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यदि A = 1/9 \(\begin{bmatrix}-8&1&4\\[0.3em]4&4&7\\[0.3em]1&-8&7 \\[0.3em] \end{bmatrix}\) तो सिद्ध कीजिए कि A-1 = AT

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 माना कि

प्रथम पंक्ति के सापेक्ष प्रसार करने पर

= – 8(16 + 56) – (16 – 7) + 4( – 32 – 4)

= – 8.72 – 9 – 4.36

= – 9(64 + 1 + 16)

= – 9 × 81

= – 93

अतः (iii) और (iv) से स्पष्ट है कि

A-1 = AT
इति सिद्धम्।

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