Using Mathematical Induction in Linear Algebra If A1, A2, · · ·, An are invertible matrices, then prove the generalization of Theorem 2.9. (A1 A2 A3 · · · An)^-1 = An^-1 · · · A3^-1A2^-1A1^-1

Question

Using Mathematical Induction in Linear Algebra

If [latex]A_1, A_2, \cdot \cdot \cdot , A_n[/latex] are invertible matrices, then prove the generalization of Theorem 2.9.

[latex](A_1A_2A_3 \cdot \cdot \cdot A_n )^{-1} = A^{-1}_{n} \cdot \cdot \cdot A^{-1}_{3}A^{-1}_{2}A^{-1}_{1}[/latex]

Accepted Answer

1. The formula is valid trivially when n = 1 because [latex]A^{-1}_{1} = A^{-1}_{1}[/latex].

2. Assuming the formula is valid for k, [latex](A_1A_2A_3 \cdot \cdot \cdot A_k )^{-1} = A^{-1}_{k} \cdot \cdot \cdot A^{-1}_{3}A^{-1}_{2}A^{-1}_{1}[/latex], you must show that it is valid for k + 1. To do this, use Theorem 2.9, which states that the inverse of a product of two invertible matrices is the product of their inverses in reverse order.

[latex]\begin{matrix} (A_1A_2A_3 \cdot \cdot \cdot A_k A_{k +1} )^{-1} = [(A_1A_2A_3 \cdot \cdot \cdot A_k )A_{k +1}]^{-1} & & \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = A^{-1}_{k +1} (A_1A_2A_3 \cdot \cdot \cdot A_k )^{-1} & ext{ Theorem }2.9 & \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =A^{-1}_{k +1}(A^{-1}_{k} \cdot \cdot \cdot A^{-1}_{3}A^{-1}_{2}A^{-1}_{1}) & \quad\quad ext{ Induction hypothesis } & \\ \quad\quad \quad \quad \quad \quad \quad \quad \quad = A^{-1}_{k +1}A^{-1}_{k} \cdot \cdot \cdot A^{-1}_{3}A^{-1}_{2}A^{-1}_{1} & \quad S_k ext{ implies } S_k+1. &\end{matrix} [/latex]

Combining the results of parts (1) and (2), you can conclude by mathematical induction that the formula is valid for all positive integers n.

Question Appendix.4: Using Mathematical Induction in Linear Algebra If A1, A2, · ...

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