# Construct a matrix whose column space contains (1, 1, 5) and (0, 3, 1) while it’s null space contains (1, 1, 2).

This question aims to understand the construction of a matrix under given constraints. To solve this question, we need to have a clear understanding of the terms column space and null space.

The space which is spanned by the column vectors of a given matrix is called its column space.

The space which is spanned by all the column vectors of a matrix ( say $A$ ) that satisfy the following condition:

$A x = 0$

In short, it’s the solution to the above system of linear equations.

Under given conditions, we can construct the following matrix:

$\left [ \begin{array}{ c c c } 1 & 0 & x \\ 1 & 3 & y \\ 5 & 1 & z \end{array} \right ]$

Since (1, 1, 2) is a solution to the null space of the given matrix, it must satisfy the following system:

$\left [ \begin{array}{ c c c } 1 & 0 & x \\ 1 & 3 & y \\ 5 & 1 & z \end{array} \right ] \left [ \begin{array}{ c } 1 \\ 1 \\ 2 \end{array} \right ] = \left [ \begin{array}{ c } 0 \\ 0 \\ 0 \end{array} \right ]$

$\left \{ \begin{array}{ c } (1)(1) + (0)(1) + (x)(2) = 0 \\ (1)(1) + (3)(1) + (y)(2) = 0 \\ (5)(1) + (1)(1) + (z)(2) = 0 \end{array} \right.$

$\left \{ \begin{array}{ c } 2x + 1 = 0 \\ 2y + 4 = 0 \\ 2z + 6 = 0 \end{array} \right.$

$\left \{ \begin{array}{ c } x = \dfrac{ -1 }{ 2 } \\ y = -2 \\ z = -3 \end{array} \right.$

Hence, the required matrix is:

$\left [ \begin{array}{ c c c } 1 & 0 & \dfrac{ -1 }{ 2 } \\ 1 & 3 & -2 \\ 5 & 1 & -3 \end{array} \right ]$

## Numerical Result

$\left [ \begin{array}{ c c c } 1 & 0 & \dfrac{ -1 }{ 2 } \\ 1 & 3 & -2 \\ 5 & 1 & -3 \end{array} \right ]$

## Example

Construct a matrix with column space comprising of (1, 2, 3) and (4, 5, 6) while its null space contains (7, 8, 9).

Under given constraints:

$\left [ \begin{array}{ c c c } 1 & 4 & x \\ 2 & 5 & y \\ 3 & 6 & z \end{array} \right ] \left [ \begin{array}{ c } 7 \\ 8 \\ 9 \end{array} \right ] = \left [ \begin{array}{ c } 0 \\ 0 \\ 0 \end{array} \right ]$

$\left \{ \begin{array}{ c } (1)(7) + (4)(8) + (x)(9) = 0 \\ (2)(7) + (5)(8) + (y)(9) = 0 \\ (3)(7) + (6)(8) + (z)(9) = 0 \end{array} \right.$

$\left \{ \begin{array}{ c } 9x + 39 = 0 \\ 9y + 54 = 0 \\ 9z + 69 = 0 \end{array} \right.$

$\left \{ \begin{array}{ c } x = – \dfrac{ 13 }{ 3 } \\ y = – 6 \\ z = – \dfrac{ 23 }{ 3 } \end{array} \right.$

Hence, the required matrix is:

$\left [ \begin{array}{ c c c } 1 & 4 & – \dfrac{ 13 }{ 3 } \\ 2 & 5 & -6 \\ 3 & 6 & – \dfrac{ 23 }{ 3 } \end{array} \right ]$