- #1
skujesco2014
- 24
- 0
Hi, all. I'm in desperate need of assistance with a matrix I can't get my head around of. I want to solve a system of equations of the type Ax=b, where
<br /> A=\begin{pmatrix}<br /> 2 & 5 & -3 \\<br /> 1 & -2 & 1 \\<br /> 7 & 4 & -3<br /> \end{pmatrix}<br />
and where
<br /> b=\begin{pmatrix}<br /> -2 \\<br /> -1 \\<br /> -7 <br /> \end{pmatrix}<br />
that is, b is the negative of the first column. Written as it is above, A has zero determinant and the determinant formed when thek^{th} column of A is substituted by the vector b is clearly zero as well. A theorem says that in this case the system has infinite solutions. If one reduces the system to reduced row-echelon form the solutions can be parameterized as, for example, x_3=t,x_2=5t/9,x_1=t−1. An immediate solution by inspection is x=(−1,0,0)^T which one obtains letting t=0.
But let's give another value of t, for example, t=1 which gives x=(0,5/9,1)^T. This is one of the parameterized solutions and yet it does not satisfy the original system. It does, however, satisfy the reduced system obtained from the original by gaussian elimination and should be equivalent, i.e.,
<br /> \begin{cases}<br /> x_1-x_3 &=-1\\<br /> 9x_2-5x_3 & = 0<br /> \end{cases}<br />
But shouldn't my parameterized solution satisfy both original and reduced systems, no matter what? Yet, the only satisfying solution for the original system seems to be x=(−1,0,0)^T. What am I not seeing here?
Thanks in advance.
<br /> A=\begin{pmatrix}<br /> 2 & 5 & -3 \\<br /> 1 & -2 & 1 \\<br /> 7 & 4 & -3<br /> \end{pmatrix}<br />
and where
<br /> b=\begin{pmatrix}<br /> -2 \\<br /> -1 \\<br /> -7 <br /> \end{pmatrix}<br />
that is, b is the negative of the first column. Written as it is above, A has zero determinant and the determinant formed when thek^{th} column of A is substituted by the vector b is clearly zero as well. A theorem says that in this case the system has infinite solutions. If one reduces the system to reduced row-echelon form the solutions can be parameterized as, for example, x_3=t,x_2=5t/9,x_1=t−1. An immediate solution by inspection is x=(−1,0,0)^T which one obtains letting t=0.
But let's give another value of t, for example, t=1 which gives x=(0,5/9,1)^T. This is one of the parameterized solutions and yet it does not satisfy the original system. It does, however, satisfy the reduced system obtained from the original by gaussian elimination and should be equivalent, i.e.,
<br /> \begin{cases}<br /> x_1-x_3 &=-1\\<br /> 9x_2-5x_3 & = 0<br /> \end{cases}<br />
But shouldn't my parameterized solution satisfy both original and reduced systems, no matter what? Yet, the only satisfying solution for the original system seems to be x=(−1,0,0)^T. What am I not seeing here?
Thanks in advance.
