(a) Check that the solutions to Ax = 0 are perpendicular to the rows: A = [ 1 0 0 2 1 0 3 4 1 ] [ 4 2 0 1 0 0 1 3 0 0 0 0 ]=ER. (b) How many independent solutions to A^T y = 0? Why is y^T the last row of E^-1?

Question

(a) Check that the solutions to Ax = 0 are perpendicular to the rows:
$A =\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{matrix} \right] \left[ \begin{matrix} 4 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{matrix} \right]=ER$ .
(b) How many independent solutions to ${A}^{T}y = 0$ ? Why is ${y}^{T}$ the last row of ${E}^{-1}$ ?

Accepted Answer

(a) Special solutions (-1, 2, 0, 0) and ([latex]-\frac { 1 }{ 4 }[/latex], 0, -3, 1) are perpendicular to the rows of R (and then ER).

(b) [latex]{A}^{T}y = 0[/latex] has 1 independent solution = last row of [latex]{E}^{-1}[/latex].

([latex]{E}^{-1}A = R[/latex] has a zero row, which is just the transpose of [latex]{A}^{T}y = 0[/latex]).

Question : (a) Check that the solutions to Ax = 0 are perpendicular to ...