Exercise 4.4 answers: Determinants

Open NCERT p. 92Matches NCERT’s answer

1. For a 2×2 matrix [[a,b],[c,d]], swap the diagonal entries and change the sign of the other two.
2. Here a=1, b=2, c=3, d=4.
3. Swap the diagonal: 4 goes to the top-left, 1 goes to the bottom-right.
4. Change sign of the other two: 2 becomes −2, 3 becomes −3.
5. Adjoint = [[4, −2], [−3, 1]].

Answer[[4, −2], [−3, 1]]

Watch this explained “The two-loop picture”, 4:31 into Building the adjoint, and why it multiplies back to a scalar times the identity

Open NCERT p. 92Matches NCERT’s answer

1. Find the cofactor of every entry of the 3×3 matrix.
2. Top row: A11 = |3 5; 0 1| = 3. A12 = −|2 5; −2 1| = −12. A13 = |2 3; −2 0| = 6.
3. Middle row: A21 = −|−1 2; 0 1| = 1. A22 = |1 2; −2 1| = 5. A23 = −|1 −1; −2 0| = 2.
4. Bottom row: A31 = |−1 2; 3 5| = −11. A32 = −|1 2; 2 5| = −1. A33 = |1 −1; 2 3| = 5.
5. This gives the cofactor matrix [[3,−12,6],[1,5,2],[−11,−1,5]].
6. The adjoint is the TRANSPOSE of the cofactor matrix (rows become columns): [[3,1,−11],[−12,5,−1],[6,2,5]].

Answer[[3, 1, −11], [−12, 5, −1], [6, 2, 5]]

Watch this explained “The construction, in two moves”, 0:47 into Building the adjoint, and why it multiplies back to a scalar times the identity

Open NCERT p. 92One way to think about it

1. |A| = 2 × (−6) − 3 × (−4) = −12 + 12 = 0.
2. For a 2×2 matrix, adj A swaps the two diagonal entries and changes the sign of the other two: adj A = [[−6, −3], [4, 2]].
3. A(adj A) = [[2 × (−6) + 3 × 4, 2 × (−3) + 3 × 2], [(−4) × (−6) + (−6) × 4, (−4) × (−3) + (−6) × 2]] = [[0, 0], [0, 0]].
4. (adj A)A = [[(−6) × 2 + (−3) × (−4), (−6) × 3 + (−3) × (−6)], [4 × 2 + 2 × (−4), 4 × 3 + 2 × (−6)]] = [[0, 0], [0, 0]].
5. |A| I = 0 × [[1, 0], [0, 1]] = [[0, 0], [0, 0]].
6. All three are the zero matrix, so A(adj A) = (adj A)A = |A| I.

In short|A| = 0, adj A = [[−6, −3], [4, 2]], and A(adj A) = (adj A)A = |A| I = [[0, 0], [0, 0]], so the result is verified.

Watch this explained “The case that earns the theorem”, 13:12 into Building the adjoint, and why it multiplies back to a scalar times the identity

Open NCERT p. 92One way to think about it

1. Find |A| by expanding along the first row: |A| = 1(0×3 − (−2)×0) − (−1)(3×3 − (−2)×1) + 2(3×0 − 0×1) = 0 + 11 + 0 = 11.
2. Find each cofactor of A: C11 = 0, C12 = −11, C13 = 0, C21 = 3, C22 = 1, C23 = −1, C31 = 2, C32 = 8, C33 = 3.
3. Write the cofactor matrix and transpose it to get adj A = [0, 3, 2; −11, 1, 8; 0, −1, 3].
4. Multiply A (adj A). Every diagonal entry works out to 11 and every off-diagonal entry works out to 0, so A (adj A) = [11, 0, 0; 0, 11, 0; 0, 0, 11].
5. Multiply (adj A) A the other way round — the same array of 11's and 0's comes out again, so (adj A) A = A (adj A).
6. |A| I is 11 times the identity matrix, which is exactly [11, 0, 0; 0, 11, 0; 0, 0, 11] — the same array both products gave.

In shortA (adj A) = (adj A) A = |A| I is verified: all three equal [11, 0, 0; 0, 11, 0; 0, 0, 11].

