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$A =\left[ {\begin{array}{cc} 1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ -2&1&1&b \\ 1&1&-2&c \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&3&-3&c-a \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c \\ \end{array} } \right] $

$A$ is the matrix. I concatenated $[A|b]$ and tried to reduce it. I end up with the last matrix, does it mean that the system is not compatible? Is there another way to solve it?
Or should I concatenate $[A|0]$ to make a homogeneous system?

UPDATE:

$A =\left[ {\begin{array}{cc} 1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&0 \\ -2&1&1&0 \\ 1&1&-2&0 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&0 \\ 0&-3&3&0 \\ 0&0&0&0 \\ \end{array} } \right] \ \ \ \left[ {\begin{array}{cc} 1&0&-1&0 \\ 0&1&-1&0 \\ 0&0&0&0 \\ \end{array} } \right] $

After solving $[A|0]$ I get $a-c=0$ and $b-c=0$, the equations of the system seem to be: $a=c,b=c$, so $a=b=c$. Is this the correct solution?

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up vote 1 down vote accepted

What you did in the Update just shows that $A{\bf x}=\bf0$ has one (independent) solution. That is the dimension of the null space of $A$ is 1.

So $A{\bf x}=\bf b$ has two independent solutions. That is, the dimension of the column space of $A$ is 2. You have an echelon form of $A$. What you need to do is find a basis for the column space of $A$. Then the set of linear combinations of elements of this basis will be the answer to your problem (that is $\bf y$ is in the column space of $A$ if and only if $A{\bf x}=\bf y$ has a solution).

That is, if I'm interpreting the problem correctly: for what vectors $\bf b$ does $A{\bf x}=\bf b$ have a solution?

You could also keep going with what you did at the start: find conditions on $a$, $b$, and $c$ so that the system represented by $\left[ {\begin{array}{cc} 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c \\ \end{array} } \right]$ has a solution.

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so, the first two columns of $A$ form a basis in the column space of $A$. $x_1-2x_2=a, \ -2x_1+x_2=b, \ x_1+x_2=c$, Is this what is needed? –  Andrew Jan 19 '12 at 11:44
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@Andrew the system $Ax=y$ has a solution if and only if $y= c_1\left[ {{1\atop -2}\atop1}\right]+c_2\left[ { {-2\atop 1}\atop1}\right]$. –  David Mitra Jan 19 '12 at 11:51
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