Question

Solve by finding the basis over R for the solution space.

(A) X + 3y -3z=0

2x - 3y + z=0

3x -2y + 2z=0

(B) X + Y + Z + W=0

2X + 3Y - Z +W=0

3X + 4Y +2W=0

Answer 1

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

Performing row operations on the matrix, we have

$\begin{pmatrix} 1 & 3 & -3\\ 2&-3&1\\ 3&-2&2 \end{pmatrix} \to^{R_{21}(-2)}_{R_{31}(-3)} \begin{pmatrix} 1&3&-3\\ 0&-9&7\\ 0&-11&11 \end{pmatrix}$

$\to R_2(\frac{-1}{9}) \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&-11&11 \end{pmatrix} \to^{R_{32}(11)} \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&0&\frac{22}{9} \end{pmatrix}$

$\to R_3(\frac{9}{22}) \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&0&1 \end{pmatrix}$

$x+3y-3z=0\\ y-\frac{7}{9}z = 0\\ z=0$

Therefore x=0,y=0, z=0. This is a trivial solution hence it is not a basis

Question B

$\begin{pmatrix} 1&1&1&1\\ 2&3&-1&1\\ 3&4&0&2 \end{pmatrix} \begin{pmatrix} x\\y\\z\\w \end{pmatrix} = \begin{pmatrix} 0\\0 \\ 0 \end{pmatrix}$

The matrix of the coefficient is

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

$x+y+z+w=0\\ y-3z-w=0$

Let $z=a, w=b$ where $a,b \in \mathbb{R}$

So, we have

$y-3a-b=0 \\ y = 3a+b \\ x = -y-z-w \\ x= -(3a+b) -a-b \\ x= -4a-2b\\ \therefore x = -4a-2b, y = 3a+b , z =a, w=b \\ (x,y,z,w) = a(-4,3,1,0)+b(-2,1,0,1)$

So a basis for the system is {(-4,3,1,0),(-2,1,0,1)}