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यदि आव्यूह A = \(\begin{bmatrix}1&-1\\[0.3em]2&-1\\[0.3em]\end{bmatrix}\)हो, तो सिद्ध कीजिए कि A-1 = A3

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दिया गया आव्यूह, A = \(\begin{bmatrix}1&-1\\[0.3em]2&-1\\[0.3em]\end{bmatrix}\) =  \(\begin{bmatrix}1&-1\\[0.3em]2&-1\\[0.3em]\end{bmatrix}\)

= -1 + 2

|A| = 1 ≠ 0

अतः A-1 का अस्तित्व है।

आव्यूह A के सहखण्ड ज्ञात करने पर,

a11 = – 1, a12 = – 2, a21 = 1, a22 = 1

आव्यूह A के सहखण्डों के बना आव्यूह

समीकरण (i) व (ii) से,

A-1 = A3.
इति सिद्धम्।

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