Watch this explained “An unrigged verification”, 14:12 into Building the adjoint, and why it multiplies back to a scalar times the identity

Open NCERT p. 92Matches NCERT’s answer

1. Find |A| = 2×3 − (−2)×4 = 6 + 8 = 14. Since |A| ≠ 0, A⁻¹ exists.
2. Find the cofactors: C11 = 3, C12 = −4, C21 = 2, C22 = 2.
3. Write the cofactor matrix [3, −4; 2, 2] and transpose it to get adj A = [3, 2; −4, 2].
4. A⁻¹ = (1/|A|) × adj A = (1/14) × [3, 2; −4, 2] = [3/14, 1/7; −2/7, 1/7].

AnswerA⁻¹ = [3/14, 1/7; −2/7, 1/7]

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. 92Matches NCERT’s answer

1. Find |A| = (−1)×2 − 5×(−3) = −2 + 15 = 13. Since |A| ≠ 0, A⁻¹ exists.
2. Find the cofactors: C11 = 2, C12 = 3, C21 = −5, C22 = −1.
3. Write the cofactor matrix [2, 3; −5, −1] and transpose it to get adj A = [2, −5; 3, −1].
4. A⁻¹ = (1/|A|) × adj A = (1/13) × [2, −5; 3, −1] = [2/13, −5/13; 3/13, −1/13].

AnswerA⁻¹ = [2/13, −5/13; 3/13, −1/13]

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. 92Matches NCERT’s answer

1. Expand along the first column: |A| = 1×(2×5 − 4×0) = 10. Since |A| ≠ 0, A⁻¹ exists.
2. Find all nine cofactors of A and transpose the cofactor matrix to get adj A = [10, −10, 2; 0, 5, −4; 0, 0, 2].
3. A⁻¹ = (1/|A|) × adj A = (1/10) × [10, −10, 2; 0, 5, −4; 0, 0, 2] = [1, −1, 1/5; 0, 1/2, −2/5; 0, 0, 1/5].

AnswerA⁻¹ = [1, −1, 1/5; 0, 1/2, −2/5; 0, 0, 1/5]

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. 92Matches NCERT’s answer

1. Expand along the first row: |A| = 1×(3×(−1) − 0×2) = −3. Since |A| ≠ 0, A⁻¹ exists.
2. Find all nine cofactors and transpose the cofactor matrix to get adj A = [−3, 0, 0; 3, −1, 0; −9, −2, 3].
3. A⁻¹ = (1/|A|) × adj A = (1/−3) × [−3, 0, 0; 3, −1, 0; −9, −2, 3] = [1, 0, 0; −1, 1/3, 0; 3, 2/3, −1].

AnswerA⁻¹ = [1, 0, 0; −1, 1/3, 0; 3, 2/3, −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. 92Matches NCERT’s answer

1. Expand along the first row: |A| = 2(−1×1 − 0×2) − 1(4×1 − 0×(−7)) + 3(4×2 − (−1)×(−7)) = −2 − 4 + 3 = −3. Since |A| ≠ 0, A⁻¹ exists.
2. Find all nine cofactors and transpose the cofactor matrix to get adj A = [−1, 5, 3; −4, 23, 12; 1, −11, −6].
3. A⁻¹ = (1/|A|) × adj A = (1/−3) × [−1, 5, 3; −4, 23, 12; 1, −11, −6] = [1/3, −5/3, −1; 4/3, −23/3, −4; −1/3, 11/3, 2].

AnswerA⁻¹ = [1/3, −5/3, −1; 4/3, −23/3, −4; −1/3, 11/3, 2]

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. 92Matches NCERT’s answer

1. Expand along the first row: |A| = 1(2×4 − (−3)×(−2)) − (−1)(0×4 − (−3)×3) + 2(0×(−2) − 2×3) = 2 + 9 − 12 = −1. Since |A| ≠ 0, A⁻¹ exists.
2. Find all nine cofactors and transpose the cofactor matrix to get adj A = [2, 0, −1; −9, −2, 3; −6, −1, 2].
3. A⁻¹ = (1/|A|) × adj A = (1/−1) × [2, 0, −1; −9, −2, 3; −6, −1, 2] = [−2, 0, 1; 9, 2, −3; 6, 1, −2].

