आव्यूह विधि से हल कीजिएः
`2x+y+z=1`
`x-2y-3z=1`
`3x+2y+4z=5`

Explanation

उक्त समीकरणों को आव्यूहों के रूप में लिखने पर
`[(2,1,1),(1,-2,-3),(3,2,4)][(x),(y),(z)]=[(1),(1),(5)]`
`impliesAX=B`(माना)
अब `|A|=-|(2,1,1),(1,-2,-3),(3,2,4)|`
`=2(-8+6)-1(4+9)+1(2+6)`
`=-4-13+8=-9!=0`
`:.A` व्युत्क्रमणीय है।
आव्यूह A के सहगुणनखण्ड-
`c_(11)=(-1)^(1+1)|(-2,-3),(2,4)|=-8+6=-2`
`c_(12)=(-1)^(1+2)|(1,-3),(3,4)|=-(4+9)=-13`
`c_(13)=(-1)^(1+3)|(1,-2),(3,2)|=2+6=8`
`c_(21)=(-1)^(2+1)|(1,1),(2,4)|=-(4-2)=-2`
`c_(22)=(-1)^(2+2)|(2,1),(3,4)|=8-3=55`
`=c_(23)=(-1)^(2+3)|(2,1),(3,2)|=-(4-3)=-1`
`c_(31)=(-1)^(3+1)|(1,1),(-2,-3)|=-3+2=-1`
`=c_(32)=(-1)^(3+2)|(2,1),(1,-3)|=-(-6-1)=7`
`c_(33)=(-1)^(3+3)|(2,1),(1,-2)|=-4-1=-5`
`:.adj.A=[(c_(11),c_(12),c_(13)),(c_(21),c_(22),c_(23)),(c_(31),c_(32),c_(33))]'=[(-2,-13,8),(-2,5,-1),(-1,7,-5)]'`
`=[(-2,-2,-1),(-13,5,7),(8,-1,-5)]`
और `A^(-1)=1/(|A|).adj.A=-1/9[(-2,-2,-1),(-13,5,7),(8,-1,-5)]`
अब `AX=B`
`impliesX=A^(-1)B`
`implies[(x),(y),(z)]=1/9[(-2,-2,-1),(-13,5,7),(8,-1,-5)][(1),(1),(5)]`
`==1/9[(-2-2-5),(-13+5+35),(8-1-25)]=-1/9[(-9),(27),(-18)]=[(1),(-3),(2)]`
`:.x=1,y=-3,z=2`