Question 8.4.5: Using the Inverse of a Matrix to Solve a System Solve the sy...
Using the Inverse of a Matrix to Solve a System
Solve the system by using A^–1, the inverse of the coefficient matrix:
\left\{\begin{aligned}x-y+z &=2 \\-2 y+z &=2 \\-2 x-3 y&=\frac{1}{2}.\end{aligned}\right.
The linear system can be written as
The solution is given by X=A^{-1} B. Consequently, we must find A^{-1}. We found the inverse of matrix A in Example 4. Using this result,
X=A^{-1} B=\left[\begin{array}{rrr}3 & -3 & 1 \\-2 & 2 & -1 \\-4 & 5 & -2\end{array}\right]\left[\begin{array}{l}2 \\2 \\\frac{1}{2}\end{array}\right]=\left[\begin{array}{r}3 \cdot 2+(-3) \cdot 2+1 \cdot \frac{1}{2} \\-2 \cdot 2+2 \cdot 2+(-1) \cdot \frac{1}{2} \\-4 \cdot 2+5 \cdot 2+(-2) \cdot \frac{1}{2}\end{array}\right]=\left[\begin{array}{r}\frac{1}{2} \\\\-\frac{1}{2} \\\\ 1\end{array}\right]
Thus, x=\frac{1}{2}, y=-\frac{1}{2}, and z = 1. The solution set is \left\{\left(\frac{1}{2},-\frac{1}{2}, 1\right)\right\}.