Question 626896
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Hi,
A={{{(matrix(3,3,a1,b1,c1,a2,b2,c2,a3,b3,c3))}}} ={{{(matrix(3,3,3,-1,0,4,1,0,0,0,4))}}}
I. find value of determinant =  a1(b2c3-c2b3) - b1(a2c3-c2a3) + c1(a2b3-b2a3) = 28
II. Then Apply formula for the Inverse:
{{{A^(-1)}}} = 1/d{{{matrix(3,3,(b2*c3-c2*b3),(c1*b3-b1*c3),(b1*c2-c1*b2),
                   (c2a3-a2c3),(a1c3-c1a3),(c1a2-a1c2),
                   (a2b3 - b2a3), (b1a3 -a1b3),(a1b2-b1a2)))}}}

{{{A^(-1)}}}  = 1/28{{{(matrix(3,3,4,4,0,-16,12,0,0,0,7))}}}