Question

5. Let 5 6 7 8 9 10 11 12 Give a basis for its row space S and a basis for its column space Se

Answer 1

$\small 5.$

Solution:

We first reduced the matrix $\small Y$ to a row echelon form

$\small \begin{pmatrix} 1 & 2 & 3&4 \\ 5 & 6 & 7 &8 \\ 9 &10 &11 &12 \end{pmatrix}$

By _{$\small R_{2}-5R_{1}\: \: \: \: ,\: \: \: R_{3}-9R_{1}$}

$\small \begin{pmatrix} 1 & 2 & 3&4 \\ 0 & -4 & -8 &-12 \\ 0 &-8 &-16 &-24 \end{pmatrix}$

By   $\small R_{3}-2R_{2}$

$\small \begin{pmatrix} 1 & 2 & 3&4 \\ 0 & -4 & -8 &-12 \\ 0 &0 &0 &0 \end{pmatrix}$

By   _{$\small R_{2}\times \frac{-1}{4}$}

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

By $\small R_{1}-2R_{2}$

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

The nonzero rows in the row echelon form of $\small Y$   are linearly independent .

$\small {\color{DarkBlue} \therefore \left \{ \left ( 1\: ,\: 0\: ,\: -1\: ,\: -2 \right )\: ,\: \left ( 0\: ,\: 1\: ,\: 2\: ,\: 3\: \right ) \right \}}$   form a basis for $\small S_{r}$

In the row reduced form of $\small Y$ the first and the second column are pivot columns.

$\small \therefore$ The first and the second column of the original matrix form a basis for column space of $\small Y$

$\small {\color{DarkRed} \therefore \left \{ \begin{bmatrix} 1\\ 5\\ 9\\ \end{bmatrix}\: ,\: \begin{bmatrix} 2\\ 6\\ 10\\ \end{bmatrix} \right \}}$   form a basis for $\small S_{c}$