silemus.

17Solving a 3×33\times 3 linear system

Solve the system {x+y+z=62y+5z=−42x+5y−z=27\displaystyle \begin{cases}x + y + z = 6\\2y + 5z = -4\\2x + 5y - z = 27\end{cases}

Eliminate x using equation 1, then solve the remaining pair in y and z. The solution is x = 5, y = 3, z = −2. Check: 2(5) + 5(3) − (−2) = 27 ✓.

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