निकाले `(AB)^(-1)` ,जहाँ
`A= [(5,0,4),(2,3,2),(1,2,1)]. B^(-1)= [(1,3,3),(1,4,3),(1,3,4)]`

Explanation

हम जानते है कि `(AB)^(-1) = B^(-1) A^(-1)` ….(1)
अब `|A|=|(5,0,4),(2,3,2),(1,2,1)| = 5 (3-4)-0(2-2) + (4 -3)= - 1 ne 0`
अतः `A^(-1)` का अस्तित्व है ।
माना कि आव्यूह A के i पंक्ति और j वे स्तम्भ के अवयव का सहखण्ड (cofactor ) `A_(if)` है , तो
`A_(11) = (-1)^(1+1) |(3,2),(2,1)| = - 1 , A_(12) = (-1)^(1 + 2) |(2,2),(1,1)| = 0 `
`A_(13) = (-1)^(1+3) |(2,3),(1,2)| = 1 , A_(21) = (-1)^(2+ 1) |(0,4),(2,1)| = 8 `
`A_(22) = (-1)^(2+2) |(5,4),(1,1)| = 1 , A_(23) = (-1)^(2+ 3) |(5,0),(1,2)| = -10 `
`A_(31) = (-1)^(3+1) |(0,4),(3,2)| =-12 , A_(32) = (-1)^( 3+2) |(5,4),(2,2)| = -2 `
`A_(33) = (-1)^(3+3) |(5,0),(2,3)|=15`
अब adj `A=[(-1,0,1),(8,1,-10),(-12,-2,15)]"का परिवर्त "=[(-1,8,-12),(0,1,-2),(1,-10,15)]`
`therefore A^(-1) = (1)/(|A|) "adj A "= (1)/((-1)) "adj A = - adj A " = [(1 ,-8,12),(0,-1,2),(-1,10,-15)]`
(1) से `(AB)^(-1) = B^(-1) A^(-1)`
`[(1,3,3),(1,4,3),(1,3,4)][(1,-8,12),(0,-1,2),(-1,10,-15)]`
`=[(1+0-3, -8 - 3 + 30, 12 + 6-45),(1 +0-3,-8-4+30,12 +8-45),(1+0-4,-8-3 + 40, 12 + 6-60)]=[(-2,19,-27),(-2,18,-25),(-3,29,-42)]`.