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Inverse of a Matrix

If A=[ -1 ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 0 - 1 & 1 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$ , Show that $A^{2} = A^{- 1}$

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 0 - 1 & 1 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ∴ A^{2} = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 0 - 1 & 1 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 0 - 1 & 1 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 2 & - 2 + 2 & 2 1 - 1 & - 2 + 1 + 1 & 1 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 0 & 2 0 & 0 & 1 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$
Now, $| A | = (- 1) (- 1 + 0) = 1 a d j A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 0 & - 1 0 & 0 & 1 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 0 & 2 0 & 0 & 1 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$
Now, $A^{- 1} = \frac{1}{| A |} a d j A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 0 & 2 0 & 0 & 1 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = A^{2} ∴ A^{- 1} = A^{2}$

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A = [\begin{matrix} - 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}]

A^{2} = A^{- 1} .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & - 1 & 0 1 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

A^{2}

A^{2} = A^{- 1}

A = ∣ ∣ ∣ \begin{matrix} 1 & - 1 - 1 & 1 \end{matrix} ∣ ∣ ∣

, then show that

A^{2} = 2 A & A^{3} = 4 A

A = [\begin{matrix} 4 & - 1 & - 4 \\ 3 & 0 & - 4 \\ 3 & - 1 & - 3 \end{matrix}]

, show that A² = I₃.

\sqrt{\frac{1 + i}{1 - i}} = (a + i b)

(a^{2} + b^{2}) = 1

.

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