Find a nonzero vector in Nul A.

mattgondek4
2021-06-13
Answered

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.

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Find a nonzero vector in Nul A.

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Willie

Answered 2021-06-14
Author has **95** answers

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.

$A=\left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]{\textstyle \phantom{\rule{1em}{0ex}}}{R}_{1}\to \frac{1}{2}{R}_{1}$

$\approx \left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ 0& 3& 9& -15\\ 0& -9& -27& 45\end{array}\right]{\textstyle \phantom{\rule{1em}{0ex}}}{R}_{2}\to {R}_{2}+8{R}_{1},{R}_{3}\to {R}_{3}-4{R}_{1}$

$\approx \left[\begin{array}{cccc}1& 3/2& 5/2& -9/2\\ 0& 3& 9& -15\\ 0& 0& 0& 0\end{array}\right]{\textstyle \phantom{\rule{1em}{0ex}}}{R}_{3}\to {R}_{3}+3{R}_{2}$

$\approx \left[\begin{array}{cccc}1& 0& -2& 3\\ 0& 1& 3& -5\\ 0& 0& 0& 0\end{array}\right]{\textstyle \phantom{\rule{1em}{0ex}}}{R}_{1}\to {R}_{1}-\frac{1}{2}{R}_{2}$

The matrix has 2 pivot columns and 2 free columns.

Then, AX=0

$\left[\begin{array}{cccc}1& 0& -2& 3\\ 0& 1& 3& -5\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

${x}_{1}-2{x}_{3}+3{x}_{4}=0\Rightarrow {x}_{1}=2{x}_{3}-3{x}_{4}$

${x}_{2}+3{x}_{3}-5{x}_{4}=0\Rightarrow {x}_{2}=-3{x}_{3}+5{x}_{4}$

So, the general solution is,

$\{\begin{array}{l}{x}_{1}=2{x}_{3}-3{x}_{4}\\ {x}_{2}=-3{x}_{3}+5{x}_{4}\\ {x}_{3}\text{}and\text{}{x}_{4}\text{are free variables}\end{array}$

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,

$(2,-3,1,0),(-3,5,0,1)$

Find a non zero vector in Col A

observe the equation (1), the first two columns of the matrix

Find a non-zero vector in Nul A.

Use row reduced method.

The matrix has 2 pivot columns and 2 free columns.

Then, AX=0

So, the general solution is,

Then, a vector parametric solution is,

Therefore, the required non-zero vectors in Nul A are,

Find a non zero vector in Col A

observe the equation (1), the first two columns of the matrix

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