Question

Show that the set of vectors {(−4,1,3),(5,1,6),(6,0,2)} does not span R 3 , but that it does span the subspace of R 3 consisting of all vectors lying in the plane with equation x + 13y − 3z = 0.

Answer 1

Answer:

The vectors does not span R3 and only span a subspace of R3 which satisfies x+13y-3z=0

Explanation:

The vectors are given as

$v_1=\left[\begin{array}{c}-4&1&3\end{array}\right] \\v_2=\left[\begin{array}{c}-5&1&6\end{array}\right] \\v_3=\left[\begin{array}{c}6&0&2\end{array}\right]$

Now if the vectors  would span the $R^3$, the rank of  the consolidated matrix will be 3 if it is not 3 this indicates that the vectors does not span the $R^3$.

So the matrix is given as

$M=\left[\begin{array}{ccc}v_1&v_2&v_3\end{array}\right] \\M=\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]$

In order to calculate the rank, the matrix is reduced to the Row Echelon form as

$\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 3&6&2\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}$

$\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 0&\frac{39}{4}&\frac{13}{2}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}$

$\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2\\ 0&\frac{9}{4}&\frac{3}{2}}\end{array}\right] R_2\:\leftrightarrow \:R_3$

$\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2}\\ 0&0&0\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}$

As the Rank is given as number of non-zero rows in the Row echelon form which are 2 so the rank is 2.

Thus this indicates that the vectors does not span $R^3$

Now for any vector the corresponding equation is formulated by using the combined matrix which is given as  for any arbitrary vector and the coordinate as

$v=\left[\begin{array}{c}x&y&z\end{array}\right]$

$c=\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]$

$Mc=v$

$\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]=\left[\begin{array}{c}x&y&z\end{array}\right]$

$M=\left[\begin{array}{ccccc}v_1&v_2&v_3& | &v\end{array}\right] \\M=\left[\begin{array}{ccccc}-4&5&6&|&x\\1&1&0&|&y\\3&6&2&|&z\end{array}\right]$

Now converting the combined matrix as

$\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 3&6&2&|&z\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}$

$\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}$

$\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\end{array}\right] R_3 \leftrightarrow R_2$

$\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&0&0&|&\frac{13y+x-3z}{13}\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}$

From this it is seen that whatever the values of the coordinates does not effect the value of the plane with equation as

$\frac{13y+x-3z}{13}=0\\or\\13y+x-3z=0$

So it is verified that the subspace of R3 such that it satisfies x+13y-3z=0 consists of all vectors.