Answer to Question #350138 in Algebra for Vickie

Question

Solve the following using matrix algebra

2X+Y-Z=5

3X-2Y+2Z=-3

X-3Y-3Z=-2

Answer 1

$$A=\begin{pmatrix} 2 & 1 & -1 \\ 3 & -2 & 2 \\ 1 & -3 & -3 \\ \end{pmatrix}, X=\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}, B=\begin{pmatrix} 5 \\ -3 \\ -2 \\ \end{pmatrix}$$

$$AX=B$$

$$A^{-1}AX=A^{-1}B$$

$$X=A^{-1}B$$

Augment the matrix with the identity matrix

$$\begin{pmatrix} 2 & 1 & -1& & 1 & 0 & 0 \\ 3 & -2 & 2 & & 0 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1/2$

$$\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 3 & -2 & 2 & & 0 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_2=R_2-3R_1$

$$\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & -7/2 & 7/2 & & -3/2 & 1 & 0 \\ 1 & -3 & -3 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & -7/2 & 7/2 & & -3/2 & 1 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}$$

$R_2=-2R_2/7$

$$\begin{pmatrix} 1 & 1/2 & -1/2& & 1/2 & 0 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1-R_2/2$

$$\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & -7/2 & -5/2 & & -1/2 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3+7R_2/2$

$$\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & 0 & -6 & & 1 & -1 & 1 \\ \end{pmatrix}$$

$R_3=-R_3/6$

$$\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & -1 & & 3/7 & -2/7 & 0 \\ 0 & 0 & 1 & & -1/6 & 1/6 & -1/6 \\ \end{pmatrix}$$

$R_2=R_2+R_3$

$$\begin{pmatrix} 1 & 0& 0 & & 2/7 & 1/7 & 0 \\ 0 & 1 & 0 & & 11/42 & -5/42 & -1/6 \\ 0 & 0 & 1 & & -1/6 & 1/6 & -1/6 \\ \end{pmatrix}$$

We are done. On the left is the identity matrix. On the right is the inverse matrix.

$$A^{-1}=\begin{pmatrix} 2/7 & 1/7 & 0 \\ 11/42 & -5/42 & -1/6 \\ -1/6 & 1/6 & -1/6 \\ \end{pmatrix}$$

$$X=\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} 2/7 & 1/7 & 0 \\ 11/42 & -5/42 & -1/6 \\ -1/6 & 1/6 & -1/6 \\ \end{pmatrix}\begin{pmatrix} 5\\ -3 \\ -2 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 10/7-3/7\\ 55/42+15/42+14/42 \\ -5/6-3/6+2/6 \\ \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ -1 \\ \end{pmatrix}$$

$(1, 2, -1)$