Miscellaneous Exercise answers: Determinants

Open NCERT p. 99One way to think about it

1. Expand the determinant along the first row.
2. x[(−x)(x) − (1)(1)] − sinθ[(−sinθ)(x) − (1)(cosθ)] + cosθ[(−sinθ)(1) − (−x)(cosθ)]
3. = x(−x² − 1) − sinθ(−x sinθ − cosθ) + cosθ(−sinθ + x cosθ)
4. = −x³ − x + x sin²θ + sinθ cosθ − sinθ cosθ + x cos²θ
5. = −x³ − x + x(sin²θ + cos²θ) = −x³ − x + x = −x³.
6. The θ terms cancel out, leaving −x³, which has no θ in it — so the value does not depend on θ.

In shortThe determinant equals −x³ for every θ, so it is independent of θ.

Watch this explained “Four steps along the top row”, 6:38 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 99Matches NCERT’s answer

1. The third column is −sin α, 0, cos α. Its middle entry is 0, so expand along the third column: only two terms survive.
2. Entry (1,3) = −sin α, sign (+). Its minor: |−sin β, cos β; sin α cos β, sin α sin β| = −sin α sin²β − sin α cos²β = −sin α(sin²β + cos²β) = −sin α.
3. Entry (3,3) = cos α, sign (+). Its minor: |cos α cos β, cos α sin β; −sin β, cos β| = cos α cos²β + cos α sin²β = cos α(cos²β + sin²β) = cos α.
4. Determinant = (−sin α)(−sin α) + 0 + (cos α)(cos α) = sin²α + cos²α = 1.

AnswerThe determinant equals 1.

Watch this explained “Counting the zeros first”, 16:29 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 99Matches NCERT’s answer

1. Use the reversal rule (AB)⁻¹ = B⁻¹A⁻¹. A⁻¹ is given, so only B⁻¹ is needed.
2. |B| = 1[(3)(1) − (0)(−2)] − 2[(−1)(1) − (0)(0)] + (−2)[(−1)(−2) − (3)(0)] = 3 + 2 − 4 = 1. Since |B| ≠ 0, B⁻¹ exists.
3. Cofactors of B: C11 = 3, C12 = 1, C13 = 2, C21 = 2, C22 = 1, C23 = 2, C31 = 6, C32 = 2, C33 = 5.
4. adj B = transpose of the cofactor matrix = [[3,2,6],[1,1,2],[2,2,5]], and B⁻¹ = adj B / |B| = [[3,2,6],[1,1,2],[2,2,5]].
5. (AB)⁻¹ = B⁻¹A⁻¹ = [[3,2,6],[1,1,2],[2,2,5]] × [[3,−1,1],[−15,6,−5],[5,−2,2]].
6. Row 1: [9 − 30 + 30, −3 + 12 − 12, 3 − 10 + 12] = [9, −3, 5]. Row 2: [3 − 15 + 10, −1 + 6 − 4, 1 − 5 + 4] = [−2, 1, 0]. Row 3: [6 − 30 + 25, −2 + 12 − 10, 2 − 10 + 10] = [1, 0, 2].

Answer(AB)⁻¹ = [[9,−3,5],[−2,1,0],[1,0,2]]

Watch this explained “Why the reversal reverses”, 13:10 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 99One way to think about it

(i) [adj A]⁻¹ = adj (A⁻¹)

1. |A| = 1[(3)(5) − (1)(1)] − 2[(2)(5) − (1)(1)] + 1[(2)(1) − (3)(1)] = 14 − 18 − 1 = −5.
2. Cofactors of A: C11 = 14, C12 = −9, C13 = −1, C21 = −9, C22 = 4, C23 = 1, C31 = −1, C32 = 1, C33 = −1. The cofactor matrix is symmetric, so adj A = [[14,−9,−1],[−9,4,1],[−1,1,−1]].
3. A⁻¹ = adj A / |A| = (−1/5) × [[14,−9,−1],[−9,4,1],[−1,1,−1]].
4. Left side: |adj A| = 14[(4)(−1) − (1)(1)] − (−9)[(−9)(−1) − (1)(−1)] + (−1)[(−9)(1) − (4)(−1)] = −70 + 90 + 5 = 25.
5. Cofactors of adj A give adj(adj A) = [[−5,−10,−5],[−10,−15,−5],[−5,−5,−25]], so [adj A]⁻¹ = (1/25) × that = [[−1/5,−2/5,−1/5],[−2/5,−3/5,−1/5],[−1/5,−1/5,−1]].
6. Right side: A⁻¹ is adj A with every entry multiplied by −1/5. Each cofactor of a 3×3 matrix is a 2×2 determinant, so each cofactor gets multiplied by (−1/5)² = 1/25. Hence adj(A⁻¹) = (1/25) × adj(adj A) = [[−1/5,−2/5,−1/5],[−2/5,−3/5,−1/5],[−1/5,−1/5,−1]].
7. Both sides are the same matrix, so [adj A]⁻¹ = adj(A⁻¹).

In shortVerified: [adj A]⁻¹ and adj(A⁻¹) both equal [[−1/5,−2/5,−1/5],[−2/5,−3/5,−1/5],[−1/5,−1/5,−1]].

(ii) (A⁻¹)⁻¹ = A

1. |A⁻¹| = (−1/5)³ × |adj A| = (−1/125) × 25 = −1/5, which is not 0, so A⁻¹ has an inverse.
2. (A⁻¹)⁻¹ = adj(A⁻¹) / |A⁻¹| = (−5) × [[−1/5,−2/5,−1/5],[−2/5,−3/5,−1/5],[−1/5,−1/5,−1]] (using adj(A⁻¹) from part (i)).
3. = [[1,2,1],[2,3,1],[1,1,5]], which is A.

