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we've got the following 4x4 Matrix

$$\begin{pmatrix} 4 & -2 & 3 & 2\\ 3 & 5 & 1 & -4\\ -1 & 6 & -4 & -7\\ -2 & 0 & -2 & 4 \end{pmatrix}$$

and I need to find $B$ from the equation: $(A-3I)B=0$.

i started to solve it by finding first $A-3I$. and I got:

$$\begin{pmatrix} 1 & -2 & 3 & 2\\ 3 & 2 & 1 & -4\\ -1 & 6 & -7 & -7\\ -2 & 0 & -2 & 1 \end{pmatrix}$$

now I know that every column $[AB]^j$ [$j$ represents column number] can be calculated by $A[B]^j$ [$j$ represents column number].

and I was trying to solve it by multiplying $A$ with a specific column in $B$ but I wasn't able to reach the zero matrix.

EDIT: almost forgot to mention that B needs to be 4x4 matrix and different form 0!

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  • $\begingroup$ Is $B$ a matrix? I can choose $B$ to be the $0$-matrix, that will certainly work. What exactly do you need to find here? $\endgroup$ – Mankind Apr 26 '15 at 13:38
  • $\begingroup$ @HowDoIMath, $B$ needs to be different than the 0-matirx and has to be a 4x4 matrix. sorry I forgot to mention that. I just edited the question. $\endgroup$ – Dan Revah Apr 26 '15 at 13:39
  • $\begingroup$ I amounts to checking $3$ is an eigenvalue for $A$ and finding $4$ eigenvectors for that eigenvalue. $\endgroup$ – Bernard Apr 26 '15 at 13:41
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by row reducing $A-3I,$ using my ti-83, i get $$\pmatrix{1&-2&3&2\\3&2&1&-4\\-1&6&-7&-7\\-2&0&-2&1}\to\pmatrix{1&0&1&-0.5\\0&1&-1&-1.25\\0&0&0&0\\0&0&0&0} $$ we can see that two linearly independent vectors $a, b$ such that $(A-3I)a = 0, (A-3I)b = 0$ where $$a=\pmatrix{0.5\\1.25\\0\\1}, b = \pmatrix{-1\\1\\1\\0}.$$ you can make a matrix $B$ made up of columns $a, b, a, b.$ in fact each column can be any linear combination of $a, b$ would do as well.

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  • $\begingroup$ only $a$ worked for me when I tried to multiply with $b$ I didn't get the zero matrix and when saying $a$ I mean 4x4 matrix of this column vector. Thanks! $\endgroup$ – Dan Revah Apr 26 '15 at 14:04
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    $\begingroup$ @DinRevah, there was a typo at $3,1$ element of the row echelon from. i had a zero instead of $1.$ i have fixed it now. $\endgroup$ – abel Apr 26 '15 at 14:18
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$$\begin{pmatrix} 1 & -2 & 3 & 2\\ 3 & 2 & 1 & -4\\ -1 & 6 & -7 & -7\\ -2 & 0 & -2 & 1 \end{pmatrix}$$

is a zero divisor since : column$(3)=$column$(1)-$column$(2)$.

So the matrix: $$B=\begin{pmatrix} 0 & 0 & -a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$ gives $AB=0$

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  • $\begingroup$ you said column(3)=column(1)−column(1) you mind explaining a bit more? I didn't understand exactly what you ment. $\endgroup$ – Dan Revah Apr 26 '15 at 14:24
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    $\begingroup$ Sorry. It was a typo.... I edit. $\endgroup$ – Emilio Novati Apr 26 '15 at 14:32

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