If A=[2 3 1 2], prove that A3-4A2+A=0 Hence find A-1

We have,A = 2312A2 = A A = 2312 2312 = 4+36+62+23+4 = 71247A3 = A2 A = 712472312 = 14+1221+248+712+14 = 26451526Now, LHS = A3 - 4A2 + A=26431526 - 471247 + 2312=26451526 + -28-48-16-28 + 2312=26-28+245-48+315-16+126-28+2 = 0000 = 0 = RHSHence,A3 - 4A2 + A = 0⇒A2 - 4A + 1 = 0⇒A2 - 4A + I = 0⇒A-1AA - 4A-1A + A-1I = 0⇒IA - 4I + A-1 = 0⇒A-1 = 4I - A⇒A-1 = 4004 - 2312 = 2-3-12

If A=[2 3 1 2], prove that A³-4A²+A=0. Hence find A^-1.

We have, A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] A^{2} = A . A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{matrix}] = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] A^{3} = A^{2} . A = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 14 + 12 & 21 + 24 \\ 8 + 7 & 12 + 14 \end{matrix}] = [\begin{matrix} 26 & 45 \\ 15 & 26 \end{matrix}] Now, LHS = A^{3} - 4 A^{2} + A = [\begin{matrix} 26 & 43 \\ 15 & 26 \end{matrix}] - 4 [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] + [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 26 & 45 \\ 15 & 26 \end{matrix}] + [\begin{matrix} - 28 & - 48 \\ - 16 & - 28 \end{matrix}] + [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 26 - 28 + 2 & 45 - 48 + 3 \\ 15 - 16 + 1 & 26 - 28 + 2 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 = RHS Hence, A^{3} - 4 A^{2} + A = 0 \Rightarrow A^{2} - 4 A + 1 = 0 \Rightarrow A^{2} - 4 A + I = 0 \Rightarrow (A^{- 1} A) A - 4 (A^{- 1} A) + A^{- 1} I = 0 \Rightarrow IA - 4 I + A^{- 1} = 0 \Rightarrow A^{- 1} = 4 I - A \Rightarrow A^{- 1} = [\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}] - [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]

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If A=[2 3 1 2], prove that A3-4A​2+A=0. Hence find A-1.

If A=[2 3 1 2], prove that A³-4A²+A=0. Hence find A^-1.