Question

20. Use an inverse matrix to solve the following system of equations.
x + y – 2z = 0
x – 2y + z = 0
x – y - z = -1

Answer 1

$$\left\{ \begin{array}{l} x + y - 2 z = 0 \\ x - 2 y + z = 0 \\ x - y - z = - 1 \end{array} \right.$$

Solution

$$A X = B$$ $$A = \left( \begin{array}{rrr} 1 & 1 & -2 \\ 1 & -2 & 1 \\ 1 & -1 & -1 \end{array} \right)$$ $$X = \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$ $$B = \left( \begin{array}{c} 0 \\ 0 \\ -1 \end{array} \right)$$ $$X = A^{-1} B$$

Find the inverse matrix $A^{-1}$

$$\begin{array}{l} det A = \left| \begin{array}{rrr} 1 & 1 & -2 \\ 1 & -2 & 1 \\ 1 & -1 & -1 \end{array} \right| = 1(2 + 1) - 1(-1 - 2) + 1(1 - 4) = 3 + 3 - 3 = 3 \\ A^{-1} = \frac{1}{det A} \left( \begin{array}{rrr} 3 & 3 & -3 \\ 2 & 1 & -3 \\ 1 & 2 & -3 \end{array} \right) \\ A^{-1} = \left( \begin{array}{rrr} 1 & 1 & -1 \\ \frac{2}{3} & \frac{1}{3} & -1 \\ \frac{1}{3} & \frac{2}{3} & -1 \end{array} \right) \\ X = \left( \begin{array}{rrr} 1 & 1 & -1 \\ \frac{2}{3} & \frac{1}{3} & -1 \\ \frac{1}{3} & \frac{2}{3} & -1 \end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\ -1 \end{array} \right) \\ X = \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) \\ x = 1, \qquad y = 1, \qquad z = 1 \\ \end{array}$$

Answer: $x = 1, y = 1, \quad z = 1$