Consider the given homogenous system of linear equations.
x 1 − x 2 + 4 x 3 − x 4 − x 5 = 0 − x 1 + x 2 − x 3 + 2 x 4 + x 5 = 0 x 3 − 3 x 4 + x 5 = 0 \begin{aligned}
x_{1}-x_{2}+4 x_{3}-x_{4}-x_{5} &=0 \\
-x_{1}+x_{2}-x_{3}+2 x_{4}+x_{5} &=0 \\
x_{3}-3 x_{4}+x_{5} &=0
\end{aligned} x 1 − x 2 + 4 x 3 − x 4 − x 5 − x 1 + x 2 − x 3 + 2 x 4 + x 5 x 3 − 3 x 4 + x 5 = 0 = 0 = 0
The above system of equations can be expressed as,
[ 1 − 1 4 − 1 − 1 − 1 1 − 1 2 1 0 0 1 − 3 1 ] [ x 2 x 2 x 3 x 4 x 5 ] = [ 0 0 0 0 0 ] . . . ( 1 ) \left[\begin{array}{ccccc}
1 & -1 & 4 & -1 & -1 \\
-1 & 1 & -1 & 2 & 1 \\
0 & 0 & 1 & -3 & 1
\end{array}\right]\left[\begin{array}{l}
x_{2} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0 \\
0
\end{array}\right]
...(1) ⎣ ⎡ 1 − 1 0 − 1 1 0 4 − 1 1 − 1 2 − 3 − 1 1 1 ⎦ ⎤ ⎣ ⎡ x 2 x 2 x 3 x 4 x 5 ⎦ ⎤ = ⎣ ⎡ 0 0 0 0 0 ⎦ ⎤ ... ( 1 )
Let W be the set of solutions of the above linear homogeneous system of equations.
[ i ] W ⊆ R 5 [i] W \subseteq \mathbb{R}^{5} [ i ] W ⊆ R 5
clearly 0 ‾ = [ 0 0 0 0 0 ] ∈ W . \overline{0}=\left[\begin{array}{l}0 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right] \in W. 0 = ⎣ ⎡ 0 0 0 0 0 ⎦ ⎤ ∈ W .
Now, let x 1 , x 2 ∈ N . x_{1}, x_{2} \in \mathbb{N}. x 1 , x 2 ∈ N .
⇒ A X 2 = 0 and A X 2 = 0 \Rightarrow A X_{2}=0 \text { and } A X_{2}=0 ⇒ A X 2 = 0 and A X 2 = 0
Consider,
A ( X 1 + X 2 ) = A X 1 + A X 2 = 0 + 0 = 0 ∴ A ( X 1 + X 2 ) = 0 ⇒ X 1 + X 2 ∈ W . \begin{gathered}
A\left(X_{1}+X_{2}\right)=A X_{1}+A X_{2}=0+0=0 \\
\therefore A\left(X_{1}+X_{2}\right)=0 \Rightarrow X_{1}+X_{2} \in W .
\end{gathered} A ( X 1 + X 2 ) = A X 1 + A X 2 = 0 + 0 = 0 ∴ A ( X 1 + X 2 ) = 0 ⇒ X 1 + X 2 ∈ W .
Thus. vector addition is satisfied.
Further. Let λ ∈ R \lambda \in \mathbb{R} λ ∈ R x ∈ N . x \in \mathbb{N}. x ∈ N .
Now.
⇒ A x = 0 . \Rightarrow \quad A x=0 \text {. } ⇒ A x = 0 .
A ( λ x ) = λ ( A x ) = λ ( 0 ) = 0 A(\lambda x)=\lambda(A x)=\lambda(0)=0 A ( λ x ) = λ ( A x ) = λ ( 0 ) = 0
Thus, A ( λ x ) = 0 ⇒ λ x ∈ W . A(\lambda x)=0 \Rightarrow \lambda x \in W. A ( λ x ) = 0 ⇒ λ x ∈ W .
therefore scalar multiplication is satisfied.
