Using the Inverse of a Matrix to Solve a System Solve the system by using A^-1, the inverse of the coefficient matrix: {x - y + z = 2 -2y + z = 2 -2x -3y = 1/2.

Question

Using the Inverse of a Matrix to Solve a System

Solve the system by using A^-1, the inverse of the coefficient matrix:

[latex]\left\{\begin{aligned}x-y+z &=2 \-2 y+z &=2 \-2 x-3 y&=\frac{1}{2}.\end{aligned} ight.[/latex]

Accepted Answer

The linear system can be written as

The solution is given by [latex]X=A^{-1} B[/latex]. Consequently, we must find [latex]A^{-1}[/latex]. We found the inverse of matrix A in Example 4. Using this result,

[latex]X=A^{-1} B=\left[\begin{array}{rrr}3 & -3 & 1 \-2 & 2 & -1 \-4 & 5 & -2\end{array} ight]\left[\begin{array}{l}2 \2 \\frac{1}{2}\end{array} ight]=\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} ight]=\left[\begin{array}{r}\frac{1}{2} \\-\frac{1}{2} \\ 1\end{array} ight][/latex]

Thus, [latex]x=\frac{1}{2}, y=-\frac{1}{2}[/latex], and z = 1. The solution set is [latex]\left\{\left(\frac{1}{2},-\frac{1}{2}, 1 ight) ight\}[/latex].

Question 8.4.5: Using the Inverse of a Matrix to Solve a System Solve the sy...

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