In shortVerified: (A⁻¹)⁻¹ = [[1,2,1],[2,3,1],[1,1,5]] = A.

Watch this explained “One inverted properly”, 11:58 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 99Matches NCERT’s answer

1. Expand along the first row (entries x, y, x + y with signs +, −, +).
2. First term: x × |x + y, x; x, y| = x[(x + y)y − x²] = x²y + xy² − x³.
3. Second term: −y × |y, x; x + y, y| = −y[y² − x(x + y)] = −y³ + x²y + xy².
4. Third term: (x + y) × |y, x + y; x + y, x| = (x + y)[xy − (x + y)²] = (x + y)(−x² − xy − y²) = −x³ − 2x²y − 2xy² − y³.
5. Add: (x²y + xy² − x³) + (−y³ + x²y + xy²) + (−x³ − 2x²y − 2xy² − y³) = −2x³ − 2y³.

AnswerThe determinant equals −2x³ − 2y³, that is −2(x³ + y³).

Watch this explained “Four steps along the top row”, 6:38 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 99Matches NCERT’s answer

1. Expand along the first row (entries 1, x, y with signs +, −, +).
2. First term: 1 × |x + y, y; x, x + y| = (x + y)² − xy = x² + xy + y².
3. Second term: −x × |1, y; 1, x + y| = −x[(x + y) − y] = −x².
4. Third term: y × |1, x + y; 1, x| = y[x − (x + y)] = −y².
5. Add: (x² + xy + y²) − x² − y² = xy.

AnswerThe determinant equals xy.

Watch this explained “Four steps along the top row”, 6:38 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 100Matches NCERT’s answer

1. Put u = 1/x, v = 1/y, w = 1/z. The equations become 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2.
2. Write this as AX = B with A = [[2,3,10],[4,−6,5],[6,9,−20]], X = [[u],[v],[w]], B = [[4],[1],[2]].
3. |A| = 2[(−6)(−20) − (5)(9)] − 3[(4)(−20) − (5)(6)] + 10[(4)(9) − (−6)(6)] = 2(75) − 3(−110) + 10(72) = 1200. Since |A| ≠ 0, A⁻¹ exists.
4. Cofactors of A: C11 = 75, C12 = 110, C13 = 72, C21 = 150, C22 = −100, C23 = 0, C31 = 75, C32 = 30, C33 = −24.
5. adj A is the transpose of the cofactor matrix: adj A = [[75,150,75],[110,−100,30],[72,0,−24]].
6. A⁻¹ = (1/|A|) adj A = (1/1200) × [[75,150,75],[110,−100,30],[72,0,−24]].
7. X = A⁻¹B = (1/1200) × [[75,150,75],[110,−100,30],[72,0,−24]] × [[4],[1],[2]] = (1/1200) × [[600],[400],[240]] = [[1/2],[1/3],[1/5]].
8. So u = 1/2, v = 1/3, w = 1/5, and x = 1/u = 2, y = 1/v = 3, z = 1/w = 5. Check in the first equation: 2/2 + 3/3 + 10/5 = 4 ✓.

Answerx = 2, y = 3, z = 5

Watch this explained “Unknowns underneath”, 16:36 into Packing three equations into AX = B and reading the solution off the inverse

Open NCERT p. 100Matches NCERT’s answer

1. |A| = x × (yz − 0) − 0 + 0 = xyz, which is not 0 because x, y, z are nonzero, so A⁻¹ exists.
2. Cofactors: C11 = yz, C22 = xz, C33 = xy, and every other cofactor is 0 (each of their 2×2 minors has a row or column of zeros). So adj A = [[yz,0,0],[0,xz,0],[0,0,xy]].
3. A⁻¹ = adj A / |A| = [[yz/xyz,0,0],[0,xz/xyz,0],[0,0,xy/xyz]] = [[x⁻¹,0,0],[0,y⁻¹,0],[0,0,z⁻¹]].
4. Check: A × A⁻¹ has diagonal x·x⁻¹ = 1, y·y⁻¹ = 1, z·z⁻¹ = 1 and zeros elsewhere, so it is I. This is option (A). Options (B), (C), (D) differ from it (for example, (D) times A gives diagonal x/xyz = 1/(yz), not 1).

Answer(A) A⁻¹ = [[x⁻¹,0,0],[0,y⁻¹,0],[0,0,z⁻¹]]

Watch this explained “One inverted properly”, 11:58 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 100Matches NCERT’s answer

1. Expand along the first row: Det(A) = 1 × |1, sin θ; −sin θ, 1| − sin θ × |−sin θ, sin θ; −1, 1| + 1 × |−sin θ, 1; −1, −sin θ|.
2. First minor: 1 − (sin θ)(−sin θ) = 1 + sin²θ.
3. Second minor: (−sin θ)(1) − (sin θ)(−1) = −sin θ + sin θ = 0.
4. Third minor: (−sin θ)(−sin θ) − (1)(−1) = sin²θ + 1.
5. So Det(A) = (1 + sin²θ) − 0 + (sin²θ + 1) = 2 + 2sin²θ.
6. For 0 ≤ θ ≤ 2π, sin²θ takes every value from 0 to 1: it is 0 at θ = 0 and 1 at θ = π/2. So 2 ≤ Det(A) ≤ 4, and both 2 and 4 are reached.
7. Hence Det(A) ∈ [2, 4], which is option (D).

Answer(D) Det(A) ∈ [2, 4]

Watch this explained “Four steps along the top row”, 6:38 into Orders one, two and three, and expanding along a chosen row