Question 863267
{{{A+B+C=2}}}
{{{2B+A+C=1}}}
{{{B+C+5A=6}}}
So then,
{{{A+B+C=2}}}
{{{A+2B+C=1}}}
{{{5A+B+C=6}}}
or in matrix form,
{{{(matrix(3,3,1,1,1,1,2,1,5,1,1))*(matrix(3,1,A,B,C))=(matrix(3,1,2,1,6))}}}
Find the matrix inverse.
{{{(matrix(3,1,A,B,C))=-(1/4)*(matrix(3,3,1,0,-1,4,-4,0,-9,4,1))*(matrix(3,1,2,1,6))}}}
{{{(matrix(3,1,A,B,C))=(matrix(3,1,1,-1,2))}}}