Question 188320
Remember, you can multiply a scalar value "k" by any row of a determinant like this:


{{{k*abs(matrix(3,3,a,b,c,d,e,f,g,h,i))=abs(matrix(3,3,k*a,k*b,k*c,d,e,f,g,h,i))}}} Note: you can pick any row. In this example, I'm using the first row.





{{{abs(matrix(3,3,x,y,z,u,v,w,1,2,3))=4}}} Start with the given equation.



Since the bottom row of the resulting determinant is 2, 4, and 6 this means that this row is 2 times the bottom row of the given determinant.




{{{2*abs(matrix(3,3,x,y,z,u,v,w,1,2,3))=2*4}}} Multiply both sides by 2



{{{abs(matrix(3,3,x,y,z,u,v,w,2*1,2*2,2*3))=2*4}}} Multiply the scalar 2 by the entire bottom row of the determinant



{{{abs(matrix(3,3,x,y,z,u,v,w,2,4,6))=8}}} Multiply




So the answer is {{{abs(matrix(3,3,x,y,z,u,v,w,2,4,6))=8}}}