Answer to Question #203407 in Linear Algebra for Rajay Myrie

Question

Given the system of equations 2 3 2 11 3 2 3 7 4 4 14 x y z x y z x y z + − = − − + = − + = , find the values of x, y, and z using matrix inversion. 

Answer 1

$$2x+3y-2z=11$$

$$-3x-2y+3z=7$$

$$4x-4y+z=14$$

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

Augment the matrix with the identity matrix:

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

$R_1=R_1/2$

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

$R_2=R_2+3R_1$

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

$R_3=R_3-4R_1$

$$\begin{pmatrix} 1 & 3/2 & -1 & & 1/2 & 0 & 0 \\ 0 & 5/2 & 0 & & 3/2 & 1 & 0 \\ 0 & -10 & 5 & & -2 & 0 & 1\\ \end{pmatrix}$$

$R_2=(2/5)R_2$

$$\begin{pmatrix} 1 & 3/2 & -1 & & 1/2 & 0 & 0 \\ 0 & 1 & 0 & & 3/5 & 2/5 & 0 \\ 0 & -10 & 5 & & -2 & 0 & 1\\ \end{pmatrix}$$

$R_1=R_1-(3/2)R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & -2/5 & -3/5 & 0 \\ 0 & 1 & 0 & & 3/5 & 2/5 & 0 \\ 0 & -10 & 5 & & -2 & 0 & 1\\ \end{pmatrix}$$

$R_3=R_3+10R_2$

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

$R_3=R_3/5$

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

$R_1=R_1+R_3$

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

On the left is the identity matrix. On the right is the inverse matrix.

$$A^{-1}=\begin{pmatrix} 2/5 & 1/5 & 1/5 \\ 3/5 & 2/5 & 0 \\ 4/5 & 4/5 & 1/5\\ \end{pmatrix}$$ $$AX=B=>A^{-1}AX=A^{-1}B=>X=A^{-1}B$$

$$X=\begin{pmatrix} x\\ y \\ z\\ \end{pmatrix}, B=\begin{pmatrix} 11\\ 7 \\ 14\\ \end{pmatrix}$$

$$A^{-1}B=\begin{pmatrix} 2/5 & 1/5 & 1/5 \\ 3/5 & 2/5 & 0 \\ 4/5 & 4/5 & 1/5\\ \end{pmatrix}\begin{pmatrix} 11\\ 7 \\ 14\\ \end{pmatrix}$$

$$=\begin{pmatrix} (22+7+14)/5\\ (33+14+0)/5 \\ (44+28+14)/5\\ \end{pmatrix}=\begin{pmatrix} 43/5\\ 47/5 \\ 86/5\\ \end{pmatrix}$$

$$x=\dfrac{43}{5}, y=\dfrac{47}{5}, z=\dfrac{86}{5}$$

$$(\dfrac{43}{5},\dfrac{47}{5}, \dfrac{86}{5})$$