AnswerA⁻¹ = [−2, 0, 1; 9, 2, −3; 6, 1, −2]

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. 92Matches NCERT’s answer

1. Expand along the first row: |A| = 1 × (cos α × (−cos α) − sin α × sin α) = −(cos²α + sin²α) = −1. Since |A| ≠ 0, A⁻¹ exists.
2. Find all nine cofactors and transpose the cofactor matrix to get adj A = [−1, 0, 0; 0, −cos α, −sin α; 0, −sin α, cos α].
3. A⁻¹ = (1/|A|) × adj A = (1/−1) × [−1, 0, 0; 0, −cos α, −sin α; 0, −sin α, cos α] = [1, 0, 0; 0, cos α, sin α; 0, sin α, −cos α].
4. So A⁻¹ = A itself, because cos²α + sin²α = 1 makes A its own inverse.

AnswerA⁻¹ = [1, 0, 0; 0, cos α, sin α; 0, sin α, −cos α] (the same as A)

Watch this explained “Three things that fall out”, 16:44 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 93One way to think about it

1. Multiply AB = [3, 7; 2, 5] × [6, 8; 7, 9] = [67, 87; 47, 61].
2. Find |AB| = 67×61 − 87×47 = 4087 − 4089 = −2. Since |AB| ≠ 0, (AB)⁻¹ exists: (AB)⁻¹ = (1/−2) × [61, −87; −47, 67] = [−61/2, 87/2; 47/2, −67/2].
3. Find |A| = 3×5 − 7×2 = 1, so A⁻¹ = [5, −7; −2, 3]. Find |B| = 6×9 − 8×7 = −2, so B⁻¹ = (1/−2) × [9, −8; −7, 6] = [−9/2, 4; 7/2, −3].
4. Multiply B⁻¹A⁻¹ = [−9/2, 4; 7/2, −3] × [5, −7; −2, 3] = [−61/2, 87/2; 47/2, −67/2].
5. This is the same matrix as (AB)⁻¹, so (AB)⁻¹ = B⁻¹A⁻¹ is verified.

In short(AB)⁻¹ = B⁻¹A⁻¹ = [−61/2, 87/2; 47/2, −67/2] is verified.

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. 93Matches NCERT’s answer

1. Find A² = [3, 1; −1, 2] × [3, 1; −1, 2] = [8, 5; −5, 3].
2. Find A² − 5A + 7I = [8, 5; −5, 3] − [15, 5; −5, 10] + [7, 0; 0, 7] = [0, 0; 0, 0] = O. This proves the equation.
3. From A² − 5A + 7I = O, write 7I = 5A − A², so 7I = A(5I − A).
4. Multiply both sides on the right by A⁻¹: 7A⁻¹ = 5I − A.
5. So A⁻¹ = (5I − A)/7 = ([5, 0; 0, 5] − [3, 1; −1, 2])/7 = [2, −1; 1, 3]/7 = [2/7, −1/7; 1/7, 3/7].

AnswerA² − 5A + 7I = O is shown, and A⁻¹ = [2/7, −1/7; 1/7, 3/7].

Watch this explained “An inverse with no cofactors in it”, 14:34 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 93Matches NCERT’s answer

1. Find A² = [3, 2; 1, 1] × [3, 2; 1, 1] = [11, 8; 4, 3].
2. Write A² + aA + bI = [11, 8; 4, 3] + a[3, 2; 1, 1] + b[1, 0; 0, 1] and set every entry to 0.
3. The (1,1) entry gives 11 + 3a + b = 0, and the (2,2) entry gives 3 + a + b = 0.
4. Subtract: (11 + 3a) − (3 + a) = 0, so 8 + 2a = 0, giving a = −4.
5. Put a = −4 into 3 + a + b = 0: 3 − 4 + b = 0, so b = 1.
6. Check with the off-diagonal entries: (1,2) gives 8 + 2a = 8 − 8 = 0 ✓, (2,1) gives 4 + a = 4 − 4 = 0 ✓.

Answera = −4 and b = 1.