Therefore, ∣ N ⊆ R 5 \mid N \subseteq \mathbb{R}^{5} ∣ N ⊆ R 5
ii) From (1) we have,
[ 1 − 1 4 − 1 − 1 − 1 1 − 1 2 1 0 0 1 − 3 1 ] [ x 2 x 2 x 3 x 4 x 5 ] = [ 0 0 0 0 0 ] \left[\begin{array}{ccccc}
1 & -1 & 4 & -1 & -1 \\
-1 & 1 & -1 & 2 & 1 \\
0 & 0 & 1 & -3 & 1
\end{array}\right]\left[\begin{array}{l}
x_{2} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0 \\
0
\end{array}\right] ⎣ ⎡ 1 − 1 0 − 1 1 0 4 − 1 1 − 1 2 − 3 − 1 1 1 ⎦ ⎤ ⎣ ⎡ x 2 x 2 x 3 x 4 x 5 ⎦ ⎤ = ⎣ ⎡ 0 0 0 0 0 ⎦ ⎤
Apply, R 2 → R 2 + R 1 . R_{2} \rightarrow R_{2}+R_{1}. R 2 → R 2 + R 1 .
[ 1 − 1 4 − 1 − 1 0 0 3 1 0 0 0 1 − 3 1 ] [ x 1 x 2 x 3 x 4 x 5 ] = [ 0 0 0 0 0 ] \left[\begin{array}{ccccc}
1 & -1 & 4 & -1 & -1 \\
0 & 0 & 3 & 1 & 0 \\
0 & 0 & 1 & -3 & 1
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0 \\
0
\end{array}\right] ⎣ ⎡ 1 0 0 − 1 0 0 4 3 1 − 1 1 − 3 − 1 0 1 ⎦ ⎤ ⎣ ⎡ x 1 x 2 x 3 x 4 x 5 ⎦ ⎤ = ⎣ ⎡ 0 0 0 0 0 ⎦ ⎤
[ R 3 → 3 R 3 − R 2 1 − 1 4 − 1 − 1 0 0 3 1 0 0 0 0 − 8 3 ] [ x 2 x 2 x 3 x 4 x 5 ] = [ 0 0 0 0 0 ] \left[\begin{array}{ccccc}
R_{3} & \rightarrow & 3 R_{3}-R_{2} & \\
1 & -1 & 4 & -1 & -1 \\
0 & 0 & 3 & 1 & 0 \\
0 & 0 & 0 & -8 & 3
\end{array}\right]\left[\begin{array}{l}
x_{2} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0 \\
0
\end{array}\right] ⎣ ⎡ R 3 1 0 0 → − 1 0 0 3 R 3 − R 2 4 3 0 − 1 1 − 8 − 1 0 3 ⎦ ⎤ ⎣ ⎡ x 2 x 2 x 3 x 4 x 5 ⎦ ⎤ = ⎣ ⎡ 0 0 0 0 0 ⎦ ⎤
⇒ x 1 − x 2 + 4 x 3 − x 4 − x 5 = 0 3 x 3 + x 4 = 0 − 8 x 4 + 3 x 5 = 0 \begin{array}{r}
\Rightarrow x_{1}-x_{2}+4 x_{3}-x_{4}-x_{5}=0 \\
3 x_{3}+x_{4}=0 \\
-8 x_{4}+3 x_{5}=0
\end{array} ⇒ x 1 − x 2 + 4 x 3 − x 4 − x 5 = 0 3 x 3 + x 4 = 0 − 8 x 4 + 3 x 5 = 0
No. of equations =3
No. of variables =5.
Therefore, No. of free variables =5-3=2.
