$$eginpmatrix4 & -2 \-8 & 4 \endpmatrix imeseginpmatrixunknown & unknown \ unrecognized & unrecognized \endpmatrix= eginpmatrix0 & 0 \ 0 & 0 \endpmatrix$$

What I tried doing is replacing every one of the unknowns by a arbitrary variables, so $x,y,s,t$. that provided me $4$ variables (unknown), and $4$ equations.

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i readjusted the top left unwell-known to x, top ideal to y, reduced left to s, reduced appropriate to t. i finished up acquiring $4x-2s=0$, $4y-2t=0$, $-8x+4s=0$, $-8y+4t=0$...

I tried isolating variables, but I ended up no wright here (retained gaining 0=0).

linear-algebra matrices
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edited Oct 14 "18 at 5:55
Bakarr
asked Oct 14 "18 at 5:44

BakarrBakarr
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This is the answer that I offered for the post that you deleted:

Let"s take into consideration

$$eginbmatrix 4 &- 2 \ - 8 & 4endbmatrixeginbmatrix x & y\ s & tendbmatrix= eginbmatrix 0 & 0\ 0 & 0endbmatrix$$

We can focus on$$eginbmatrix 4 &- 2 \ - 8 & 4endbmatrixeginbmatrix x \ s endbmatrix= eginbmatrix 0 \ 0 endbmatrix$$

Keep in mind that the 2 equations are dependent, hence

$$eginbmatrix 2 &- 1 \ 0 & 0endbmatrixeginbmatrix x \ s endbmatrix= eginbmatrix 0 \ 0 endbmatrix$$

That is we have actually $$2x-s=0$$

that is $s=2x$ is a solution.

Similarly, $t=2y$ is a solution, that is whenever before, we decide $x$ and $y$, we can recover our $s$ and also $t$.

There are non-trivial services to the system.

You simply have to pick $x=y=1$, then you acquire $s=t=2$ to acquire a feasible solution.

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answered Oct 14 "18 at 5:53

Siong Thye GohSiong Thye Goh
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Let $$eginpmatrixa & b \c & d \endpmatrix$$ signify the unknown matrix you want to find.

You acquire four equations,

$$-8a+4c=0$$$$4a-2c=0$$$$4b-2d=0$$$$-8a+4d=0$$

First equation say that $-8a+4c=0$ in other words $8a=4c$ which even more implies that $c=2a$. similarly you deserve to usage the 3rd equation to conclude that $d=2b$

So you have these 2 relation $$c=2a$$ and also $$d=2b$$It suggests you are totally free to pick $a$ and also then for $c$ you have to just double the value of $a$ you picked. Similarly for $b$ and $d$.

In specific,

Take $a=1$ then 2 times of $1$ is $2$ so $c=2$,

similarly $b=1,d=2$ and check out whether they satisfy the equation or not.

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edited Oct 14 "18 at 6:04
answered Oct 14 "18 at 5:58

Shweta AggrawalShweta Aggrawal
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