Applications of Matrices and Determinants — Important Questions

(1/14)(9I − A)

Step 1: Start with A² − 9A + 14I = O. We want to isolate A⁻¹. Step 2: Post-multiply both sides by A⁻¹: A²·A⁻¹ − 9A·A⁻¹ + 14I·A⁻¹ = O·A⁻¹. Step 3: This gives A − 9I + 14A⁻¹ = O. Step 4: Rearranging: 14A⁻¹ = 9I − A → A⁻¹ = (1/14)(9I − A). Final Answer: A⁻¹ = (1/14)(9I − A). Common mistake: Students post-multiply incorrectly or forget the sign change when isolating A⁻¹.

8

Step 1: Use the property |adj A| = |A|^(n−1) for an n×n matrix. Step 2: For n = 3: |adj A| = |A|^(3−1) = |A|² = 64. Step 3: So |A|² = 64, giving |A| = ±8. Step 4: However, the problem states matrix is non-singular (|A| ≠ 0). From the note in the textbook, for a 3×3 non-singular matrix, |adj A| = |A|² is always positive, so |A| could be ±8. But by the theorem that |adj A| is positive for odd-order matrices, |A|^(2m) > 0 holds for both signs. The standard answer given is 8 (positive root) unless sign is specified. Final Answer: |A| = 8 (taking positive value as standard). Note: ±8 is technically

Infinitely many solutions forming a two-parameter family

Step 1: Notice equations 2, 3 are multiples of equation 1: 6x−3y+3z=6 is 3× eq.1, and 4x−2y+2z=4 is 2× eq.1. Step 2: Form augmented matrix and reduce. All three rows reduce to the same equation: 2x − y + z = 2. Step 3: Row-echelon form gives only 1 non-zero row. So ρ(A) = ρ([A|B]) = 1. Step 4: Number of unknowns = 3. Since ρ = 1 < 3, the solution has 3 − 1 = 2 free parameters → two-parameter family. Final Answer: Infinitely many solutions forming a two-parameter family. Common mistake: Counting parameters incorrectly as n − ρ.

adj(AB) = (adj A)(adj B)

Step 1: We check each option against known theorems. Step 2: (AB)⁻¹ = B⁻¹A⁻¹ is TRUE by the Reversal Law for Inverses. Step 3: |A⁻¹| = 1/|A| is TRUE since |A||A⁻¹| = |I| = 1. Step 4: (Aᵀ)⁻¹ = (A⁻¹)ᵀ is TRUE by Theorem 1.4. Step 5: adj(AB) = (adj B)(adj A) — note the ORDER is REVERSED, like the reversal law. So adj(AB) = (adj A)(adj B) is INCORRECT; the correct formula is adj(AB) = (adj B)(adj A). Final Answer: adj(AB) = (adj A)(adj B) is incorrect.