Answer to Question #350968 in Linear Algebra for Nhm

Question

Solve the system of linear equations using matrix inverse method.

x-y+z=1

2y-z=1

2x+3y=1

Answer 1

$$A=\begin{pmatrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{pmatrix}, X=\begin{pmatrix} x \\ y\\ z \end{pmatrix}, B=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}$$

$$AX=B$$

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

augment the matrix with the identity matrix:

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

$R_3=R_3-2R_1$

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

$R_2=R_2/2$

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

$R_1=R_1+R_2$

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

$R_3=R_3-5R_2$

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

$R_3=2R_3$

$$\begin{pmatrix} 1 & 0 & 1/2 & & 1& 1/2 & 0 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}$$

$R_1=R_1-R_3/2$

$$\begin{pmatrix} 1 & 0 & 0 & & 3 & 3 & -1 \\ 0 & 1 & -1/2 & & 0& 1/2 & 0 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}$$

$R_2=R_2+R_3/2$

$$\begin{pmatrix} 1 & 0 & 0 & & 3 & 3 & -1 \\ 0 & 1 & 0 & & -2 & -2 & 1 \\ 0 & 0 & 1 & & -4 & -5 & 2 \\ \end{pmatrix}$$

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

$$A^{-1}=\begin{pmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{pmatrix}$$

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

$$=\begin{pmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{pmatrix}\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}$$

$$=\begin{pmatrix} 3+3-1\\ -2-2+1\\ -4-5+2 \end{pmatrix}=\begin{pmatrix} 5\\ -3\\ -7 \end{pmatrix}$$

$(5, -3, -7)$