3. Verify: A A^-1 = diag(x· 1/x, y· 1/y, z· 1/z) = I.
4. This matches option (A).

Option (A).

Concept used. Expand the determinant; obtain det(A) as a function of θ; determine its range over 0θ≤ 2π.

1. Let s = sinθ for brevity. Expand along R₁: det(A) = 1vmatrix 1 & s -s & 1 vmatrix - svmatrix -s & s -1 & 1 vmatrix + 1vmatrix -s & 1 -1 & -s vmatrix.
2. Evaluate each minor. vmatrix 1 & s -s & 1 vmatrix = 1 - (-s²) = 1 + s². vmatrix -s & s -1 & 1 vmatrix = -s - (-s) = -s + s = 0. vmatrix -s & 1 -1 & -s vmatrix = (-s)(-s) - 1·(-1) = s² + 1.
3. Substitute: det(A) = (1 + s²) - s· 0 + (s² + 1) = 2 + 2s².
4. Range of s = sinθ over 0θ≤ 2π is [-1,1], so s²∈[0,1].
5. Hence det(A) = 2 + 2s² ∈ [2 + 0, 2 + 2] = [2,4].
6. Match: option (D).

Option (D): det(A)∈[2,4].