Watch this explained “The same trick, run backwards”, 15:43 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 93Matches NCERT’s answer

1. Find A² = A × A = [4, 2, 1; −3, 8, −14; 7, −3, 14].
2. Find A³ = A² × A = [8, 7, 1; −23, 27, −69; 32, −13, 58].
3. Write the other terms: 6A² = [24, 12, 6; −18, 48, −84; 42, −18, 84], 5A = [5, 5, 5; 5, 10, −15; 10, −5, 15] and 11I = [11, 0, 0; 0, 11, 0; 0, 0, 11].
4. Add entry by entry. First row: 8 − 24 + 5 + 11 = 0, 7 − 12 + 5 + 0 = 0, 1 − 6 + 5 + 0 = 0. Second row: −23 + 18 + 5 + 0 = 0, 27 − 48 + 10 + 11 = 0, −69 + 84 − 15 + 0 = 0. Third row: 32 − 42 + 10 + 0 = 0, −13 + 18 − 5 + 0 = 0, 58 − 84 + 15 + 11 = 0. So A³ − 6A² + 5A + 11I = O.
5. |A| = 1(2×3 − (−3)×(−1)) − 1(1×3 − (−3)×2) + 1(1×(−1) − 2×2) = 3 − 9 − 5 = −11 ≠ 0, so A⁻¹ exists.
6. Multiply A³ − 6A² + 5A + 11I = O on the right by A⁻¹: A² − 6A + 5I + 11A⁻¹ = O, so A⁻¹ = −(1/11)(A² − 6A + 5I).
7. A² − 6A + 5I = [4 − 6 + 5, 2 − 6, 1 − 6; −3 − 6, 8 − 12 + 5, −14 + 18; 7 − 12, −3 + 6, 14 − 18 + 5] = [3, −4, −5; −9, 1, 4; −5, 3, 1].
8. So A⁻¹ = −(1/11) × [3, −4, −5; −9, 1, 4; −5, 3, 1] = [−3/11, 4/11, 5/11; 9/11, −1/11, −4/11; 5/11, −3/11, −1/11].

AnswerA³ − 6A² + 5A + 11I = O is shown, and A⁻¹ = [−3/11, 4/11, 5/11; 9/11, −1/11, −4/11; 5/11, −3/11, −1/11].

Watch this explained “An inverse with no cofactors in it”, 14:34 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 93Matches NCERT’s answer

1. Find A² = A × A = [6, −5, 5; −5, 6, −5; 5, −5, 6].
2. Find A³ = A² × A = [22, −21, 21; −21, 22, −21; 21, −21, 22].
3. Substitute into A³ − 6A² + 9A − 4I: every entry cancels to 0, so A³ − 6A² + 9A − 4I = O is verified.
4. Rearrange: 4I = A³ − 6A² + 9A = A(A² − 6A + 9I), so 4I = A × (A² − 6A + 9I).
5. Multiply both sides on the right by A⁻¹: 4A⁻¹ = A² − 6A + 9I.
6. So A⁻¹ = (A² − 6A + 9I)/4 = [3/4, 1/4, −1/4; 1/4, 3/4, 1/4; −1/4, 1/4, 3/4].

AnswerA³ − 6A² + 9A − 4I = O is verified, and A⁻¹ = [3/4, 1/4, −1/4; 1/4, 3/4, 1/4; −1/4, 1/4, 3/4].

Watch this explained “An inverse with no cofactors in it”, 14:34 into Singular against non-singular, and the one test an inverse has to pass

Open NCERT p. 93Matches NCERT’s answer

1. For a square matrix of order n, the rule is |adj A| = |A|^(n−1).
2. Here n = 3, so |adj A| = |A|^(3−1) = |A|².
3. This matches option (B).

Answer(B) |A|²

Watch this explained “The determinant of the adjoint”, 14:54 into Building the adjoint, and why it multiplies back to a scalar times the identity

Open NCERT p. 93Matches NCERT’s answer

1. For any invertible matrix, |A| × |A⁻¹| = |A A⁻¹| = |I| = 1.
2. So |A⁻¹| = 1/|A|.
3. This matches option (B).

Answer(B) 1/det(A)

Watch this explained “Three things that fall out”, 16:44 into Singular against non-singular, and the one test an inverse has to pass