Question

# For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A. A=\begin{bmatrix}2&3&5&-9\\-8&-9&-11&21\\4&-3&-17&27\end{bmatrix} Find a nonzero vector in Nul A. A=\begin{bmatrix}-3\\2\\0\\1\end{bmatrix}

Matrix transformations
For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.
$$A=\begin{bmatrix}2&3&5&-9\\-8&-9&-11&21\\4&-3&-17&27\end{bmatrix}$$
Find a nonzero vector in Nul A.
$$A=\begin{bmatrix}-3\\2\\0\\1\end{bmatrix}$$

2021-06-14
Consider the following matrix,
$$A=\begin{bmatrix}2&3&5&-9\\-8&-9&-11&21\\4&-3&-17&27\end{bmatrix}$$
Find a non-zero vector in Nul A.
Use row reduced method.
$$A=\begin{bmatrix}1&3/2&5/2&-9/2\\-8&-9&-11&21\\4&-3&-17&27\end{bmatrix}\quad R_1\to\frac{1}{2}R_1$$
$$\approx\begin{bmatrix}1&3/2&5/2&-9/2\\0&3&9&-15\\0&-9&-27&45\end{bmatrix}\quad R_2\to R_2+8R_1,R_3\to R_3-4R_1$$
$$\approx\begin{bmatrix}1&3/2&5/2&-9/2\\0&3&9&-15\\0&0&0&0\end{bmatrix}\quad R_3\to R_3+3R_2$$
$$\approx\begin{bmatrix}1&0&-2&3\\0&1&3&-5\\0&0&0&0\end{bmatrix}\quad R_1\to R_1-\frac{1}{2}R_2$$
The matrix has 2 pivot columns and 2 free columns.
Then, AX=0
$$\begin{bmatrix}1&0&-2&3\\0&1&3&-5\\0&0&0&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
$$x_1-2x_3+3x_4=0\Rightarrow x_1=2x_3-3x_4$$
$$x_2+3x_3-5x_4=0\Rightarrow x_2=-3x_3+5x_4$$
So, the general solution is,
$$\begin{cases}x_1=2x_3-3x_4\\x_2=-3x_3+5x_4\\x_3\ and\ x_4 \text{ are free variables}\end{cases}$$
Then, a vector parametric solution is,
$$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}2x_3-3x_4\\-3x_3+5x_4\\x_3\\x_4\end{bmatrix}$$
$$=\begin{bmatrix}2x_3\\-3x_3\\x_3\\x_4\end{bmatrix}+\begin{bmatrix}-3x_4\\5x_4\\0\\x_4\end{bmatrix}$$
$$=\begin{bmatrix}2\\-3\\1\\0\end{bmatrix}x_3+\begin{bmatrix}-3\\5\\0\\1\end{bmatrix}x_4$$
Therefore, the required non-zero vectors in Nul A are,
$$(2,-3,1,0),(-3,5,0,1)$$
Find a non zero vector in Col A
observe the equation (1), the first two columns of the matrix
$$\begin{bmatrix}1&0&-2&3\\0&1&3&-5\\0&0&0&0\end{bmatrix}$$ are pivot columns
these columns in the original matrix define the Column Space of the matrix.
Column Space $$=p\begin{bmatrix}2\\-8\\4\end{bmatrix}+q\begin{bmatrix}3\\-9\\-3\end{bmatrix}$$ where p and q are any real numbers.
Therefore, the required nonzero vectors in Col A are,
$$(2,-8,4),(3,-9,3)$$