Find an explicit description of $\text{Nul } A$ by listing vectors that span the null space.

$A = \begin{bmatrix} 1 & 3 & 6 & 0 \\ 0 & 1 & 4 & -5 \end{bmatrix}$

Question :

Find an explicit description of $\text{nul } a$ by listing vectors that span the null space.

$a = \begin{bmatrix} 1 & 3 & 6 & 0 \\ 0 & 1 & 4 & -5 \end{bmatrix}$

$Find an explicit description of \text{nul } a by listing vectors that span the | Doubtlet.com$

Solution:

Neetesh Kumar | October 15, 2024

Linear Algebra Homework Help

This is the solution to Math2B Course: Linear Algebra
Assignment: Ch4 Section 2-3 Question Number 1
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Step-by-step solution:

To find the null space of matrix $A$ , we solve the equation $A \mathbf{x} = \mathbf{0}$ , where $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ and $A \mathbf{x} = 0$

Means solving:

$\begin{bmatrix} 1 & 3 & 6 & 0 \\ 0 & 1 & 4 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This gives the system of linear equations :

$x_1 + 3x_2 + 6x_3 = 0$

$x_2 + 4x_3 - 5x_4 = 0$

Step 1: Solve the system

From the second equation, solve for $x_2$ :

$x_2 = -4x_3 + 5x_4$

Substitute this into the first equation:

$x_1 + 3(-4x_3 + 5x_4) + 6x_3 = 0$

$x_1 - 12x_3 + 15x_4 + 6x_3 = 0$

$x_1 - 6x_3 + 15x_4 = 0$

$x_1 = 6x_3 - 15x_4$

Step 2: Express the solution in parametric vector form

Let $x_3 = t$ and $x_4 = s$ , where $t$ and $s$ are free parameters. Then:

$x_1 = 6t - 15s$

$x_2 = -4t + 5s$

$x_3 = t, \quad x_4 = s$

Thus, the general solution for the null space is:

$\mathbf{x} = t \begin{bmatrix} 6 \\ -4 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} -15 \\ 5 \\ 0 \\ 1 \end{bmatrix}$

Step 3: Conclusion

The null space of $A$ , $\text{Nul } A$ , is spanned by the vectors:

$\left\{ \begin{bmatrix} 6 \\ -4 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -15 \\ 5 \\ 0 \\ 1 \end{bmatrix} \right\}$

Final Answer:

A spanning set for $\text{Nul } A$ is: