Class 12th Determinants NCERT Solutions Chapter 4 Ex 4.1

Concept used. For a 2× 2 matrix A=pmatrix a & b c & d pmatrix, the determinant is defined as |A| = vmatrix a & b c & d vmatrix = ad - bc. This is the difference of the products of the two diagonals: the principal diagonal a· d minus the anti-diagonal b· c.

1. Identify the entries with a=2, b=4, c=-5, d=-1.
2. Substitute in ad-bc: vmatrix 2 & 4 -5 & -1 vmatrix = (2)(-1) - (4)(-5).
3. Evaluate each product separately: (2)(-1) = -2, (4)(-5) = -20.
4. Combine, taking care of the sign: -2 - (-20) = -2 + 20 = 18.

vmatrix 2 & 4 -5 & -1 vmatrix = 18.

Concept used. For a 2× 2 matrix the determinant is ad-bc. We also recall the Pythagorean identity sin²θ + cos²θ = 1 for part (i) and the difference-of-squares a²-b²=(a-b)(a+b) for part (ii).

Part (i).

1. Identify a=cosθ, b=-sinθ, c=sinθ, d=cosθ.
2. Apply ad-bc: vmatrix cosθ & -sinθ sinθ & cosθ vmatrix = (cosθ)(cosθ) - (-sinθ)(sinθ).
3. Simplify the two products: = cos²θ + sin²θ.
4. Use sin²θ + cos²θ = 1: = 1.

Part (ii).

1. Here a=x²-x+1, b=x-1, c=x+1, d=x+1.
2. Apply ad-bc: vmatrix x²-x+1 & x-1 x+1 & x+1 vmatrix = (x²-x+1)(x+1) - (x-1)(x+1).
3. Expand the first product. Multiply (x²-x+1) by (x+1) term by term: aligned (x²-x+1)(x+1) &= x²· x + x²· 1 - x· x - x· 1 + 1· x + 1· 1
   &= x³ + x² - x² - x + x + 1
   &= x³ + 1. aligned (This is the sum-of-cubes factorisation x³+1=(x+1)(x²-x+1) read backwards.)
4. Expand the second product using a²-b²=(a-b)(a+b): (x-1)(x+1) = x² - 1.
5. Subtract: (x³+1) - (x²-1) = x³ + 1 - x² + 1 = x³ - x² + 2.

Concept used. Two facts.
(a) For a 2× 2 matrix A=pmatrix a & b c & d pmatrix, |A|=ad-bc.
(b) If every entry of an n× n matrix is multiplied by a scalar k, the determinant is multiplied by kⁿ. That is, |kA| = kⁿ |A|. Here n=2, so we expect |2A| = 2² |A| = 4 |A|.

1. Compute |A| directly: |A| = vmatrix 1 & 2 4 & 2 vmatrix = (1)(2) - (2)(4) = 2 - 8 = -6.
2. Form 2A by doubling every entry: 2A = pmatrix 2· 1 & 2· 2 2· 4 & 2· 2 pmatrix = pmatrix 2 & 4 8 & 4 pmatrix.
3. Compute |2A| directly: |2A| = vmatrix 2 & 4 8 & 4 vmatrix = (2)(4) - (4)(8) = 8 - 32 = -24.
4. Compute 4 |A|: 4 |A| = 4 × (-6) = -24.
5. Compare: |2A| = -24 = 4 |A|. The required equality holds.

|2A| = -24 = 4 |A|.

Concept used. For an n× n matrix and any scalar k, |kA| = kⁿ |A|. Here n=3, so |3A| = 3³ |A| = 27 |A|. We also use the fact that the determinant of an upper-triangular matrix is the product of its diagonal entries (since cofactor expansion along the column with the zeros collapses to a single term repeatedly).

1. Inspect A. The entries below the diagonal are all zero, so A is upper-triangular. Its determinant is the product of diagonal entries: |A| = 1 × 1 × 4 = 4. (One can verify by expanding along column 1, which has only one non-zero entry a₁₁=1.)
2. Form 3A by multiplying every entry by 3: 3A = pmatrix 3 & 0 & 3 0 & 3 & 6 0 & 0 & 12 pmatrix. This is also upper-triangular.
3. Compute |3A| as a product of diagonal entries: |3A| = 3 × 3 × 12 = 108.
4. Compute 27 |A|: 27 |A| = 27 × 4 = 108.
5. Compare: |3A| = 108 = 27 |A|. The required equality holds.

|3A| = 108 = 27 |A|.

Concept used. For a 3× 3 matrix A=pmatrix a₁₁ & a₁₂ & a₁₃ a₂₁ & a₂₂ & a₂₃ a₃₁ & a₃₂ & a₃₃ pmatrix, the determinant expanded along the first row is |A| = a₁₁ vmatrix a₂₂ & a₂₃ a₃₂ & a₃₃ vmatrix - a₁₂ vmatrix a₂₁ & a₂₃ a₃₁ & a₃₃ vmatrix + a₁₃ vmatrix a₂₁ & a₂₂ a₃₁ & a₃₂ vmatrix. The signs follow the chessboard pattern + - +
- + -
+ - +. Expansion along any row or column gives the same value; choosing the row/column with the most zeros saves work.

Part (i). Two entries of R₂ are zero, so expand along R₂.

1. Along R₂, |A| = -a₂₁ M₂₁ + a₂₂ M₂₂ - a₂₃ M₂₃ where M_ij is the 2× 2 minor.
2. With a₂₁=0, a₂₂=0, a₂₃=-1, only the last term survives: |A| = -(-1) M₂₃ = M₂₃ = vmatrix 3 & -1 3 & -5 vmatrix.
3. Compute the minor: vmatrix 3 & -1 3 & -5 vmatrix = (3)(-5) - (-1)(3) = -15 + 3 = -12.

Part (ii). Expand along R₁.

1. Write the expansion: |A| = 3 vmatrix 1 & -2 3 & 1 vmatrix - (-4) vmatrix 1 & -2 2 & 1 vmatrix + 5 vmatrix 1 & 1 2 & 3 vmatrix.
2. Each minor: vmatrix 1 & -2 3 & 1 vmatrix = (1)(1)-(-2)(3) = 1+6 = 7. vmatrix 1 & -2 2 & 1 vmatrix = (1)(1)-(-2)(2) = 1+4 = 5. vmatrix 1 & 1 2 & 3 vmatrix = (1)(3)-(1)(2) = 3-2 = 1.
3. Combine: |A| = 3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46.

Part (iii). The corner entry a₁₁=0. Expand along R₁.

1. Expand: |A| = 0· M₁₁ - 1 vmatrix -1 & -3 -2 & 0 vmatrix + 2 vmatrix -1 & 0 -2 & 3 vmatrix.
2. First minor: vmatrix -1 & -3 -2 & 0 vmatrix = (-1)(0) - (-3)(-2) = 0 - 6 = -6.
3. Second minor: vmatrix -1 & 0 -2 & 3 vmatrix = (-1)(3) - (0)(-2) = -3 - 0 = -3.
4. Combine: |A| = 0 - 1×(-6) + 2×(-3) = 6 - 6 = 0.

Part (iv). Expand along R₁.

1. Expand: |A| = 2 vmatrix 2 & -1 -5 & 0 vmatrix - (-1) vmatrix 0 & -1 3 & 0 vmatrix + (-2) vmatrix 0 & 2 3 & -5 vmatrix.
2. First minor: vmatrix 2 & -1 -5 & 0 vmatrix = (2)(0) - (-1)(-5) = 0 - 5 = -5.
3. Second minor: vmatrix 0 & -1 3 & 0 vmatrix = (0)(0) - (-1)(3) = 0 + 3 = 3.
4. Third minor: vmatrix 0 & 2 3 & -5 vmatrix = (0)(-5) - (2)(3) = 0 - 6 = -6.
5. Combine: |A| = 2(-5) + 1(3) + (-2)(-6) = -10 + 3 + 12 = 5.

Concept used. For a 3× 3 matrix, the determinant by expansion along the first row is |A| = a₁₁(a₂₂a₃₃-a₂₃a₃₂) - a₁₂(a₂₁a₃₃-a₂₃a₃₁) + a₁₃(a₂₁a₃₂-a₂₂a₃₁).

1. Read off the entries: a₁₁=1, a₁₂=1, a₁₃=-2, a₂₁=2, a₂₂=1, a₂₃=-3, a₃₁=5, a₃₂=4, a₃₃=-9.
2. First minor (cofactor of a₁₁): M₁₁ = vmatrix 1 & -3 4 & -9 vmatrix = (1)(-9) - (-3)(4) = -9 + 12 = 3.
3. Second minor: M₁₂ = vmatrix 2 & -3 5 & -9 vmatrix = (2)(-9) - (-3)(5) = -18 + 15 = -3.
4. Third minor: M₁₃ = vmatrix 2 & 1 5 & 4 vmatrix = (2)(4) - (1)(5) = 8 - 5 = 3.
5. Combine with the alternating signs: |A| = 1· M₁₁ - 1· M₁₂ + (-2)· M₁₃. |A| = (1)(3) - (1)(-3) + (-2)(3). |A| = 3 + 3 - 6 = 0.

|A| = 0.

Concept used. A 2× 2 determinant equals ad - bc. Setting the two determinants equal gives a polynomial equation in x that we then solve.

Part (i).

1. Compute the LHS: vmatrix 2 & 4 5 & 1 vmatrix = (2)(1) - (4)(5) = 2 - 20 = -18.
2. Compute the RHS in terms of x: vmatrix 2x & 4 6 & x vmatrix = (2x)(x) - (4)(6) = 2x² - 24.
3. Equate and solve: -18 = 2x² - 24. Add 24 to both sides: 6 = 2x². Divide by 2: x² = 3. Take the square root (both signs): x = ±√3.

Part (ii).

1. LHS: vmatrix 2 & 3 4 & 5 vmatrix = (2)(5) - (3)(4) = 10 - 12 = -2.
2. RHS: vmatrix x & 3 2x & 5 vmatrix = (x)(5) - (3)(2x) = 5x - 6x = -x.
3. Equate: -2 = -x x = 2.

Concept used. A 2× 2 determinant equals ad - bc. Once both sides are reduced to numbers (or to a polynomial in x), the equation is straightforward.

1. Evaluate the RHS: vmatrix 6 & 2 18 & 6 vmatrix = (6)(6) - (2)(18) = 36 - 36 = 0.
2. Evaluate the LHS as a polynomial in x: vmatrix x & 2 18 & x vmatrix = (x)(x) - (2)(18) = x² - 36.
3. Set LHS = RHS: x² - 36 = 0. Factorise as a difference of squares: (x-6)(x+6) = 0. Hence x = 6 or x = -6, i.e. x = ± 6.
4. Match with options: option (B) ± 6 is correct.

Option (B): x = ± 6.