More practice questions

5. Q5. For A with rows 1, −1, 2; 3, 0, −2; 1, 0, 3, verify that A(adj A) = (adj A)A = |A| I (NCERT Exercise 4.4, Q4).

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   Along column 2, which has two zeros: |A| = −(−1)(9 + 2) = 11. adj A has rows 0, 3, 2; −11, 1, 8; 0, −1, 3. Row 1 of A times the columns of adj A gives 0 + 11 + 0 = 11, 3 − 1 − 2 = 0 and 2 − 8 + 6 = 0; the other rows give 0, 11, 0 and 0, 0, 11 in the same way, and (adj A)A works out to 11I as well.

6. Q6. Find the inverse of the matrix with rows 1, 0, 0; 3, 3, 0; 5, 2, −1 (NCERT Exercise 4.4, Q8).

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   It is lower triangular, so |A| = 1 × 3 × (−1) = −3. adj A has rows −3, 0, 0; 3, −1, 0; −9, −2, 3, so A^⁻¹ = −(1/3) adj A, with rows 1, 0, 0; −1, 1/3, 0; 3, 2/3, −1.

7. Q7. Find the inverse of the matrix with rows 2, 1, 3; 4, −1, 0; −7, 2, 1 (NCERT Exercise 4.4, Q9).

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   |A| = 2(−1 − 0) − 1(4 − 0) + 3(8 − 7) = −2 − 4 + 3 = −3. adj A has rows −1, 5, 3; −4, 23, 12; 1, −11, −6, so A^⁻¹ = −(1/3) adj A, with rows 1/3, −5/3, −1; 4/3, −23/3, −4; −1/3, 11/3, 2.

8. Q8. Find the inverse of the matrix with rows 1, 0, 0; 0, cos α, sin α; 0, sin α, −cos α (NCERT Exercise 4.4, Q11).

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   |A| = 1 × (−cos² α − sin² α) = −1. adj A has rows −1, 0, 0; 0, −cos α, −sin α; 0, −sin α, cos α, so A^⁻¹ = −adj A, with rows 1, 0, 0; 0, cos α, sin α; 0, sin α, −cos α. This is A itself: A² = I.

9. Q9. A has rows 3, 7 and 2, 5, and B has rows 6, 8 and 7, 9. Verify that (AB)^⁻¹ = B^⁻¹A^⁻¹ (NCERT Exercise 4.4, Q12).

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   AB has rows 67, 87 and 47, 61, with |AB| = 4087 − 4089 = −2, so (AB)^⁻¹ = −½ times the matrix with rows 61, −87 and −47, 67. Also |A| = 1 and |B| = −2, so A^⁻¹ has rows 5, −7 and −2, 3, and B^⁻¹ has rows −9/2, 4 and 7/2, −3. Their product B^⁻¹A^⁻¹ has rows −61/2, 87/2 and 47/2, −67/2, the same as (AB)^⁻¹.

10. Q10. For A with rows 2, −1, 1; −1, 2, −1; 1, −1, 2, verify that A³ − 6A² + 9A − 4I = O and hence find A^⁻¹ (NCERT Exercise 4.4, Q16).

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    A² has rows 6, −5, 5; −5, 6, −5; 5, −5, 6, and A³ = A A² has rows 22, −21, 21; −21, 22, −21; 21, −21, 22; the combination is then O element by element. The polynomial identity factors as A(A² − 6A + 9I) = 4I, with the factors also commuting, so it establishes invertibility. Multiplying by A^⁻¹ gives A² − 6A + 9I = 4A^⁻¹, so A^⁻¹ = ¼(A² − 6A + 9I), with rows 3/4, 1/4, −1/4; 1/4, 3/4, 1/4; −1/4, 1/4, 3/4.

11. Q11. For A with rows 3, 1 and −1, 2, show that A² − 5A + 7I = O, and hence find A^⁻¹ (NCERT Exercise 4.4, Q13, page 93).

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    A² has rows 8, 5 and −5, 3, so A² − 5A + 7I has (1, 1) element 8 − 15 + 7 = 0, and the others are 0 likewise. Since |A| = 3 × 2 − 1 × (−1) = 7 ≠ 0, multiplying A² − 5A = −7I by A^⁻¹ gives A − 5I = −7A^⁻¹, so A^⁻¹ = (5I − A)/7, that is (1/7) times the matrix with rows 2, −1 and 1, 3.

12. Q12. For A with rows 1, 1, 1; 1, 2, −3; 2, −1, 3, show that A³ − 6A² + 5A + 11I = O, and hence find A^⁻¹ (NCERT Exercise 4.4, Q15, page 93).

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    A² has rows 4, 2, 1; −3, 8, −14; 7, −3, 14, and A³ has rows 8, 7, 1; −23, 27, −69; 32, −13, 58; the combination is O element by element. The identity A(A² − 6A + 5I) = −11I, whose factors commute, establishes invertibility. Multiplying by A^⁻¹: A² − 6A + 5I = −11A^⁻¹, so A^⁻¹ = (1/11) times the matrix with rows −3, 4, 5; 9, −1, −4; 5, −3, −1.