Answer:
![\left[\begin{array}{ccc}x1\\x2\\x3\\x4\end{array}\right] =\left[\begin{array}{ccc}4\\2\\2\\3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5C%5Cx4%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C2%5C%5C3%5Cend%7Barray%7D%5Cright%5D)
4KO2 + 2CO2 → 2K2CO3 + 3O2
Step-by-step explanation:
First we write the reaction in the asked form :
x1 KO2 + x2 CO2 → x3 K2CO3 + x4 O2
In a chemical equation, tha amount of substance that react is the same amount that is formed :
On each side of the equation we must have the same amount of K,C and O.
Let's write this in equations :
K balance : 
C balance : 
O balance : 
The linear equations system is :

Let's equal to 0 to obtain a homogeneous equations system (for convenience) :

Let's work with the extended system matrix :
![\left[\begin{array}{ccccc}1&0&-2&0&0\\2&2&-3&-2&0\\0&1&-1&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C2%262%26-3%26-2%260%5C%5C0%261%26-1%260%260%5Cend%7Barray%7D%5Cright%5D)
Working with the matrix :![\left[\begin{array}{ccccc}1&0&-2&0&0\\0&2&1&-2&0\\0&1&-1&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C0%262%261%26-2%260%5C%5C0%261%26-1%260%260%5Cend%7Barray%7D%5Cright%5D)
→
![\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&1/2&-1&0\\0&1&-1&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C0%261%261%2F2%26-1%260%5C%5C0%261%26-1%260%260%5Cend%7Barray%7D%5Cright%5D)
→![\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&1&1/2&-1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C0%261%26-1%260%260%5C%5C0%261%261%2F2%26-1%260%5Cend%7Barray%7D%5Cright%5D)
→![\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&3/2&-1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C0%261%26-1%260%260%5C%5C0%260%263%2F2%26-1%260%5Cend%7Barray%7D%5Cright%5D)
→![\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&1&-2/3&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%26-2%260%260%5C%5C0%261%26-1%260%260%5C%5C0%260%261%26-2%2F3%260%5Cend%7Barray%7D%5Cright%5D)
→![\left[\begin{array}{ccccc}1&0&0&-4/3&0\\0&1&0&-2/3&0\\0&0&1&-2/3&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%260%26-4%2F3%260%5C%5C0%261%260%26-2%2F3%260%5C%5C0%260%261%26-2%2F3%260%5Cend%7Barray%7D%5Cright%5D)
The matrix is extended to the following linear system :


![\left[\begin{array}{c}x1&x2&x3&x4\end{array}\right] =\left[\begin{array}{c}\frac{4}{3} x4&\frac{2}{3}x4 &\frac{2}{3}x4 &x4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx1%26x2%26x3%26x4%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%5Cfrac%7B4%7D%7B3%7D%20x4%26%5Cfrac%7B2%7D%7B3%7Dx4%20%26%5Cfrac%7B2%7D%7B3%7Dx4%20%26x4%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}\frac{4}{3}&\frac{2}{3}&\frac{2}{3}&1 \end{array}\right] x4=\left[\begin{array}{c}\frac{4}{3}&\frac{2}{3}&\frac{2}{3}&1 \end{array}\right]t](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%5Cfrac%7B4%7D%7B3%7D%26%5Cfrac%7B2%7D%7B3%7D%26%5Cfrac%7B2%7D%7B3%7D%261%20%20%20%5Cend%7Barray%7D%5Cright%5D%20x4%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%5Cfrac%7B4%7D%7B3%7D%26%5Cfrac%7B2%7D%7B3%7D%26%5Cfrac%7B2%7D%7B3%7D%261%20%20%20%5Cend%7Barray%7D%5Cright%5Dt)
t∈ IR
Let's choose t=3 to eliminate the fractional numbers →
![\left[\begin{array}{c}x1&x2&x3&x4\end{array}\right] = \left[\begin{array}{c}4&2&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx1%26x2%26x3%26x4%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%262%262%263%5Cend%7Barray%7D%5Cright%5D)