Solve a linear equation system
{4x+2y−z=15x+3y−2z=23x+2y−3z=0\begin{cases} 4x+2y-z= 1\\ 5x+3y-2z= 2\\ 3x+2y-3z = 0 \end{cases}⎩⎨⎧4x+2y−z=15x+3y−2z=23x+2y−3z=0
In the answer field write the values of x,y,zx, y, zx,y,z, divided by spaces.
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