Let us treat, x 2 x_{2} x 2 x 4 x_{4} x 4
Thus. we have,
x 5 = 8 3 x 4 x 3 = − 1 3 x 4 x_{5}=\frac{8}{3} x_{4} \quad x_{3}=\frac{-1}{3} x_{4} x 5 = 3 8 x 4 x 3 = 3 − 1 x 4
And,
x 1 = x 2 − 4 x 3 + x 4 + x 5 = x 2 + 4 3 x 4 + x 4 + 8 3 x 4 x 1 = x 2 + 5 x 4 \begin{aligned}
x_{1} &=x_{2}-4 x_{3}+x_{4}+x_{5} \\
&=x_{2}+\frac{4}{3} x_{4}+x_{4}+\frac{8}{3} x_{4} \\
x_{1} &=x_{2}+5 x_{4}
\end{aligned} x 1 x 1 = x 2 − 4 x 3 + x 4 + x 5 = x 2 + 3 4 x 4 + x 4 + 3 8 x 4 = x 2 + 5 x 4
X = [ x 1 x 2 x 3 x 4 x 5 ] = [ x 2 + 5 x 4 x 2 − 1 / 3 x 4 x 4 8 / 3 x 4 ] = [ x 2 x 2 0 0 0 ] + [ 5 x 4 0 − 1 / 3 x 4 x 4 8 / 3 x 4 ] = x 2 [ 1 1 0 0 0 ] + x 4 [ 5 0 − 1 / 3 1 8 / 3 ] ∴ [ x 1 x 2 x 3 x 4 x 5 ] = x 2 [ 1 1 0 0 0 ] + x 4 [ 15 0 − 1 3 8 ] where x 2 , x 4 ∈ R \begin{aligned} X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5}\end{array}\right] &=\left[\begin{array}{c}x_{2}+5 x_{4} \\ x_{2} \\ -1 / 3 x_{4} \\ x_{4} \\ 8 / 3 x_{4}\end{array}\right]=\left[\begin{array}{l}x_{2} \\ x_{2} \\ 0 \\ 0 \\ 0\end{array}\right]+\left[\begin{array}{c}5 x_{4} \\ 0 \\ -1 / 3 x_{4} \\ x_{4} \\ 8 / 3 x_{4}\end{array}\right]=x_{2}\left[\begin{array}{l}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]+x_{4}\left[\begin{array}{c}5 \\ 0 \\ -1 / 3 \\ 1 \\ 8 / 3\end{array}\right] \\ & \therefore\left[\begin{array}{c}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5}\end{array}\right]=x_{2}\left[\begin{array}{c}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]+x_{4}\left[\begin{array}{c}15 \\ 0 \\ -1 \\ 3 \\ 8\end{array}\right] \text { where } x_{2}, x_{4} \in \mathbb{R} \end{aligned} X = ⎣ ⎡ x 1 x 2 x 3 x 4 x 5 ⎦ ⎤ = ⎣ ⎡ x 2 + 5 x 4 x 2 − 1/3 x 4 x 4 8/3 x 4 ⎦ ⎤ = ⎣ ⎡ x 2 x 2 0 0 0 ⎦ ⎤ + ⎣ ⎡ 5 x 4 0 − 1/3 x 4 x 4 8/3 x 4 ⎦ ⎤ = x 2 ⎣ ⎡ 1 1 0 0 0 ⎦ ⎤ + x 4 ⎣ ⎡ 5 0 − 1/3 1 8/3 ⎦ ⎤ ∴ ⎣ ⎡ x 1 x 2 x 3 x 4 x 5 ⎦ ⎤ = x 2 ⎣ ⎡ 1 1 0 0 0 ⎦ ⎤ + x 4 ⎣ ⎡ 15 0 − 1 3 8 ⎦ ⎤ where x 2 , x 4 ∈ R
Thus the solution space W is generated by two linearly independent vectors, namely
[ 1 1 0 0 0 ] and [ 15 0 − 1 3 8 ] \left[\begin{array}{l}
1 \\
1 \\
0 \\
0 \\
0
\end{array}\right] \text { and }\left[\begin{array}{c}
15 \\
0 \\
-1 \\
3 \\
8
\end{array}\right] ⎣ ⎡ 1 1 0 0 0 ⎦ ⎤ and ⎣ ⎡ 15 0 − 1 3 8 ⎦ ⎤
Therefore,
Basis for W is,
B = { [ 1 1 0 0 0 ] , [ 15 0 − 1 3 8 ] } B=\left\{\left[\begin{array}{c}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right],\left[\begin{array}{c}15 \\ 0 \\ -1 \\ 3 \\ 8\end{array}\right]\right\} B = ⎩ ⎨ ⎧ ⎣ ⎡ 1 1 0 0 0 ⎦ ⎤ , ⎣ ⎡ 15 0 − 1 3 8 ⎦ ⎤ ⎭ ⎬ ⎫
o r ‾ B = { ( 1 , 1 , 0 , 0 , 0 ) , ( 15 , 0 , − 1 , 3 , 8 ) } ⊆ R 5 \underline{or} \quad B=\{(1,1,0,0,0),(15,0,-1,3,8)\} \subseteq \mathbb{R}^{5} or B = {( 1 , 1 , 0 , 0 , 0 ) , ( 15 , 0 , − 1 , 3 , 8 )} ⊆ R 5
(iii) Thus, dim ( W ) = 2. \operatorname{dim}(W)=2. dim ( W ) = 2.
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