Question

Given the homogeneous system of linear equations:

x1 + 2x2 − 2x3 + 2x4 − x5 = 0

x1 + 2x2 − x3 + 3x4 − 2x5 = 0

2x1 + 4x2 − 7x3 + x4 + x5 = 0

4.1. Write out the augmented matrix for the system of equations.

2.2. Solve the system by Gauss elimination method to the augmented matrix and determine a basis and the dimension of the solution space S of the homogeneous system.

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Answer 1

• The augmented matrix for the system of equation above is

$\begin{pmatrix} 1&2&-2&2&-1&0\\1&2&-1&3&-2&0\\2&4&-7&4&1&0 \end{pmatrix}$

• Using Gaus Elimination, we reduce our augmented matrix to row echelon form by performing some row operations

$R_{21}(-1)\implies\begin{pmatrix} 1&2&-2&2&-1&0\\0&0&1&1&-1&0\\2&4&-7&1&1&0 \end{pmatrix}\\R_{31}(-2)\implies \begin{pmatrix} 1&2&-2&2&-1&0\\0&0&1&1&-1&0\\0&0&-3&-3&3&0\\ \end{pmatrix}\\R_{32}(3)\implies\begin{pmatrix} 1&2&-2&2&-1&0\\0&0&1&1&-1&0\\0&0&0&0&0&0 \end{pmatrix}$

Using back substitution,

$x_{3}+x_{4}-x_{5} =0\text{ let }x_{4}=s \text{ and }x_{5}=t\\\implies x_{3}=t-s$

Also,

$x_{1}+2x_{2}-2x_{3}+2x_{4}-x_{5} =0\text{, let }x_{2}= l\\\implies x_{1}=-2x_{2}+2x_{3}-2x_{4}+x_{5} =0\\\text{Therefore, } x_{1}=-2l+2(t-s)-2s+t\\x_2=l\\x_3=t-s\\x_4=s\\x_5=t\\\text{The basis for the solution space can be wriitten as follows }(-2l,l,0,0,0)+(3t,0,t,0,t)+(-4s,0,-s,s,0)\implies l(-2,1,0,0,0)+t(3,0,1,0,1)+s(-4,0,-1,1,0)$

Therefore the basis for the solution space S is $\\(-2,1,0,0,0),(3,0,1,0,1),(-4,0,-1,1,0)$

and the dimension is 3