Answer to Question #350512 in Linear Algebra for Busi

Question

Use the matrix method (together with elementary row transformations) to solve the following:

2x −y +3z = 2

x +2y −z = 4

−4x +5y +z = 10

Answer 1

$A=\begin{pmatrix} 2 & -1 & 3 \\ 1 & 2 & -1\\ -4 & 5 & 1 \\ \end{pmatrix}, X=\begin{pmatrix} x \\ y \\ z \end{pmatrix}, B=\begin{pmatrix} 2\\ 4\\ 10 \end{pmatrix}$

$AX=B$

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

$X=A^{-1}B$

Augment the matrix with the identity matrix:

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

$R_1=R_1/2$

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

$R_2=R_2-R_1$

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

$R_3=R_3+4R_1$

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

$R_2=2R_2/5$

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

$R_1=R_1+R_2/2$

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

$R_3=R_3-3R_2$

$\begin{pmatrix} 1 & 0 & 1 && 2/5 & 1/5 & 0 \\ 0 & 1 & -1 && -1/5 & 2/5 & 0\\ 0 & 0 & 10 && 13/5 & -6/5 & 1 \\ \end{pmatrix}$

$R_3=R_3/10$

$\begin{pmatrix} 1 & 0 & 1 && 2/5 & 1/5 & 0 \\ 0 & 1 & -1 && -1/5 & 2/5 & 0\\ 0 & 0 & 1 && 13/50 & -6/50 & 1/10 \\ \end{pmatrix}$

$R_1=R_1-R_3$

$\begin{pmatrix} 1 & 0 & 0 && 7/50 & 8/25 & -1/10 \\ 0 & 1 & -1 && -1/5 & 2/5 & 0\\ 0 & 0 & 1 && 13/50 & -6/50 & 1/10 \\ \end{pmatrix}$

$R_2=R_2+R_3$

$\begin{pmatrix} 1 & 0 & 0 && 7/50 & 8/25 & -1/10 \\ 0 & 1 & 0 && 3/50 & 7/25 & 1/10\\ 0 & 0 & 1 && 13/50 & -6/50 & 1/10 \\ \end{pmatrix}$

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

$A^{-1}B=\begin{pmatrix} 0.14 & 0.32 & -0.1 \\ 0.06 & 0.28 & 0.1 \\ 0.26 & -0.12 & 0.1 \\ \end{pmatrix}\begin{pmatrix} 2\\ 4\\ 10 \end{pmatrix}$

$=\begin{pmatrix} 0.28+1.28-1\\ 0.12+1.12+1\\ 0.52-0.48+1 \end{pmatrix}=\begin{pmatrix} 0.56\\ 2.24\\ 1.04 \end{pmatrix}$

$(x,y,z)=(0.56, 2.24, 1.04)$