## Comparison of solutions

By elementary transformations decide which of the following systems of equations have the same sets of solutions. It is not necessary to compute these sets.
\begin{alignat}{8} x_1 &\ +\ & 3x_2 & & &\ +\ & 4x_4 &\ =\ &0\tag{A}\\ 2x_1 &\ +\ & x_2 &\ +\ & 3x_3 & & &\ =\ &0\notag\\ 3x_1 & & &\ +\ & x_3 &\ +\ & 2x_4 &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} x_1 &\ -\ & 2x_2 &\ +\ & 3x_3 &\ -\ & 4x_4 &\ =\ &0\tag{B}\\ x_1 &\ +\ & 3x_2 & & &\ +\ & 4x_4 &\ =\ &0\notag\\ 9x_1 & & &\ +\ & 3x_3 &\ +\ & 6x_4 &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} 3x_1 &\ +\ & 4x_2 &\ +\ & 3x_3 &\ +\ & 4x_4 &\ =\ &0\tag{C}\\ x_1 &\ +\ & 3x_2 & & &\ +\ & 4x_4 &\ =\ &0\notag\\ 2x_1 &\ +\ & x_2 &\ +\ & 3x_3 & & &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} 3x_1 & & &\ +\ & x_3 &\ +\ & 2x_4 &\ =\ &0\tag{D}\\ 3x_1 &\ +\ & 9x_2 & & &\ +\ &12x_4 &\ =\ &0\notag\\ 3x_1 &\ +\ & 4x_2 &\ +\ & 3x_3 &\ +\ & 4x_4 &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} 2x_1 &\ +\ & 6x_2 & & &\ +\ & 8x_4 &\ =\ &0\tag{E}\\ x_1 &\ -\ & x_2 &\ -\ & 2x_3 &\ +\ & 2x_4 &\ =\ &0\notag\\ 4x_1 &\ +\ & 3x_2 &\ +\ & 1x_3 &\ +\ & 6x_4 &\ =\ &0\notag\\ 2x_1 &\ +\ & x_2 &\ +\ & 3x_3 & & &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} 4x_1 &\ +\ & 2x_2 &\ +\ & 6x_3 & & &\ =\ &0\tag{F}\\ 3x_1 & & &\ +\ & x_3 &\ +\ & 2x_4 &\ =\ &0\notag\\ 3x_1 &\ -\ & x_2 &\ +\ & 6x_3 &\ +\ & 4x_4 &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} x_1 &\ +\ & 3x_2 & & &\ +\ & 4x_4 &\ =\ &0\tag{G}\\ &\ -\ & 5x_2 &\ + \ & 3x_3 &\ -\ & 8x_4 &\ =\ &0\notag\\ &\ -\ & 9 x_2&\ + \ & x_3 &\ -\ &10x_4 &\ =\ &0\notag \end{alignat}
\begin{alignat}{8} 6x_1 & & &\ +\ & 2x_3 &\ +\ & 4x_4 &\ =\ &0\tag{H}\\ & & x_2 &\ -\ & 5x_3 &\ -\ & 2x_4 &\ =\ &0\notag\\ 2x_1 &\ +\ & x_2 &\ +\ & 3x_3 & & &\ =\ &0\notag \end{alignat}

• #### Hint

One way is to compare ranks of matrices, however we did not yet explore this method.

Alternatively, it is possible to compute a unique echelon form, e.g. by Gauss-Jordan elimination (all pivots 1, in columns above pivots all 0).

Guess elementary transformations that would transform one matrix to another.

• #### Resolution

$$A=\begin{pmatrix} 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 2 & 1 & 3 & 0 \\ 1 & 3 & 0 & 4 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 2 & 1 & 3 & 0 \\ 1 & 3 & 0 & 4 \\ 9 & 0 & 3 & 6 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & 3 & -4 \\ 1 & 3 & 0 & 4 \\ 9 & 0 & 3 & 6 \\ \end{pmatrix} =B$$

$$A=\begin{pmatrix} 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 3 & 0 & 1 & 2 \\ 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 3 & 0 & 1 & 2 \\ 1 & 3 & 0 & 4 \\ 3 & 4 & 3 & 4 \\ \end{pmatrix} \sim \begin{pmatrix} 3 & 0 & 1 & 2 \\ 3 & 9 & 0 & 12 \\ 3 & 4 & 3 & 4 \\ \end{pmatrix} =D$$

$$A=\begin{pmatrix} 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ 3 & 0 & 1 & 2 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 3 & 0 & 4 \\ 1 &-1 &-2 & 2 \\ 3 & 0 & 1 & 2 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 2 & 6 & 0 & 8 \\ 1 &-1 &-2 & 2 \\ 4 & 3 & 1 & 6 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} =E$$

$$A=\begin{pmatrix} 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 3 & 0 & 4 \\ 0 &-5 & 3 &-8 \\ 3 & 0 & 1 & 2 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 3 & 0 & 4 \\ 0 &-5 & 3 &-8 \\ 0 &-9 & 1 &-10\\ \end{pmatrix} =G$$

$$F= \begin{pmatrix} 4 & 2 & 6 & 0 \\ 3 & 0 & 1 & 2 \\ 3 &-1 & 6 & 4 \\ \end{pmatrix} \sim \begin{pmatrix} 3 & 0 & 1 & 2 \\ 3 &-1 & 6 & 4 \\ 4 & 2 & 6 & 0 \\ \end{pmatrix}\sim \begin{pmatrix} 3 & 0 & 1 & 2 \\ 0 & 1 &-5 &-2 \\ 4 & 2 & 6 & 0 \\ \end{pmatrix}\sim \begin{pmatrix} 6 & 0 & 2 & 4 \\ 0 & 1 &-5 &-2 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} =H$$

$$C= \begin{pmatrix} 3 & 4 & 3 & 4 \\ 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 4 \\ 2 & 1 & 3 & 0 \\ \end{pmatrix}$$

To show that sets of solutions are distinct it is necessary to show that some row of the target matrix cannot be combined from original ones. This yields another system of equations which should not have a solution.

In our case we show that the last row of the matrix $$F$$ cannot be combined from rows of $$A$$. Otherwise would exist $$x,y,z$$ such that
$$(3,-1{,}6,4)=x(1{,}3,0{,}4)+y(2{,}1,3{,}0)+z(3{,}0,1{,}2)$$,

This system has no solution.

$$\begin{pmatrix} 1 & 2 & 3 & 3 \\ 3 & 1 & 0 &-1 \\ 0 & 3 & 1 & 6 \\ 4 & 0 & 2 & 4 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 & 3 \\ 3 & 1 & 0 &-1 \\ -4 & 0 &-2 & 4 \\ 4 & 0 & 2 & 4 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 & 3 \\ 3 & 1 & 0 &-1 \\ -4 & 0 &-2 & 4 \\ 0 & 0 & 0 & 8 \\ \end{pmatrix}$$

• #### Result

Systems A, B, D, E and G have equal solutions, and also F and H.

System C is only a "part" of A, hence it has more solutions.