# 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}

For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.
$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$
Find a nonzero vector in Nul A.
$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$
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Willie
Consider the following matrix,
$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$
Find a non-zero vector in Nul A.
Use row reduced method.
$A=\left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]\phantom{\rule{1em}{0ex}}{R}_{1}\to \frac{1}{2}{R}_{1}$
$\approx \left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ 0& 3& 9& -15\\ 0& -9& -27& 45\end{array}\right]\phantom{\rule{1em}{0ex}}{R}_{2}\to {R}_{2}+8{R}_{1},{R}_{3}\to {R}_{3}-4{R}_{1}$
$\approx \left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ 0& 3& 9& -15\\ 0& 0& 0& 0\end{array}\right]\phantom{\rule{1em}{0ex}}{R}_{3}\to {R}_{3}+3{R}_{2}$
$\approx \left[\begin{array}{cccc}1& 0& -2& 3\\ 0& 1& 3& -5\\ 0& 0& 0& 0\end{array}\right]\phantom{\rule{1em}{0ex}}{R}_{1}\to {R}_{1}-\frac{1}{2}{R}_{2}$
The matrix has 2 pivot columns and 2 free columns.
Then, AX=0
$\left[\begin{array}{cccc}1& 0& -2& 3\\ 0& 1& 3& -5\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
${x}_{1}-2{x}_{3}+3{x}_{4}=0⇒{x}_{1}=2{x}_{3}-3{x}_{4}$
${x}_{2}+3{x}_{3}-5{x}_{4}=0⇒{x}_{2}=-3{x}_{3}+5{x}_{4}$
So, the general solution is,

Then, a vector parametric solution is,
$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}2{x}_{3}-3{x}_{4}\\ -3{x}_{3}+5{x}_{4}\\ {x}_{3}\\ {x}_{4}\end{array}\right]$
$=\left[\begin{array}{c}2{x}_{3}\\ -3{x}_{3}\\ {x}_{3}\\ {x}_{4}\end{array}\right]+\left[\begin{array}{c}-3{x}_{4}\\ 5{x}_{4}\\ 0\\ {x}_{4}\end{array}\right]$
$=\left[\begin{array}{c}2\\ -3\\ 1\\ 0\end{array}\right]{x}_{3}+\left[\begin{array}{c}-3\\ 5\\ 0\\ 1\end{array}\right]{x}_{4}$
Therefore, the required non-zero vectors in Nul A are,
$\left(2,-3,1,0\right),\left(-3,5,0,1\right)$
Find a non zero vector in Col A
observe the equation (1), the first two columns of the matrix