## Systems with the same matrix

Solve systems $$\mathbf A\mathbf x=\mathbf 0$$, $$\mathbf A\mathbf x=\mathbf b^1$$, $$\mathbf A\mathbf x=\mathbf b^2$$ and $$\mathbf A\mathbf x=\mathbf b^3$$ for:

$\mathbf A= \begin{pmatrix} 6 & 3 & 2 & 3 & 4 \\ 4 & 2 & 1 & 2 & 3 \\ 4 & 2 & 3 & 2 & 1 \\ 2 & 1 & 7 & 3 & 2 \\ \end{pmatrix},\ \mathbf b^1= \begin{pmatrix} 5 \\ 4 \\ 0 \\ 1 \\ \end{pmatrix},\ \mathbf b^2= \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{pmatrix},\ \mathbf b^3= \begin{pmatrix} 3 \\ 2 \\ 2 \\ 3 \\ \end{pmatrix}$

What is the relation between th geometric interpretations of these systems?

• #### Resolution

We solve all four systems simultaneously by reducing $$(\mathbf A|\mathbf 0\ \mathbf b^1\ \mathbf b^2\ \mathbf b^3)$$ into echelon form:

$$\begin{pmatrix} \begin{array}{ccccc|cccc} 6 & 3 & 2 & 3 & 4 & 0 & 5 & 1 & 3 \\ 4 & 2 & 1 & 2 & 3 & 0 & 4 & 1 & 2 \\ 4 & 2 & 3 & 2 & 1 & 0 & 0 & 1 & 2 \\ 2 & 1 & 7 & 3 & 2 & 0 & 1 & 1 & 3 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{ccccc|cccc} 2 & 1 & 7 & 3 & 2 & 0 & 1 & 1 & 3 \\ 4 & 2 & 1 & 2 & 3 & 0 & 4 & 1 & 2 \\ 0 & 0 & 2 & 0 &-2 & 0 &-4 & 0 & 0 \\ 0 & 0 & 1 & 0 &-1 & 0 &-2 &-1 & 0 \\ \end{array} \end{pmatrix}\sim$$

$$\begin{pmatrix} \begin{array}{ccccc|cccc} 2 & 1 & 7 & 3 & 2 & 0 & 1 & 1 & 3 \\ 0 & 0 &-13&-4 &-1 & 0 & 2 &-1 &-4 \\ 0 & 0 & 1 & 0 &-1 & 0 &-2 &-1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{ccccc|cccc} 2 & 1 & 7 & 3 & 2 & 0 & 1 & 1 & 3 \\ 0 & 0 & 1 & 0 &-1 & 0 &-2 &-1 & 0 \\ 0 & 0 & 0 &-4 &-14& 0 &-24&-14&-4 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ \end{array} \end{pmatrix}\sim$$

$$\begin{pmatrix} \begin{array}{ccccc|cccc} 2 & 1 & 7 & 3 & 2 & 0 & 1 & 1 & 3 \\ 0 & 0 & 1 & 0 &-1 & 0 &-2 &-1 & 0 \\ 0 & 0 & 0 & 2 & 7 & 0 &12 & 7 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ \end{array} \end{pmatrix}$$

The homogeneous system $$\mathbf A\mathbf x=\mathbf 0$$ has solution $$\mathbf x^0=p_1(3{,}0,4,-14{,}4)^T+p_2(-1{,}2,0{,}0,0)^T$$,
the system $$\mathbf A\mathbf x=\mathbf b^1$$ has solution $$\mathbf x=(0,-3,-2{,}6,0)^T+\mathbf x^0$$,
the system $$\mathbf A\mathbf x=\mathbf b^2$$ has no solution, and
the system $$\mathbf A\mathbf x=\mathbf b^3$$ has solution $$\mathbf x=(0{,}0,0{,}1,0)^T+\mathbf x^0$$.

Geometrically, each system are yields an intersection of hyperplanes, the corresponding hyperplanes in these systems are parallel. The vector $$\mathbf b$$ detrmines the shift of these hyperplanes.

The left sides of the first three equations are dependent, hence the corresponding pairwise intersections should coincide. Otherwise the system has no solution as in the case of $$\mathbf b^2$$.