Exercise 4.1 answers: Determinants

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1. For a 2×2 determinant |a b; c d|, the value is a×d − b×c.
2. Here a = 2, b = 4, c = −5, d = −1.
3. a×d = 2 × (−1) = −2.
4. b×c = 4 × (−5) = −20.
5. Value = −2 − (−20) = 18.

Answer18

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(i) cos θ −sin θ; sin θ cos θ

1. For a 2×2 determinant, the value is (top-left × bottom-right) − (top-right × bottom-left).
2. Main diagonal: cos θ × cos θ = cos²θ.
3. Other diagonal: (−sin θ) × sin θ = −sin²θ.
4. Value = cos²θ − (−sin²θ) = cos²θ + sin²θ.
5. Since cos²θ + sin²θ = 1 for every θ, the value is 1.

Answer1

(ii) x² − x + 1 x − 1; x + 1 x + 1

1. Main diagonal: (x² − x + 1)(x + 1) = x³ + x² − x² − x + x + 1 = x³ + 1.
2. Other diagonal: (x − 1)(x + 1) = x² − 1.
3. Value = (x³ + 1) − (x² − 1) = x³ + 1 − x² + 1 = x³ − x² + 2.

Answerx³ − x² + 2

Watch this explained “A number, and a vanishing”, 2:06 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 81One way to think about it

1. First, find |A|: |A| = 1(2) − 2(4) = 2 − 8 = −6
2. Next, find 2A = [2 4; 8 4]
3. Calculate |2A| = 2(4) − 4(8) = 8 − 32 = −24
4. Verify: 4|A| = 4(−6) = −24
5. Therefore |2A| = 4|A|, which confirms the general property |kA| = k^n|A| for an n×n matrix

In shortShown

Watch this explained “A zero entry, and scaling every entry”, 18:02 into Orders one, two and three, and expanding along a chosen row

Open NCERT p. 81One way to think about it

1. First, find |A| by expanding along the first column (which has zeros)
2. |A| = 1 · |1 2; 0 4| = 1(1·4 − 2·0) = 4
3. Next, find 3A = [3 0 3; 0 3 6; 0 0 12]
4. Calculate |3A| by expanding along the first column
5. |3A| = 3 · |3 6; 0 12| = 3(3·12 − 6·0) = 3(36) = 108
6. Verify: 27|A| = 27(4) = 108
7. Therefore |3A| = 27|A|, confirming the property |kA| = k³|A| for a 3×3 matrix

In shortShown

Watch this explained “A zero entry, and scaling every entry”, 18:02 into Orders one, two and three, and expanding along a chosen row

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(i) 3 −1 −2; 0 0 −1; 3 −5 0

1. The second row has two zeros, so expand along it to save work: row 2 is 0, 0, −1.
2. Only the last entry, −1, contributes a term.
3. Cross out its row and column; the remaining 2×2 is |3 −1; 3 −5| = 3×(−5) − (−1)×3 = −15 + 3 = −12.
4. Its position (row 2, col 3) has an odd sign, so its cofactor is −1 × (−12) = 12.
5. Value = (−1) × 12 = −12.

Answer−12

(ii) 3 −4 5; 1 1 −2; 2 3 1

1. Expand along the top row: 3, −4, 5.
2. 3 × |1 −2; 3 1| = 3 × (1 + 6) = 21.
3. −(−4) × |1 −2; 2 1| = 4 × (1 + 4) = 20.
4. 5 × |1 1; 2 3| = 5 × (3 − 2) = 5.
5. Add them: 21 + 20 + 5 = 46.

Answer46

(iii) 0 1 2; −1 0 −3; −2 3 0

1. Expand along the top row: 0, 1, 2.
2. 0 × (anything) = 0.
3. −1 × |−1 −3; −2 0| = −1 × (0 − 6) = 6.
4. 2 × |−1 0; −2 3| = 2 × (−3 − 0) = −6.
5. Add them: 0 + 6 − 6 = 0.

Answer0

(iv) 2 −1 −2; 0 2 −1; 3 −5 0

1. Expand along the top row: 2, −1, −2.
2. 2 × |2 −1; −5 0| = 2 × (0 − 5) = −10.
3. −(−1) × |0 −1; 3 0| = 1 × (0 + 3) = 3.
4. −2 × |0 2; 3 −5| = −2 × (0 − 6) = 12.
5. Add them: −10 + 3 + 12 = 5.

Answer5

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1. Expand along the top row: 1, 1, −2.
2. 1 × |1 −3; 4 −9| = 1 × (−9 + 12) = 3.
3. −1 × |2 −3; 5 −9| = −1 × (−18 + 15) = 3.
4. −2 × |2 1; 5 4| = −2 × (8 − 5) = −6.
5. Add them: 3 + 3 − 6 = 0.

Answer0

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(i) 2 4; 5 1 = 2x 4; 6 x

1. Left side: |2 4; 5 1| = 2×1 − 4×5 = 2 − 20 = −18.
2. Right side: |2x 4; 6 x| = 2x×x − 4×6 = 2x² − 24.
3. Set them equal: 2x² − 24 = −18, so 2x² = 6, so x² = 3.
4. Since x is squared, there are two answers: x = √3 or x = −√3.

Answerx = ±√3

(ii) 2 3; 4 5 = x 3; 2x 5

1. Left side: |2 3; 4 5| = 2×5 − 3×4 = 10 − 12 = −2.
2. Right side: |x 3; 2x 5| = x×5 − 3×2x = 5x − 6x = −x.
3. Set them equal: −x = −2, so x = 2.

Answerx = 2

Watch this explained “What one reduction leaves you able to do”, 20:07 into Orders one, two and three, and expanding along a chosen row

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1. Left side: |x 2; 18 x| = x² − 2×18 = x² − 36.
2. Right side: |6 2; 18 6| = 6×6 − 2×18 = 36 − 36 = 0.
3. Set them equal: x² − 36 = 0, so x² = 36.
4. Since x is squared, x = 6 or x = −6, which is option (B), ±6.

Answer(B) ±6

Watch this explained “What one reduction leaves you able to do”, 20:07 into Orders one, two and three, and expanding along a chosen row