Solution 12) A=-8524Therefore,A2=74-20-826andA2+4A-42I=74-20-826+4-8524-421001=74-20-826+-3220816-420042=0000=0Thus A satisfies the equation x2+4x-42 Now,A2+4A-42I=0⇒A2+4A=42I⇒A-1A2+4A-1A=42IA-1 Pre-multiplyng both sides by A-1⇒A+4I=42A-1⇒A-1=142A+4I⇒A-1=142-8524+4004=142-4528

Plz tell que 12 ???

Solution
12)

A = [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] T h e r e f o r e, A^{2} = [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] a n d A^{2} + 4 A - 42 I = [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] + 4 [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] - 42 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] + [\begin{matrix} - 32 & 20 \\ 8 & 16 \end{matrix}] - [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 T h u s A s a t i s f i e s t h e e q u a t i o n x^{2} + 4 x - 42 . N o w, A^{2} + 4 A - 42 I = 0 \Rightarrow A^{2} + 4 A = 42 I \Rightarrow A^{- 1} A^{2} + 4 A^{- 1} A = 42 I A^{- 1} [P r e - m u l t i p l y n g b o t h s i d e s b y A^{- 1}] \Rightarrow A + 4 I = 42 A^{- 1} \Rightarrow A^{- 1} = \frac{1}{42} (A + 4 I) \Rightarrow A^{- 1} = \frac{1}{42} \{[\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] + [\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}]\} = \frac{1}{42} [\begin{matrix} - 4 & 5 \\ 2 & 8 \end{matrix}]

View Full Answer